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Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8
thth
ed. ed. Chapter Chapter 44
Chapter 4Chapter 4
4-14-1 (a)(a) The The millimolemillimole is an amount of a chemical species, such as an atom, an ion, a molecule is an amount of a chemical species, such as an atom, an ion, a molecule
or or an an electron. electron. A A millimolmillimole e containscontains
millimolemillimole
particlesparticles
10100202..66
millimolemillimole
molemole
1010
molemole
particlesparticles
10100202..66 2020332323 ××==∗∗×× −−
(b)(b) The The molar massmolar mass is the mass in grams of one mole of a chemical species. is the mass in grams of one mole of a chemical species.
(c)(c) TheThe millimolar massmillimolar mass is the mass in grams of one millimole of a chemical species. is the mass in grams of one millimole of a chemical species.
(d)(d) Parts per million, c Parts per million, cppmppm, is a term expressing the concentration of very dilute solutions., is a term expressing the concentration of very dilute solutions.
Thus,Thus,
ccppmppm ppmppm1010
solutionsolutionof of massmass
solutesoluteof of massmass 66
××==
The units of mass in the numerator and the denominator must be the same.The units of mass in the numerator and the denominator must be the same.
4-24-2 The species molarity of a solution expresses the equilibrium concentration of a chemicalThe species molarity of a solution expresses the equilibrium concentration of a chemical
species in terms of species in terms of moles per liter. moles per liter. The analytical molarity of a The analytical molarity of a solution gives the totalsolution gives the total
number of moles of number of moles of a solute in one a solute in one liter. liter. The species molarity takes into accountThe species molarity takes into account
chemical reactions that occur chemical reactions that occur in solution. in solution. The analytical molarity specifies how The analytical molarity specifies how thethe
solution was prepared, but does not account for any subsequent reactions.solution was prepared, but does not account for any subsequent reactions.
4-34-3 3333
33
33
mm1010
cmcm100100
mm
mLmL
cmcm11
LL11
mLmL10001000
1L1L −−==
  
  

  
  
××××==
33333333 mm1010
molemole11
mm1010
LL
LL
molemole11
MM11
−−−−
==××==
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8
thth
ed. ed. Chapter Chapter 44
4-44-4 (a)(a) kHzkHz320320
HzHz10001000
kHzkHz
HzHz101022..33 55 ==××××
(b)(b) ngng66..4545
gg
ngng1010
gg10105656..44
99
88
==××××
−−
(c)(c)  mmol  mmol843843
molmol1010
mmolmmol
molmol10104343..88
33
55
==
µµ
××µµ××
(d)(d) MsMs55..66
ss1010
MsMs
ss101055..66
66
66
==××××
(e)(e) mm66..8989
nmnm1010
mm
nmnm10109696..88
33
44
µµ==
µµ
××××
(f)(f) kgkg7272
gg10001000
kgkg
gg000000,,7272 ==××
4-54-5 ++
++
++++
××==
××
×××××× NaNa10109898..55
NaNamolmol
NaNa10100202..66
POPONaNamolmol
NaNamolmol33
gg9494..163163
POPONaNamolmol11
POPONaNagg4343..55 2222
2323
4433
4433
4433
4-64-6 ++
++
++++
××==
××
×××× KK10102222..11
KKmolmol
KK10100202..66
POPOKKmolmol
KKmolmol33
POPOKKmolmol7676..66 2525
2323
4433
4433
4-74-7 (a)(a) 3322
3322
3322
3322 OOBBmolmol07120712..00
OOBBgg6262..6969
OOBBmolmol
OOBBgg9696..44 ==××
(b)(b)
OO10H10HOOBBNaNamolmol10107373..88
381.37g381.37g
OO10H10HOOBBNaNamolmol
mgmg10001000
gg
OOHH1010OOBBNaNamgmg333333
22774422
44
22774422
22774422
⋅⋅××==
⋅⋅
×××ו•
−−
(c)(c) 4433
4433
4433
4433 OOMnMnmolmol03820382..00
OOMnMngg8181..228228
OOMnMnmolmol
OOMnMngg7575..88 ==××
(d)(d) 4422
33
4422
4422
4422 OOCaCCaCmolmol10103131..11
OOCaCCaCgg128.10128.10
OOCaCCaCmolmol
mgmg10001000
gg
OOCaCCaCmgmg22..167167 −−××==××××
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8
thth
ed. ed. Chapter Chapter 44
4-84-8 (a)(a) 5522
5522
5522
5522 OOPPmmolmmol4040..00
molmol
mmolmmol10001000
OOPPgg9494..141141
OOPPmolmol
mgmg10001000
gg
OOPPmgmg5757 ==××××××
(b)(b) 22
22
22
22 COCOmmolmmol66..293293
molmol
mmolmmol10001000
COCOgg0101..4444
COCOmolmol
COCOgg9292..1212 ==××××
(c)(c) 33
33
33
33 NaHCONaHCOmmolmmol476476
molmol
mmolmmol10001000
NaHCONaHCOgg0101..8484
NaHCONaHCOmolmol
NaHCONaHCOgg00..4040 ==××××
(d)(d)
4444
4444
4444
4444
POPOMgNHMgNHmmolmmol22..66
molmol
mmolmmol10001000
POPOMgNHMgNHgg137.32137.32
POPOMgNHMgNHmolmol
mgmg10001000
gg
POPOMgNHMgNHmgmg850850
==
××××××
4-94-9 (a)(a)
44
44
33
44
33
KMnOKMnOmmolmmol5050..66
LL0000..22
molmol
mmolmmol10001000
LL
KMnOKMnOmolmol10102525..33
M KMnOM KMnO10102525..33
==
××××
××
≡≡××
−−
−−
(b)(b)
KSCNKSCNmmolmmol66..4141
mLmL750750
mLmL10001000
LL
molmol
mmolmmol10001000
LL
KSCNKSCNmolmol05550555..00
 M KSCN M KSCN05550555..00
==
××××××≡≡
(c)(c)
44
33
44
4444
44
CuSOCuSOmmolmmol10104747..88mLmL250250
mLmL10001000
LL
molmol
mmolmmol10001000
CuSOCuSOgg6161..159159
CuSOCuSOmolmol
mgmg10001000
gg
LL
CuSOCuSOmgmg4141..55
CuSOCuSOppmppm4141..55
−−
××==××××
××××××≡≡
(d)(d) KClKClmmolmmol66..11651165LL5050..33
molmol
mmolmmol10001000
LL
KClKClmolmol333333..00
 M KCl M KCl333333..00 ==××××≡≡
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8
thth
ed. ed. Chapter Chapter 44
4-104-10 (a)(a)
44
44
44
HClOHClOmmolmmol00..5656
mLmL175175
mLmL10001000
LL
molmol
mmolmmol10001000
LL
HClOHClOmolmol320320..00
HClOHClOMM320320..00
==
××××××≡≡
(b)(b)
4422
4422
33
4422
33
CrOCrOKKmmolmmol121121
LL00..1515
molmol
mmolmmol10001000
LL
CrOCrOKKmolmol10100505..88
CrOCrOKKMM10100505..88
==
××××
××
≡≡××
−−
−−
(c)(c)
33
33
3333
33
AgNOAgNOmmolmmol199199..00LL0000..55
molmol
mmolmmol10001000
AgNOAgNOgg8787..169169
AgNOAgNOmolmol
mgmg10001000
gg
LL
AgNOAgNOmgmg7575..66
AgNOAgNOppmppm7575..66
==
××××××××≡≡
(d)(d)
KOHKOHmmolmmol00..1717
mLmL851851
mLmL10001000
LL
molmol
mmolmmol10001000
LL
KOHKOHmolmol02000200..00
KOHKOHMM02000200..00
==
××××××≡≡
4-114-11 (a)(a) 33
44
33
33
33 HNOHNOmgmg10109090..44
gg
mgmg10001000
HNOHNOmolmol
HNOHNOgg0101..6363
HNOHNOmolmol777777..00 ××==××××
(b)(b) MgOMgOmgmg1010015015..22
gg
mgmg10001000
MgOMgOmolmol
MgOMgOgg40.3040.30
mmolmmol10001000
molmol
MgOMgOmmolmmol500500 44××==××××××
(c)(c) 3344
66
3344
3344
3344 NONONHNHmgmg10108080..11
gg
mgmg10001000
NONONHNHmolmol
NONONHNHgg0404..8080
NONONHNHmolmol55..2222 ××==××××
(d)(d)
66332244
66
66332244
66332244
66332244
))NONO((CeCe))NHNH((mgmg10103737..22
gg
mgmg10001000
))NONO((CeCe))NHNH((molmol
))NONO((CeCe))NHNH((gg2323..548548
))NONO((CeCe))NHNH((molmol3232..44
××==
××××
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8
thth
ed. ed. Chapter Chapter 44
4-124-12 (a)(a) KBrKBrgg840840
KBrKBrmolmol
KBrKBrgg00..119119
KBrKBrmolmol11..77 ==××
(b)(b) PbOPbOgg4949..44
PbOPbOmolmol
PbOPbOgg2020..223223
mmolmmol10001000
molmol
PbOPbOmmolmmol11..2020 ==××××
(c)(c) 44
44
44
44 MgSOMgSOgg452452
MgSOMgSOmolmol
MgSOMgSOgg3737..120120
MgSOMgSOmolmol7676..33 ==××
(d)(d)
OOHH66))SOSO(())NHNH((FeFegg88..33
OOHH66))SOSO(())NHNH((FeFemolmol
OOHH66))SOSO(())NHNH((FeFegg2323..392392
mmolmmol10001000
molmol
OOHH66))SOSO(())NHNH((FeFemmolmmol66..99
2222442244
2222442244
2222442244
2222442244
⋅⋅==
⋅⋅
⋅⋅
××××⋅⋅
4-134-13 (a)(a)
sucrosesucrosemgmg10102222..22
mLmL00..2626
gg
mgmg10001000
sucrosesucrosemolmol
sucrosesucrosegg342342
mLmL10001000
LL
LL
sucrosesucrosemolmol250250..00
sucrosesucroseMM250250..00
33
××==
××××××××≡≡
(b)(b)
2222
2222
22222222
33
2222
33
OOHHmgmg88..472472
LL9292..22
gg
mgmg10001000
OOHHmolmol
OOHHgg0202..3434
LL
OOHHmolmol10107676..44
OOHHMM10107676..44
==
××××××
××
≡≡××
−−
−−
(c)(c)
2233
2233
2233 ))NONO((PbPbmgmg2525..33mLmL656656
mLmL10001000
LL
LL
))NONO((PbPbmgmg9696..44
))NONO((PbPbppmppm9696..44 ==××××≡≡
(d)(d)
33
3333
33
KNOKNOmgmg22..4242
mLmL7575..66
mLmL10001000
LL
gg
mgmg10001000
molmol
KNOKNOgg1010..101101LL
KNOKNOmolmol06190619..00
KNOKNOMM06190619..00
==
××××××××≡≡
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8
thth
ed. ed. Chapter Chapter 44
4-144-14 (a)(a)
2222
2222
22222222
2222
OOHHgg5151..22
mLmL450450
OOHHmolmol
OOHHgg34.0234.02
mLmL10001000
LL
LL
OOHHmolmol0.1640.164
OOHHMM164164..00
==
××××××≡≡
(b)(b)
acidacidbenzoicbenzoicgg10108888..22mLmL00..2727
acidacidbenzoicbenzoicmolmol
acidacidbenzoicbenzoicgg122122
mLmL10001000
LL
LL
acidacidbenzoicbenzoicmolmol10107575..88
acidacidbenzoicbenzoicMM10107575..88
33
44
44
−−
−−
−−
××==
××××××
××
≡≡××
(c)(c) 22
22
22 SnClSnClgg07600760..00LL5050..33
mgmg10001000
gg
LL
SnClSnClmgmg77..2121
SnClSnClppmppm77..2121 ==××××≡≡
(d)(d)
33
33
3333
33
KBrOKBrOgg04530453..00
mLmL77..2121
KBrOKBrOmolmol
KBrOKBrOgg167167
mLmL10001000
LL
LL
KBrOKBrOmolmol01250125..00
KBrOKBrOMM01250125..00
==
××××××≡≡
4-154-15 (a)(a)
077077..11))22((923923..00
))1010log(log())3838..88log(log(
))MM10103838..88log(log())MM08380838..00log(log())MM05030503..00MM03350335..00log(log(pNapNa
22
22
==−−−−−−==
−−−−==
××−−==−−==++−−==
−−
−−
475475..11))22((525525..00
))1010log(log())3535..33log(log(
))MM10103535..33log(log())MM03350335..00log(log(pClpCl
22
22
==−−−−−−==
−−−−==
××−−==−−==
−−
−−
298298..11))22((702702..00
))1010log(log())0303..55log(log(
))MM10100303..55log(log())MM05030503..00log(log(pOHpOH
22
22
==−−−−−−==
−−−−==
××−−==−−==
−−
−−
Fundamentals of Analytical Chemistry: 8
th
ed. Chapter 4
(b)
116.2)3(884.0
)10log()65.7log(
)M1065.7log(pBa
3
3
=−−−=
−−=
×−=
−
−
188.0)M54.1log(pMn −=−=
490.0)M08.3log())M54.12(M1065.7log(pCl 3 −=−=×+×−= −
(c)
222.0)1(778.0
)10log()00.6log(
)M1000.6log()M600.0log(pH
1
1
=−−−=
−−=
×−=−=
−
−
096.0)1(904.0
)10log()02.8log(
)M1002.8log()M802.0log())M101.02(M600.0log(pCl
1
1
=−−=
−−=
×−=−=×+−=
−
−
996.0)1(00432.0
)10log()01.1log(
)M1001.1log()M101.0log(pZn
1
1
=−−−=
−−=
×−=−=
−
−
(d)
320.1)2(679.0
)10log()78.4log(
)M1078.4log(pCu
2
2
=−−−=
−−=
×−=
−
−
983.0)1(0170.0
)10log()04.1log(
)M1004.1log()M104.0log(pZn
1
1
=−−−=
−−=
×−=−=
−
−
517.0)1(483.0
)10log()04.3log(
)M1004.3log()M304.0log())M104.02()M0478.02log((pNO
1
1
3
=−−−=
−−=
×−=−=×+×−=
−
−
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4
(e)
836.5)6(164.0
)10log()46.1log(
)M1046.1log()M1012.4))M1062.2(4log((pK
6
677
=−−−=
−−=
×−=×+××−=
−
−−−
385.6)7(615.0
)10log()12.4log(
)M1012.4log(pOH
7
7
=−−−=
−−=
×−=
−
−
582.6)7(418.0
)10log()62.2log(
)M1062.2log()CN(pFe
7
7
6
=−−−=
−−=
×−=
−
−
(f)
171.3)4(829.0
)10log()75.6log(
)M1075.6log(pH
4
4
=−−−=
−−=
×−=
−
−+
475.3)4(525.0
)10log()35.3log(
)M1035.3log(pBa
4
4
=−−−=
−−=
×−=
−
−
873.2)3(127.0
)10log()34.1log(
)M1034.1log()M1075.6)M1035.3(2log(pClO
3
344
4
=−−−=
−−=
×−=×+××−=
−
−−−
4-16 (a) M107.1)5(antilog)240.0(antilog]OH[ 53
−+
×=−×=
as in part (a)
(b) M106.2]OH[ 53
−+
×=
(c) M30.0]OH[ 3 =
+
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4
(d) M104.2]OH[ 143
−+
×=
(e) M108.4]OH[ 83
−+
×=
(f) M107.1]OH[ 63
−+
×=
(g) M04.2]OH[ 3 =
+
(h) M3.3]OH[ 3 =
+
4-17 (a)
699.1)2(301.0
)10log()00.2log()M0200.0log(pBrpNa 2
=−−−=
−−=−==
−−
pH = pOH = - log(1.0×10-7M) = 7.00
(b)
000.2)2(000.0
)10log()00.1log()M0100.0log(pBa 2
=−−=
−−=−=
−
699.1)2(301.0
)10log()00.2log()M0100.02log(pBr 2
=−−−=
−−=×−=
−
pH = pOH = - log(1.0×10
-7M) = 7.00
(c)
46.2)3(54.0
)10log()5.3log()M105.3log(pBa 33
=−−−=
−−=×−=
−−
15.2)3(84.0
)10log()0.7log()M100.7log()M105.3(2log(pOH 333
=−−−=
−−=×−=××−=
−−−
pH = 14.00 – pOH = (14.00 – 2.15) = 11.85
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4
(d)
40.1)2(60.0
)10log()0.4log()M100.4log()M040.0log(pH 22
=−−−=
−−=×−=−=
−−
70.1)2(30.0
)10log()0.2log()M100.2log()M020.0log(pNa 22
=−−−=
−−=×−=−=
−−
22.1)2(78.0
)10log()0.6log(
)M100.6log()M060.0log()M020.0M040.0log(pCl
2
2
=−−−=
−−=
×−=−=+−=
−
−
pOH = 14.00 – 1.40 = 12.60
(e)
17.2)3(83.0
)10log()7.6log()M107.6log(pCa 33
=−−−=
−−=×=
−−
12.2)3(88.0
)10log()6.7log()M106.7log(pBa 33
=−−−=
−−=×−=
−−
54.1)2(46.0
)10log()9.2log(
)M109.2log()))M106.7(2())M107.6(2log((pCl
2
233
=−−−=
−−=
×−=××+××−=
−
−−−
pH = pOH = - log(1.0×10
-7M) = 7.00
(f)
32.7)8(68.0
)10log()8.4log()M108.4log(pZn 88
=−−−=
−−=×−=
−−
25.6)7(75.0
)10log()6.5log()M106.5log(pCd 77
=−−−=
−−=×−=
−−
Fundamentals of Analytical Chemistry: 8
th
ed. Chapter 4
4-38 A balanced chemical equation can be written as shown below.
−+
++→+ 3
3
434223 NO6Al2BaSO3)SO(Al)NO(Ba3
(a)
342
3
342
342
23
3
23
623
)SO(Almol1018.6
mL0.200
mL1000
L
L
)SO(Almol03090.0
)SO(AlM03090.0
)NO(Bamol1038.1
g34.261
)NO(Bamol
mL0.750
mL
g
ppm10
1
)NO(Bappm4.480
−
−
×=
××≡
×=
××××
The Ba(NO3)2 is the limiting reagent. Thus,
formedBaSOg322.0
mol
BaSOg39.233
)NO(Bamol3
BaSOmol3
)NO(Bamol1038.1 4
4
23
4
23
3
=×××
−
(b)
342
3342
3
342
333
342
)SO(AlM1002.6
L
mL1000
mL0.950
)SO(Almol1072.5
)SO(Almol1072.5))1038.1(
3
1(1018.6(unreacted)SO(Almol
−
−
−−−
×=×
×
×=××−×=
Fundamentals of Analytical Chemistry: 8
th
ed. Chapter 5
Chapter 5
5-1 (a) Constant errors are the same magnitude regardless of sample size. Proportional
errors are proportional in size to sample size.
(b) Random error  causes data to be scattered more or less symmetrically around a mean
value. Systematic error  causes the mean of a data set to differ from the accepted value.
(c) The mean is the sum of the results in a set divided by the number of results. The
median is the central value for a set of data.
(d) The absolute error  of a measurement is the difference between the measured value
and the true value. The relative error  is the absolute error divided by the true value.
5-2 (1) Random temperature fluctuations causing random changes in the length of the metal
rule; (2) uncertainties from moving and positioning the rule twice; (3) personal judgment
in reading the rule; (4) vibrations in the table and/or rule; (5) uncertainty in locating the
rule perpendicular to the edge of the table.
5-3 (1) Instrumental errors
(2) Method errors
(3) Personal errors
5-4 (1) The analytical balance is miscalibrated; (2) after weighing an empty vial, fingerprints
are placed on the vial while adding sample to the vial; (3) a hygroscopic sample absorbs
water from the atmosphere while placing it in a weighing vial.
5-5 (1) Incorrect calibration of the pipet; (2) temperature different from calibration
temperature; (3) incorrect filling of the pipet (overshooting or undershooting the mark).
Fundamentals of Analytical Chemistry: 8
th
ed. Chapter 5
5-6 Systematic method errors are detected by application of the method to the analysis of a
standard reference material having one or more analytes at exactly known concentrations.
5-7 Constant and proportional errors.
5-8 (a) (– 0.4/700) × 100% = – 0.06%
As in part (a)
(b) – 0.09%
(c) – 0.2%
(d) – 1%
5-9 (a) First determine how much gold is needed to achieve the desired relative error.
(– 0.4/– 0.2%) × 100% = 200 mg gold
Then determine how much ore is needed to yield the required amount of gold.
(200/1.2%) × 100% = 17,000 mg ore or 17 g ore
(b) 7 g ore
(c) 4 g ore
(d) 3 g ore
5-10 (a) (0.04/50.00) ×  100% = 0.08%
As in part (a)
(b)  0.4%
(c)  0.16%
(d)  0.1%
Fundamentals of Analytical Chemistry: 8
th
ed. Chapter 5
5-11 (a) (– 0.4/40) × 100% = – 1.0%
As in part (a)
(b) – 0.23%
(c) – 0.10%
(d) – 0.07%
5-12 mean = 0106.001063.0
3
0105.00104.00110.0
≈=
 
 

 
  ++
Arranging the numbers in increasing value the median is:
0.0104
0.0105 ←  median
0.0110
The deviations from the mean are:
0004.00106.00110.0
0001.00106.00105.0
0002.00106.00104.0
=−
=−
=−
mean deviation = 0002.000023.0
3
0004.00001.00002.0
≈=
 
 

 
  ++
Fundamentals of Analytical Chemistry: 8th ed. Chapter 7
reject H 0 andconclude that the laboratories differ at 95% confidence. We can also be
99% confident that the laboratories differ, but we cannot be 99.9% confident that the
laboratories differ.
(c) Based on the calculated LSD value laboratories A, C and E differ from laboratory D,
but laboratory B does not. Laboratories E and A differ from laboratory B, but laboratory
C does not. No significant difference exists between laboratories E and A.
Fundamentals of Analytical Chemistry: 8th ed. Chapter 7
7-28 (a)  H 0:  µ Analyst1 =  µ Analyst2 =  µ Analyst3 =  µ Analyst4; H a: at least two of the means differ.
(b)
A B C D E F
1 Result No. A1 A2 A3 A4
2 1 10.24 10.14 10.19 10.19
3 2 10.26 10.12 10.1 10.15
4 3 10.29 10.04 10.15 10.16
5 4 10.23 10.07 10.12 10.1
6
7 Average = 10.26 10.09 10.14 10.15
8 Stan. Dev. = 0.02646 0.04573 0.03594 0.03742
9 Variance = 0.00070 0.00209 0.00129 0.00140
10
11 Grand Mean = 10.160
12 SSF = 0.05595
13 SSE = 0.01645
14 SST = 0.07240 10.26 - 10.09 = 0.17 (a significant difference)
15 10.15 - 10.09 = 0.06 (a significant difference)
16 MSF = 0.01865 10.14 - 10.09 = 0.05 (no significant difference)
17 MSE = 0.00137
18 10.26 - 10.14 = 0.12 (a significant difference)
19 F = 13.60486 10.15 - 10.14 = 0.01 (no significant difference)
20
21 LSD = 0.057 10.26 - 10.15 = 0.11 (a significant difference)
22
23
24 Spreadsheet Documentation
25 B7 = AVERAGE(B2:B5)
26 B8 = STDEV(B2:B5)
27 B9 = B8^2
28 B11 = SUM(B2:E5)/16
29 B12 = 4*((B7-B11)^2+(C7-B11)^2+(D7-B11)^2+(E7-B11)^2)
30 B13 = 3*SUM(B9:F9)
31 B14 = B12+B13
32 B16 = B12/3
33 B17 = B13/12
34 B19 = B16/B17
35 B21 = 2.19*SQRT(2*B17/4)
From Table 7-4 the F  value for 3 degrees of freedom in the numerator and 12 degrees of
freedom in the denominator at 95% is 3.49. Since F  calculated exceeds F  tabulated we
Fundamentals of Analytical Chemistry: 8th ed. Chapter 7
reject H 0 and conclude that the analysts differ at 95% confidence. We can also be 99%
and 99.9% confident that the analysts differ.
(c) Based on the calculated LSD value there is a significant difference between analyst 2
and analysts 1 and 4, but not analyst 3. There is a significant difference between analyst
3 and analyst 1, but not analyst 4. There is a significant difference between analyst 1 and
analyst 4.
Fundamentals of Analytical Chemistry: 8th ed. Chapter 7
7-29 (a)  H 0:  µ Des1 =  µ Des2 =  µ Des3 =  µ Des4; H a: at least two of the means differ.
(b)
A B C D E
1 Result No. Des 1 Des 2 Des 3 Des 4
2 1 72 93 96 100
3 2 93 88 95.0 84
4 3 76 97 79 91
5 4 90 74 82 94
6
7 Average = 82.75 88.00 88.00 92.25
8 Stan. Dev. = 10.30776 10.03328 8.75595 6.65207
9 Variance = 106.2500 100.6667 76.6667 44.2500
10
11 Grand Mean = 87.750
12 SSF = 181.5
13 SSE = 983.5
14 SST = 1165
15
16 MSF = 60.50000
17 MSE = 81.95833
18
19 F = 0.73818
20
21
22 Spreadsheet Documentation
23 B7 = AVERAGE(B2:B5)
24 B8 = STDEV(B2:B5)
25 B9 = B8^2
26 B11 = SUM(B2:E5)/16
27 B12 = 4*((B7-B11)^2+(C7-B11)^2+(D7-B11)^2+(E7-B11)^2)
28 B13 = 3*SUM(B9:F9)
29 B14 = B12+B13
30 B16 = B12/3
31 B17 = B13/12
32 B19 = B16/B17
33 B21 = 2.19*SQRT(2*B17/4)
From Table 7-4 the F  value for 3 degrees of freedom in the numerator and 12 degrees of
freedom in the denominator at 95% is 3.49. Since F  calculated is less than F  tabulated
we accept H 0 and conclude that 4 flow cell designs give the same results at 95%
Fundamentals of Analytical Chemistry: 8th ed. Chapter 7
confidence.
(c) No differences.
7-30 (a)  H 0:  µ Colorimetry =  µ EDTA Tit =  µ AA; H a: at least two of the means differ.
(b)
A B C D E F G
1 Result No. Color. EDTA AA
2 1 3.92 2.99 4.40
3 2 3.28 2.87 4.92
4 3 4.18 2.17 3.51
5 4 3.53 3.40 3.97
6 5 3.35 3.92 4.59
7
8 Average = 3.65 3.07 4.28
9 Stan. Dev. = 0.38571 0.64958 0.54979
10 Variance = 0.1488 0.42195 0.3023
11
12 Grand Mean = 3.667
13 SSF = 3.649773
14 SSE = 3.49196
15 SST = 7.141733
16
17 MSF = 1.82489 4.28 - 3.07 = 1.21 (a significant difference)
18 MSE = 0.29100 3.65 - 3.07 = 0.58 (no significant difference)
19
20 F = 6.27116 4.28 - 3.65 = 0.63 (no significant difference)
21
22 LSD = 0.747
23
24
25 Spreadsheet Documentation
26 B8 = AVERAGE(B2:B6)
27 B9 = STDEV(B2:B6)
28 B10 = B9^2
29 B12 = SUM(B2:D6)/15
30 B13 = 5*((B8-B12)^2+(C8-B12)^2+(D8-B12)^2)
31 B14 = 4*SUM(B10:D10)
32 B15 = B13+B14
33 B17 = B13/2
34 B18 = B14/12
35 B20 = B17/B18
36 B22 = 2.19*SQRT(2*B18/5)
Fundamentals of Analytical Chemistry: 8th ed. Chapter 7
From Table 7-4 the F  value for 2 degrees of freedom in the numerator and 12 degrees of
freedom in the denominator at 95% is 3.89. Since F  calculated is greater than F  tabulated
we reject H 0 and conclude that the 3 methods give different results at 95% confidence.
(c) Based on the calculated LSD value there is a significant difference between the
atomic absorption method and the EDTA titration. There is no significant difference
between the EDTA titration method and the colorimetry method and there is no
significant difference between the atomic absorption method and the colorimetry method.
7-31 (a) 596.0
27.4184.41
61.4127.41
=
−
−
=Q  and Qcrit for 4 observations at 95% confidence = 0.829.
Since Q < Qcrit the outlier value 41.27 cannot be rejected with 95% confidence.
(b) 894.0
284.7388.7
295.7388.7
=
−
−
=Q  and Qcrit for 4 observations at 95% confidence = 0.829.
Since Q > Qcrit the outlier value 7.388 can be rejected with 95% confidence.
7-32 (a) 833.0
62.8410.85
70.8410.85
=
−
−
=Q  and Qcrit for 3 observations at 95% confidence = 0.970.
Since Q < Qcrit the outlier value 85.10 cannot be rejected with 95% confidence.
(b) 833.0
62.8410.85
70.8410.85
=
−
−
=Q  and Qcrit for 4 observations at 95% confidence = 0.829.
Since Q > Qcrit the outlier value 85.10 can be rejected with 95% confidence.
7-33 5.0
40.460.4
50.460.4
=
−
−
=Q  and Qcrit for 5 observations at 95% confidence = 0.710.
Since Q < Qcrit the outlier value 4.60 ppm cannot be rejected with 95% confidence.
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9
(c)
M101.1)M102.2(
2
1
]F[
2
1
S
M102.2
050.0
105.2
]F[
]M)[F050.0(]][F[Sr 105.2
44
4
9
2229
sp
−−−
−
−
−
−−+−
×=××==
×=
×
=
==×=K 
(d)
M104.9)M108.3(
4
1
]OH[
4
1
S
M108.3
050.0
100.1
]OH[
]M)[OH050.0(]][OHTh[100.1
54
44
15
44415
sp
−−−
−
−
−
−−+−
×=××==
×=
×
=
==×=K 
9-12 (a)
M100.4
050.0
1002.2
]Cu[S
)](0.050MCu[]][SeOCu[1002.2
7
8
2
22
3
28
sp
−
−
+
+−+−
×=
×
==
==×=K 
(b)
M103.1
)050.0(
102.3
]Pb[S
)](0.050MPb[]][IOPb[102.3
10
2
13
2
222
3
213
sp
−
−
+
+−+−
×=
×
==
==×=K 
(c)
K sp = 2.5×10-9 = [Sr 2+][F-]2 = [Sr 2+](0.050 M)2
S = [Sr 2+] = M100.1
)050.0(
105.2 6
2
9
−
−
×=
×
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9
(d)
K sp = 1.0×10
-15 = [Th4+][OH-]4 = [Th4+](0.050 M)4
S = [Th4+] = M106.1
)050.0(
100.1 10
4
15
−
−
×=
×
9-13 ][CrO]Ag[102.1CrO2Ag)(CrOAg
2
4
212
sp
2
442
−+−−+
←
→ =×=+ K s
(a) M100.1
)1041.3(
102.1
]CrO[ 9
22
12
2
4
−
−
−
− ×=
×
×
=
(b) M30.0
)1000.2(
102.1
]CrO[
26
12
2
4 =×
×
= −
−
−
9-14 3334sp
3
3 ]][OHAl[100.33OHAl)(Al(OH)
−+−−+
←
→ =×=+ K s
3432
sp
23
3
100.3][OH][OH
3
1
M105.0
][OH
3
1
M105.0][Al
][OH
3
1
Al(OH)of solubilityMolar 
−−−−
−−+
−
×=⎟
 ⎠
 ⎞
⎜
⎝ 
⎛  +×=
+×=
=
K 
(a) Because the K sp is so small, we can assume the solubility of Al(OH)3 is not large.
Therefore, it is reasonable that the [Al3+] ≈ 5.0 X 10
-2 M. The K sp equation then
simplifies to
K sp = (5.0×10
-2
 M)[OH
-
]
3
 = 3.0×10-34
[OH-] = M100.2
M100.5
100.3 113
1
2
34
−
−
−
×=⎟⎟
 ⎠
 ⎞
⎜⎜
⎝ 
⎛ 
×
×
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9
(b) As in part (a),
M101.1
M1000.2
100.3
]OH[ 9
3
1
7
34
−
−
−
− ×=⎟⎟
 ⎠
 ⎞
⎜⎜
⎝ 
⎛ 
×
×
=
9-15 10333
3
sp 102.3)S3(S]][IOCe[
−−+ ×===K 
(a)
M0125.0
L
mL1000
)mL0.50mL0.50(
mole1025.1
mole1025.1mL0.50
mL1000
L
L
mole
105.2CeM0250.0
3
323
=×
+
×
×=×××≡
−
−−+
(b) mole100.2mL0.50mL1000
L
L
mole
100.4IOM040.0 323
−−− ×=×××≡
( ) mole10833.5mole100.2
3
1
mole1025.1unreactedCemoles 4333 −−−+ ×=×−×=
( )
SM10833.5S
L
mL1000
mL0.50mL0.50
mole10833.5
]Ce[ 3
4
3 +×=+⎟⎟
 ⎠
 ⎞
⎜⎜
⎝ 
⎛ 
×
+
×
= −
−
+
where S is derived iteratively using the equation
33
sp )S3()S10833.5( +×=
−K 
We start by solving for S assuming no contribution to [Ce
3+
] other than from the
dissociation of Ce(IO3)3. In this case, S equals 1.855 × 10-3. Now, we substitute 1.855 ×
10-3 back into the K sp equation above and solve for K iterative. K iterative equals 1.3259 × 10-9.
S is too large; we choose a smaller S (i.e., 1 × 10-3) and recalculate K iterative. Iteration
continues until K iterative ≈ K sp. The results of this approach are shown in the Table below.
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9
Iteration S  K iterative
1 1.855×10-3 1.3259×10-9
2 1×10-3 1.8449×10-10
3 1.2×10-3 3.2813×10-10
4 1.19×10-3 3.1954×10-10
We now substitute S = 1.19 × 10-3 into the equation below,
M100.71019.1M10833.5]Ce[ 3333 −−−+ ×=×+×=
(c)
mole0125.0mL0.50
mL1000
L
L
mole
105.2IOM250.0 13 =×××≡
−−
The Ce3+ is completely consumed; thus,
103
sp
3
3
33
3
102.3)S3M0875.0(S
S3M0875.0S3
L
mL1000
mL0.100
mole1075.8
]IO[
mole1075.8)mole1025.1(3mole0125.0unreactedIOmoles
−
−
−
−−−
×=+=
+=+⎟⎟
 ⎠
 ⎞
⎜⎜
⎝ 
⎛ 
×
×
=
×=×−=
K 
Make the assumption that 3S << 0.0375 M.
M108.4
)M0875.0(
102.3
S]Ce[
102.3)M0875.0(S
7
3
10
3
103
sp
−
−
+
−
×=⎟⎟
 ⎠
 ⎞
⎜⎜
⎝ 
⎛  ×
==
×==K 
(d)
mole105.7mL0.50
mL1000
L
L
mole
105.1IOM150.0 313
−−− ×=×××≡
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 99
The CeThe Ce3+3+ is completely consumed; thus, is completely consumed; thus,
101033
spsp
33
33
333333
33
101022..33))SS33MM03750375..00((SS
SS33MM03750375..00SS33
LL
mLmL10001000
mLmL00..100100
molemole10107575..33
]]IOIO[[
molemole10107575..33))molemole10102525..11((33molemole101055..77unreactedunreactedIOIOmolesmoles
−−
−−
−−
−−−−−−−−
××==++==
++==++⎟⎟⎟⎟
 ⎠ ⎠
 ⎞ ⎞
⎜⎜⎜⎜
⎝ ⎝ 
⎛ ⎛ 
××
××
==
××==××−−××==
K K 
Make the assumption that 3S << 0.0375 M.Make the assumption that 3S << 0.0375 M.
MM101011..66
))MM03750375..00((
101022..33
SS]]CeCe[[
101022..33))MM03750375..00((SS
66
33
1010
33
101033
spsp
−−
−−
++
−−
××==⎟⎟⎟⎟
 ⎠ ⎠
 ⎞ ⎞
⎜⎜⎜⎜
⎝ ⎝ 
⎛ ⎛  ××
====
××====K K 
9-169-16 6622
22
66
22
spsp 101000..66))SS(())SS22((]][PdCl[PdCl]]K K [[
−−−−++ ××======K K 
MM100100..00
LL
mLmL10001000
))mLmL00..5050mLmL00..5050((
molemole10100000..11
molemole01000100..00mLmL00..5050
mLmL10001000
LL
LL
molemole
101000..22K K MM200200..00
22
11
==××
++
××
==××××××≡≡
−−
−−++
(a)(a) molemole101055..22mLmL00..5050
mLmL10001000
LL
LL
molemole
101000..55PdClPdClMM05000500..00 3322
22
66
−−−−−− ××==××××××≡≡
))
MM050050..00
LL
mLmL10001000
))mLmL00..5050mLmL00..5050((
molemole101000..55
molemole101000..55molemole101055..2222molemole10100000..11unreactedunreactedK K molesmoles
33
333322
==××
++
××
××==××−−××==
−−
−−−−−−++
The [K The [K ++] is 0.05 M plus the con] is 0.05 M plus the contribution from the dissociation of the precipitate,tribution from the dissociation of the precipitate, x x, or, or
[K [K ++] = 0.05 M + 2] = 0.05 M + 2 x x
Substituting the equation above into Substituting the equation above into the equilibrium expression, we findthe equilibrium expression, we find
66332233
2222
66
22
spsp
101000..66))101055..22((2020..0044
))220505..00((]]PdClPdCl[[]]K K [[
−−−−
−−++
××==××++++
++====
 x x x x x x
 x x x xK K 
We use an iterative approach to solve forWe use an iterative approach to solve for x x. . We begin by We begin by ignoring the contributiignoring the contribution to [K on to [K ++]]
from the dissociation of the precipitate.from the dissociation of the precipitate.
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 99
011447011447..00
44
101000..66
33
66
==
××
==
−−
 x x
Substituting this value back into theSubstituting this value back into theK K spsp equation above, we obtain equation above, we obtainK K iterativeiterative equal to equal to
6.08246.0824 ×× 10 10-5-5,, which is too lwhich is too large. arge. We now proceed by We now proceed by choosing a smaller choosing a smaller value ofvalue of x x equal equal
11 ×× 10 10-3-3 and obtain and obtain K K iterativeiterative equals to 2.7040 equals to 2.7040×× 10 10-6-6. . We continue We continue the iterative the iterative approachapproach
untiluntil K K iterativeiterative ≈≈ K K spsp as shown in the table below. as shown in the table below.
IterationIteration  x  x K K iterativeiterative
1 0.0114471 0.011447 6.08146.0814 ×× 10 10-5-5
22 11 ×× 10 10-3-3 2.70402.7040 ××1010-6-6
33 22 ×× 10 10-3-3 5.83205.8320 ××1010-6-6
SubstitutingSubstituting x x equals 2 equals 2 ×× 10 10-3-3 into the [K into the [K ++] equation, we find] equation, we find
[K [K ++] = 0.05 M + 2(2] = 0.05 M + 2(2 ×× 10 10-3-3M) = 0.054 MM) = 0.054 M
(b)(b)
molemole10100000..55mLmL00..5050
mLmL10001000
LL
LL
molemole
101000..11PdClPdClMM100100..00 3311
22
66
−−−−−− ××==××××××≡≡
The [K The [K ++] is determined directly from] is determined directly from K K spsp; thus,; thus,
MM022022..00))MM011011..00((22SS22]]K K [[
011011..00
44
101000..66
SS
101000..66SS44SS))SS22((
SS]]PdClPdCl[[SS22]]K K [[
33
11
66
663322
spsp
22
66
======
==⎟⎟⎟⎟
 ⎠ ⎠
 ⎞ ⎞
⎜⎜⎜⎜
⎝ ⎝ 
⎛ ⎛  ××
==
××======
====
++
−−
−−
−−++
K K 
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 99
(c)(c)
molemole010010..00mLmL00..5050
mLmL10001000
LL
LL
molemole
101000..22PdClPdClMM200200..00 11
22
66 ==××××××≡≡
−−−−
The K The K ++ is completely consumed; thus, is completely consumed; thus,
MM011011..00))MM1010477477..55((22SS22]]K K [[
MM1010477477..55
))0505..00((44
101000..66
SS
101000..66))MM0505..00((SS44
MM0505..00]]PdClPdCl[[
MM0505..00SSthatthatAssumeAssume
SSMM0505..00SS
LL
mLmL10001000
mLmL100.0100.0
molemole10105.05.0
]][PdCl[PdCl
molemole10105.05.0mole)mole)1010(1.00(1.00
22
11
molemole10101.01.0unreactedunreactedPdClPdClmolesmoles
33
33
22
11
66
6622
spsp
22
66
33
22
66
33--222222
66
==××====
××==⎟⎟⎟⎟
 ⎠ ⎠
 ⎞ ⎞
⎜⎜⎜⎜
⎝ ⎝ 
⎛ ⎛  ××
==
××====
==
<<<<
++==++⎟⎟⎟⎟
 ⎠ ⎠
 ⎞ ⎞
⎜⎜⎜⎜
⎝ ⎝ 
⎛ ⎛ 
××
××
==
××==××−−××==
−−++
−−
−−
−−
−−
−−
−−
−−−−−−
K K 
9-179-17 CuI(CuI(ss)) Cu Cu++ + I + I-- K K spsp = [Cu = [Cu
++][I][I--] = 1] = 1××1010-12-12
AgI(AgI(ss)) Ag Ag++ + I + I-- K K spsp = [Ag = [Ag
++][I][I--] = 8.3] = 8.3××1010-17-17
PbIPbI22((ss)) Pb Pb
2+2+ + 2I + 2I-- K K spsp = [Pb = [Pb
2+2+][I][I--]]22 = 7.1 = 7.1××1010-9-9 = = S(2S)S(2S)22 = 4S = 4S33
BiIBiI33((ss)) Bi Bi
3+3+
 + 3I + 3I
--
K K spsp = [Bi = [Bi
3+3+
][I][I
--
]]
33
 = 8.1 = 8.1××1010-19-19 = = S(3S)S(3S)33 = 27S = 27S44
(a)(a)
water water ininAgIAgICuICuIBiIBiIPbIPbI
101033..11
2727
))101011..88((
SS]]II[[
33
11
]]BiBi[[SS
101022..11
44
))101011..77((
SS]]II[[
22
11
]]PbPb[[SS
101011..99101033..88SS]][I[I]]AgAg[[SS
101011101011SS]][I[I]]CuCu[[SS
3322
5544
1919
33
BiIBiI
3333
99
22
PbIPbI
991717
AgIAgI
661212
CuICuI
33
22
>>>>>>
××==
××
======
××==
××
======
××==××======
××==××======
−−
−−
−−++
−−
−−
−−++
−−−−−−++
−−−−−−++
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 99
(b)(b)
 NaI NaIMM1010..00ininBiIBiIAgIAgICuICuIPbIPbI
101011..88
))MM1010..00((
101011..88
SS
101011..77
))MM1010..00((
101011..77
SS
101033..88
MM1010..00
101033..88
SS
101011
MM1010..00
101011
SS
3322
1616
33
1919
BiIBiI
77
22
99
PbIPbI
1616
1717
AgIAgI
1111
1212
CuICuI
33
22
>>>>>>
××==
××
==
××==
××
==
××==
××
==
××==
××
==
−−
−−
−−
−−
−−
−−
−−
−−
(c)(c)
cationcationsolutesolutethetheof of solutionsolutionMM0.0100.010aaininAgIAgICuICuIBiIBiIPbIPbI101044..11
MM010010..00
101011..88
33
11
SS
101022..44
MM010010..00
101011..77
22
11
SS
101033..88
MM010010..00
101033..88
SS
101011
MM010010..00
101011
SS
3322
66
33
1919
BiIBiI
44
99
PbIPbI
1515
1717
AgIAgI
1010
1212
CuICuI
33
22
>>>>>>
××==
××
==
××==
××
==
××==
××
==
××==
××
==
−−
−−
−−
−−
−−
−−
−−
−−
9-189-18 BiOOH(BiOOH(ss)) BiO BiO++ + OH + OH-- K K spsp = [BiO = [BiO
++][OH][OH--] = 4.0] = 4.0××1010-10-10
Be(OH)Be(OH)22((ss)) Be Be
2+2+ + 2OH + 2OH-- K K spsp = [Be = [Be
2+2+][OH][OH--]]22 = 7.0 = 7.0××1010-22-22 = S(2S) = S(2S)22 = 4S = 4S33
Tm(OH)Tm(OH)33((ss)) Tm Tm
3+3+ + 3OH + 3OH-- K K spsp = [Tm = [Tm
3+3+][OH][OH--]]33 = 3.0 = 3.0××1010-24-24 = S(3S) = S(3S)33 = 27S = 27S44
Hf(OH)Hf(OH)44((ss)) Hf  Hf 
4+4+ + 4OH + 4OH-- K K spsp = [Hf  = [Hf 
4+4+][OH][OH--]]44 = 4.0 = 4.0××1010-26-26 = S(4S) = S(4S)44 = 256S = 256S55
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 99
(a)(a)
water.water.ininsolubilitysolubilitylowestlowestthethehashasBe(OH)Be(OH)
101077..22
256256
101000..44
SS]][OH[OH
44
11
]]Hf Hf [[SS
101088..55
2727
101000..33
SS]][OH[OH
33
11
]]TmTm[[SS
101066..55
44
101000..77
SS]][OH[OH
22
11
]]BeBe[[SS
101000..22101000..44SS]][OH[OH]]BiOBiO[[SS
22
6655
2626
44
))OHOH((Hf Hf 
7744
2424
33
))OHOH((TmTm
8833
2222
22
))OHOH((BeBe
551010
BiOOHBiOOH
44
33
22
−−
−−
−−++
−−
−−
−−++
−−
−−
−−++
−−−−−−++
××==
××
======
××==
××
======
××==
××
======
××==××======
(b)(b)
.. NaOH NaOHMM0.100.10ininsolubilitysolubilitylowestlowestthethehashasHf(OH)Hf(OH)
101000..44
))MM1010..00((
101000..44
SS
101000..33
))MM1010..00((
101000..33
SS
101000..77
))MM1010..00((
101000..77
SS
101000..44
MM1010..00
101000..44
SS
44
2222
44
2626
))OHOH((Hf Hf 
2121
33
2424
))OHOH((TmTm
2020
22
2222
))OHOH((BeBe
99
1010
BiOOHBiOOH
44
33
22
−−
−−
−−
−−
−−
−−
−−
−−
××==
××
==
××==
××
==
××==
××
==
××==
××
==
9-199-19 At 0ºC,At 0ºC,
7.4727.472M)M)1010log(3.38log(3.38]]OOlog[Hlog[H pH pH
MM10103.383.3810100.1440.144]][OH[OH]]OO[H[H10100.1140.114]]][OH][OHOOHH[[
88
33
881414
33
1414
33spsp
==××−−==−−==
××==××====××====
−−++
−−−−−−++−−−−++K K 
At 100ºC,At 100ºC,
155155..66M)M)1010log(7.00log(7.00]]OOlog[Hlog[H pH pH
MM10107.007.0010104949]][OH[OH]]OO[H[H10104949]]][OH][OHOOHH[[
77
33
771414
33
1414
33spsp
==××−−==−−==
××==××====××====
−−++
−−−−−−++−−−−++K K 
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 99
9-209-20 (a)(a)
MM10103.33.3
MM10103.03.0
10101.01.0
]][OH[OH
MM10103.03.0
22
))10104(9.04(9.0))1010(3.0(3.010103.03.0
]]OO[H[H
0010109.09.0]]OO[H[H10103.03.0]]OO[H[H
])])OO[H[HMM(0.0300(0.030010103.03.0]]OO[H[H10103.03.0
])])OO[H[HMM(0.0300(0.0300
]]OO[H[H
]]OO[H[HMM0.03000.0300[HOCl][HOCl]]][OCl[OCl]]OOHH[[
101000..33
[HOCl][HOCl]
]]OO][H][H[OCl[OCl
OOHHOClOClOOHHHOClHOCl
1010
55
1414
55
1010228888
33
1010
33
8822
33
33
8822
33
88
33
22
33
3333
8833
aa3322
−−
−−
−−
−−
−−
−−−−−−
++
−−++−−++
++−−++−−
++
++
++−−++
−−
++−−
++−−
←←
→→
××==
××
××
==
××==
××++××++××
−−==
==××−−××++
−−××==××==
−−
−−====
××====++++ K K 
(b)(b)
MM10101.061.06
MM10109.479.47
10101.01.0
]][OH[OH
MM10109.479.47
22
))10104(9.124(9.12))1010(1.52(1.5210101.521.52
]]OO[H[H
0010109.129.12]]OO[H[H10101.521.52]]OO[H[H
])])OO[H[HMM(0.0600(0.060010101.521.52]]OO[H[H10101.521.52
])])OO[H[HMM(0.0600(0.0600
]]OO[H[H
]]OO[H[HMM0.06000.0600COOH]COOH]CHCHCHCH[CH[CH]]COOCOOCHCHCHCH[CH[CH]]OO[H[H
10105252..11
COOH]COOH]CHCHCHCH[CH[CH
]]OO][H][HCOOCOOCHCHCHCH[CH[CH
OOHHCOOCOOCHCHCHCHCHCHOOHHCOOHCOOHCHCHCHCHCHCH
1111
44
1414
44
77225555
33
77
33
5522
33
33
5522
33
55
33
22
33
3322223322223333
55
222233
33222233
aa
3322223322222233
−−
−−
−−
−−
−−
−−−−−−
++
−−++−−++
++−−++−−
++
++
++−−++
−−
++−−
++−−
←←
→→
××==
××
××
==
××==
××++××++××
−−==
==××−−××++
−−××==××==
−−
−−====
××====
++++
K K 
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 99
(c)(c)
MM10101.61.6
MM10106.376.37
10101.01.0
]]OO[H[H
MM10106.46.4
22
))10104(4.334(4.33))1010(4.33(4.3310104.334.33
]][OH[OH
0010104.334.33]][OH[OH10104.334.33]][OH[OH
])])[OH[OHMM(0.100(0.10010104.334.33]][OH[OH10104.334.33
])])[OH[OHMM(0.100(0.100
]][OH[OH
]][OH[OHMM0.1000.100]] NH NHHH[C[C]] NH NHHH[C[C]][OH[OH
10103333..44
10103131..22
101000..11
]] NH NHHH[C[C
]]][OH][OH NH NHHH[C[C
OHOH NH NHHHCCOOHH NH NHHHCC
1212
33
1414
33
33
55224444
554422
442244
22
225522335522
44
1111
1414
aa
ww
225522
335522
 b b
33552222225522
−−
−−
−−
++
−−
−−−−−−
−−
−−−−−−−−
−−−−−−−−
−−
−−
−−++−−
−−
−−
−−−−++
−−++
←←
→→
××==
××
××
==
××==
××++××++××
−−==
==××−−××++
−−××==××==
−−
−−====
××==
××
××
======
++++
K K 
K K 
K K 
(d)(d)
MM10102.832.83
MM10103.533.53
10101.01.0
]]OO[H[H
MM10103.533.53
22
))10104(1.274(1.27))1010(6.33(6.3310106.336.33
]][OH[OH
0010101.271.27]][OH[OH10106.336.33]][OH[OH
])])[OH[OHMM(0.200(0.20010106.336.33]][OH[OH10106.336.33
])])[OH[OHMM(0.200(0.200
]][OH[OH
]][OH[OHMM0.2000.200 N] N]))[(CH[(CH]] NH NH))[(CH[(CH]][OH[OH
10103333..66
10105858..11
101000..11
 N] N]))[(CH[(CH
]]][OH][OH NH NH))[(CH[(CH
OHOH NH NH))(CH(CHOOHH N N))(CH(CH
1212
33
1414
33
33
55225555
555522
552255
22
33333333
55
1010
1414
aa
ww
3333
3333
 b b
3333223333
−−
−−
−−
++
−−
−−−−−−
−−
−−−−−−−−
−−−−−−−−
−−
−−
−−++−−
−−
−−
−−−−++
−−++
←←
→→
××==
××
××
==
××==
××++××++××
−−==
==××−−××++
−−××==××==
−−
−−====
××==
××
××
======
++++
K K 
K K 
K K 
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 99
(e)(e)
MM10103.93.9
MM10102.62.6
10101.01.0
]]OO[H[H
MM10102.62.6
22
))10104(6.74(6.7))1010(3.3(3.310103.33.3
]][OH[OH
0010106.76.7]][OH[OH10103.33.3]][OH[OH
])])[OH[OHMM(0.200(0.20010103.33.3]][OH[OH10103.33.3
])])[OH[OHMM(0.200(0.200
]][OH[OH
]][OH[OHMM0.2000.200]][OCl[OCl[HOCl][HOCl]]][OH[OH
101033..33
101000..33
101000..11
]][OCl[OCl
]][HOCl][OH[HOCl][OH
OHOHHOClHOClOOHHOClOCl
1111
44
1414
33
44
88227777
887722
772277
22
77
88
1414
aa
ww
 b b22
−−
−−
−−
++
−−
−−−−−−
−−
−−−−−−−−
−−−−−−−−
−−
−−
−−−−−−
−−
−−
−−
−−
−−
−−
←←
→→−−
××==
××
××
==
××==
××++××++××
−−==
==××−−××++
−−××==××==
−−
−−====
××==
××
××
======++++
K K 
K K 
K K 
(f)(f)
MM10101.251.25
MM10108.018.01
10101.01.0
]]OO[H[H
MM10108.018.01
22
))10104(6.424(6.42))1010(7.42(7.4210107.427.42
]][OH[OH
0010106.426.42]][OH[OH10107.427.42]][OH[OH
])])[OH[OHMM(0.0860(0.086010107.427.42]][OH[OH10107.427.42
])])[OH[OHMM(0.0860(0.0860
]][OH[OH
]][OH[OHMM0.08600.0860]]COOCOOHH[C[CCOOH]COOH]HH[C[C]][OH[OH
10104242..77
10103434..11
101000..11
]]COOCOOHH[C[C
]]COOH][OHCOOH][OHHH[C[C
OHOHCOOHCOOHHHCCOOHHCOOCOOHHCC
99
66
1414
33
66
11112210101010
1111101022
1010221010
22
55225522
1010
55
1414
aa
ww
5522
5522
 b b
5522225522
−−
−−
−−
++
−−
−−−−−−
−−
−−−−−−−−
−−−−−−−−
−−
−−
−−−−−−
−−
−−
−−
−−
−−
−−
←←
→→−−
××==
××
××
==
××==
××++××++××
−−==
==××−−××++
−−××==××==
−−
−−====
××==
××
××
======
++++
K K 
K K 
K K 
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 99
(g)(g)
MM10101.911.91
MM10105.245.24
10101.01.0
]][OH[OH
MM10105.245.24
22
))10104(2.754(2.75))1010(1.10(1.1010101.101.10
]]OO[H[H
0010102.752.75]]OO[H[H10101.101.10]]OO[H[H
])])OO[H[HMM(0.250(0.25010101.101.10]]OO[H[H10101.101.10
])])OO[H[HMM(0.250(0.250
]]OO[H[H
]]OO[H[HMM0.2500.250]][HONH[HONH]][HONH[HONH]]OO[H[H
10101010..11
]][HONH[HONH
]]OO][H][H[HONH[HONH
OOHHHONHHONHOOHHHONHHONH
1111
44
1414
44
77226666
33
77
33
6622
33
33
6622
33
66
33
22
33
33332233
66
33
3322
aa33222233
−−
−−
−−
−−
−−
−−−−−−
++
−−++−−++
++−−++−−
++
++
++++++
−−
++
++++
←←
→→++
××==
××
××
==
××==
××++××++××
−−==
==××−−××++
−−××==××==
−−
−−====
××====++++ K K 
(h)(h)
MM10103.553.55
MM10102.822.82
10101.01.0
]][OH[OH
MM10102.822.82
22
))10104(7.954(7.95))1010(3.18(3.1810103.183.18
]]OO[H[H
0010107.957.95]]OO[H[H10103.183.18]]OO[H[H
])])OO[H[HMM(0.0250(0.025010103.183.18]]OO[H[H10103.183.18
])])OO[H[HMM(0.0250(0.0250
]]OO[H[H
]]OO[H[HMM0.02500.0250]] NH NHHH[HOC[HOC]] NH NHHH[HOC[HOC]]OO[H[H
10101818..33
]] NH NHHH[HOC[HOC
]]OO][H][H NH NHHH[OHC[OHC
OOHH NH NHHHHOCHOCOOHH NH NHHHHOCHOC
99
66
1414
66
12122210101010
33
1212
33
101022
33
33
101022
33
1010
33
22
33
3333442222442233
1010
334422
33224422
aa3322442222334422
−−
−−
−−
−−
−−
−−−−−−
++
−−++−−++
++−−++−−
++
++
++++++
−−
++
++
++
←←
→→++
××==
××
××
==
××==
××++××++××
−−==
==××−−××++
−−××==××==
−−
−−====
××====++++ K K 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9
9-21 (a)
M1100.0
2
)104(1.36)10(1.36101.36
]O[H
0101.36]O[H101.36]O[H
])O[HM(0.100101.36]O[H101.36
])O[HM(0.100
]O[H
]O[HM0.100COOH][ClCH]COO[ClCH]O[H
1036.1
COOH][ClCH
]O][HCOO[ClCH
OHCOOClCHOHCOOHClCH
4233
3
4
3
32
3
3
32
3
3
3
2
3
3223
3
2
32
a3222
=
×+×+×
−=
=×−×+
−×=×=
−
−==
×==++
−−−
+
−+−+
+−+−
+
+
+−+
−
+−
+−
←
→ K 
(b)
M101.17
M108.57
101.0
]O[H
M108.57
2
)104(7.35)10(7.35107.35
][OH
0107.35][OH107.35][OH
])[OHM(0.100107.35][OH107.35
])[OHM(0.100
][OH
][OHM0.100]COO[ClCHCOOH][ClCH][OH
1035.7
1036.1
100.1
]COO[ClCH
]COOH][OH[ClCH
OHCOOHClCHOHCOOClCH
8
7
14
3
7
1321212
13122
12212
2
22
12
3
14
a
w
2
2
 b
222
−
−
−
+
−
−−−
−
−−−−
−−−−
−
−
−−−
−
−
−
−
−
−
←
→−
×=
×
×
=
×=
×+×+×
−=
=×−×+
−×=×=
−
−==
×=
×
×
===
++
K 
K 
K 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9
(c)
M105.32
M101.88
101.0
]O[H
M101.88
2
)104(4.35)10(4.35104.35
][OH
0104.35][OH104.35][OH
])[OHM(0.0100104.35][OH104.35
])[OHM(0.0100
][OH
][OHM0.0100] NH[CH] NH[CH][OH
1035.4
103.2
100.1
] NH[CH
]][OH NH[CH
OH NHCHOH NHCH
12
3
14
3
3
6244
642
424
2
2333
4
11
14
a
w
23
33
 b
33223
−
−
−
+
−
−−−
−
−−−−
−−−−
−
−
−+−
−
−
−−+
−+
←
→
×=
×
×
=
×=
×+×+×
−=
=×−×+
−×=×=
−
−==
×=
×
×
===
++
K 
K 
K 
(d)
M104.8
2
)104(2.3)10(2.3102.3
]O[H
0102.3]O[H102.3]O[H
])O[HM(0.0100102.3]O[H102.3
])O[HM(0.0100
]O[H
]O[HM0.0100] NH[CH] NH[CH]O[H
103.2
] NH[CH
]O][H NH[CH
OH NHCHOH NHCH
7
1321111
3
13
3
112
3
3
112
3
11
3
2
3
333233
11
33
323
a323233
−
−−−
+
−+−+
+−+−
+
+
+++
−
+
+
+
←
→+
×=
×+×+×
−=
=×−×+
−×=×=
−
−==
×==++ K 
(e)
M101.46
2
)104(2.51)10(2.51102.51
]O[H
0102.51]O[H102.51]O[H
])O[HM(0.0010102.51]O[H102.51
])O[HM(0.0010
]O[H
]O[HM0.0010] NHH[C] NHH[C]O[H
1051.2
] NHH[C
]O][H NHH[C
OH NHHCOH NHHC
4
8-255
3
8
3
52
3
3
52
3
5
3
2
3
33562563
5
356
3256
a32562356
−
−−
+
−+−+
+−+−
+
+
+++
−
+
+
+
←
→+
×=
×+×+×
−=
=×−×+
−×=×=
−
−==
×==++ K 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9
(f)
0.12M
2
4(0.034))10(1.7101.7
]O[H
00.034]O[H101.7]O[H
])O[HM(0.200101.7]O[H101.7
])O[HM(0.200
]O[H
]O[HM0.200][HIO][IO]OH[
107.1
][HIO
]O][H[IO
OHIOOHHIO
211
3
3
12
3
3
12
3
1
3
2
3
3333
1
3
33
a3323
=
+×+×
−=
=−×+
−×=×=
−
−==
×==++
−−
+
+−+
+−+−
+
+
+−+
−
+−
+−
←
→ K 
9-22 A buffer solution resists changes in pH with dilution or with addition of acids or bases. A
 buffer is composed of a mixture of a weak acid and its conjugate base.
9-23  Buffer capacity of a solution is defined as the number of moles of a strong acid (or a
strong base) that causes 1.00 L of a buffer to undergo a 1.00-unit change in pH.
9-24 (a) 943.8
(0.200M)
(0.100M)
log)107.5log(
][NH
][NH
log p pH 10
4
3
a =+×−=+=
−
+K 
(b) 943.8
(0.100M)
(0.050M)
log)107.5log(
][NH
][NH
log p pH 10
4
3
a =+×−=+=
−
+K 
The solutions have identical pH values, but the solution in part (a) has the greater buffer
capacity because it has the higher concentration of weak acid and conjugate base.
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9
9-30
( )
 NaOHmL3.89
L
mL1000
mole00.2
L
mole179.0 NaOHvolume
mole179.0
47.2
)mole300.0(47.1
47.2
)COOHOCHmole(47.1
 NaOHmole
) NaOHmoleCOOHHOCHmole(47.1 NaOHmole
47.1
) NaOHmoleCOOHHOCHmole(
 NaOHmole
47.1
COOHHOCHmole
COOHOCHmole
47.110
]COOHHOCH[
]COOHOCH[
167.0
]COOHHOCH[
]COOHOCH[
log
]COOHHOCH[
]COOHOCH[
log83.3
]COOHHOCH[
]COOHOCH[
log1047.1log
]COOHHOCH[
]COOHOCH[
log pK 00.4 pH
COOHHOCHmole300.0mL0.300
mL1000
L
L
mole00.1
COOHHOCHM00.1
2
2
2
2
2
1067.1
2
2
2
2
2
2
2
24
2
2
a
22
1
=××=
=
×
=
×
=
−×=
=
−
=
===
+=+×−=
+==
=××≡
−
−
×
−−
−−
−
−
−
 x
 x x
 x
 x
9-31 The statement “A buffer maintains the pH of a solution constant” is false. The change in
 pH of a buffered solution is relatively small with the addition of a small volume of acid
or base as shown in the example below.
]HA[
] NaA[
log p pH a += K 
mL of 0.050M NaOH
]HA[
] NaA[
∆ pH
1.48 0.170
100 1.59 0.200
200 1.70 0.230
300 1.83 0.262
Fundamentals of Analytical Chemistry: 8th ed. Chapter 10
Chapter 10
10-1 (a) Activity, aA, is the effective concentration of a chemical species A in solution. The
activity coefficient , γA, is the numerical factor necessary to convert the molar
concentration of the chemical species A to activity as shown below:
aA=γA[A]
(b) The thermodynamic equilibrium constant  refers to an ideal system within which each
chemical species is unaffected by any others. A concentration equilibrium constant  takes
into account the influence exerted by solute species upon one another. The
thermodynamic equilibrium constant is numerically constant and independent of ionic
strength; the concentration equilibrium constant depends upon molar concentrations of
reactants and products as well as other chemical species that may not participate in the
equilibrium.
10-2
Activity coefficients have the following properties:
1. The activity coefficient depends on the solution ionic strength.
2. In very dilute solutions, the activity coefficient approaches unity.
3. For a given ionic strength, the activity coefficient becomes smaller as the charge of the
chemical species increases.
4. At any ionic strength, the activity coefficients are approximately equal for chemical
species having the same charge state.
10-3 (a)  NaCl2)()OH(Mg NaOH2MgCl 22 ++ ←
→ s
Replacing divalent-Mg
2+
 with Na
+
 causes the activity coefficient to increase (i.e.,
Fundamentals of Analytical Chemistry: 8th ed. Chapter 10
approach 1.0). Thus, the ionic strength decreases.
(b)  NaClOH NaOHHCl 2 ++ ←
→
The activity coefficient remains relatively constant when the NaOH (strong base) is
added to HCl (strong acid). There is no change in the charge states of the ions present in
the solution equilibria. The ionic strength is unchanged.
(c)  NaOAcOH NaOHHOAc 2 ++ ←
→
The activity coefficient will decrease when the NaOH (strong base) is added to the acetic
acid (weak acid) generating water, Na+ and OAc- (conjugate base). Thus, the ionic
strength increases.
10-4 (a) The ionic strength will increase when FeCl3 is added to HCl.
(b)  NaCl3)()OH(FeFeCl NaOH3 33 ++ ←
→ s
Addition of FeCl3 to NaOH replaces a univalent OH
- with univalent Cl-; thus, the ionic
strength is unchanged.
(c) )(AgCl3) NO(FeFeClAgNO3 3333 s++ ←
→
Addition of FeCl3 replaces the univalent Ag
+ with the trivalent Fe3+; thus, the ionic
strength increases.
10-5 The initial slope of the change of activity coefficient for Ca
2+
 is steeper than that for K 
+
 because activity coefficients for multiply-charged ions deviate from ideality (unity) more
than activity coefficients for singly-charged ions.
10-6 The chemical species NH3 is not charged; therefore, the activity coefficient is unity.
Fundamentals of Analytical Chemistry: 8th ed. Chapter 10
%46%100
103.2
103.2102.1
error relative
M102.1
27
106
S)2(M103.2
27
103.7
S
109.2S)S3(
6
66
6
4
1
23
6
4
1
22
223
−=×
×
×−×
=
×=⎟⎟
 ⎠
 ⎞
⎜⎜
⎝ 
⎛  ×
=
×=⎟⎟
 ⎠
 ⎞
⎜⎜
⎝ 
⎛  ×
=
×=
−
−−
−
−
−
−
−
10-15 (a) ( )( ) 200.0)1(200.01200.0
2
1 22 =+=µ
%32%100
100.1
100.11032.1
error relative
M1032.1)10.0)(1075.1(]OH[
M1000.1)10.0)(1001.1(]OH[
1001.1)1075.1)(724.0)(798.0(
]HOAc[
]OAc[]OH[
'
724.0140.0
20.0)425.0)(3.3(1
20.0)1)(51.0(
log
798.00979.0
20.0)9.0)(3.3(1
20.0)1)(51.0(
log
3
33
35
3
35
3
55
aOAcOH
OAc3OH
a
OAc
2
OAc
OH
2
OH
3
3
33
=×
×
×−×
=
×=×=
×=×=
×=×=γγ=
γγ
=
=γ=
+
=γ−
=γ=
+
=γ−
−
−−
−−+
−−+
−−
−+
−+
−+
−−
++
K K 
(b) ( )( ) 200.0)1(200.01200.0
2
1 22 =+=µ
708.0150.0
20.0)35.0)(3.3(1
20.0)1)(51.0(
log
681.0167.0
20.0)25.0)(3.3(1
20.0)1)(51.0(
log
OAc
2
OH
OH
2
 NH 34
=γ=
+
=γ−
=γ=
+
=γ−
−−
++
Fundamentals of Analytical Chemistry: 8th ed. Chapter 10
%27%100
1046.1
1046.11007.1
error relative
M1007.1
1036.9
100.1
]OH[M1036.9)050.0)(1075.1(]OH[
M1046.1
1083.6
100.1
]OH[M1083.6)050.0)(1032.9(]OH[
1032.9
1070.5
100.1
)708.0)(681.0(
] NH[
]OH[] NH[
'
11
1111
11
4
14
3
45
11
4
14
3
46
6
10
14
 bOH NH
3
OH4 NH
 b
4
4
=×
×
×−×
=
×=
×
×
=×=×=
×=
×
×
=×=×=
×=⎟⎟
 ⎠
 ⎞
⎜⎜
⎝ 
⎛ 
×
×
=γγ=
γγ
=
−
−−
−
−
−
+−−−
−
−
−
+−−−
−
−
−−
+
−+
−+
K K 
(c)
( )( ) 060.0)1(0600.010600.0
2
1 22 =+=µ
807.00930.0
060.0)425.0)(3.3(1
060.0)1)(51.0(
log
847.007232.0
060.0)9.0)(3.3(1
060.0)1)(51.0(
log
COOClCH
2
COOClCH
OH
2
OH
22
33
=γ=
+
=γ−
=γ=
+
=γ−
−−
++
%21%100
1005.3
1005.31069.3
error relative
M1069.3)01.0)(1036.1(]OH[
M1005.3)01.0)(103.9(]OH[
1030.9)1036.1)(807.0)(847.0(
]COOHClCH[
]COOClCH[]OH[
'
3
33
33
3
34
3
43
aCOOClCHOH
2
2OAc3OH
a
23
3
=×
×
×−×
=
×=×=
×=×=
×=×=
γγ=
γγ
=
−
−−
−−+
−−+
−−
−+
−+
−+
K K 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 10
10-16
 A B C D E F G H I
1 Prob lem 10-16(a)
2
3 Ksp(AgIO3)
4 3.10E-08
5
6 c(Ba(NO3)2) pc(Ba(NO3)2) s(conc) ps(conc) s(act) ps(act)
+) )
7 0.0001 4.00 1.76E-04 3.75 1.81E-04 3.74 4.76E-04 0.975 0.975
8 0.0002 3.70 1.76E-04 3.75 1.82E-04 3.74 7.76E-04 0.969 0.969
9 0.0004 3.40 1.76E-04 3.75 1.84E-04 3.74 1.38E-03 0.959 0.959
10 0.0008 3.10 1.76E-04 3.75 1.86E-04 3.73 2.58E-03 0.944 0.946
11 0.001 3.00 1.76E-04 3.75 1.87E-04 3.73 3.18E-03 0.939 0.941
12 0.002 2.70 1.76E-04 3.75 1.92E-04 3.72 6.18E-03 0.917 0.920
13 0.004 2.40 1.76E-04 3.75 1.98E-04 3.70 1.22E-02 0.888 0.894
14 0.008 2.10 1.76E-04 3.75 2.06E-04 3.69 2.42E-02 0.851 0.861
15 0.01 2.00 1.76E-04 3.75 2.09E-04 3.68 3.02E-02 0.837 0.849
16 0.02 1.70 1.76E-04 3.75 2.21E-04 3.66 6.02E-02 0.787 0.807
17 0.04 1.40 1.76E-04 3.75 2.37E-04 3.63 1.20E-01 0.729 0.760
18 0.08 1.10 1.76E-04 3.75 2.56E-04 3.59 2.40E-01 0.664 0.711
19 0.1 1.00 1.76E-04 3.75 2.64E-04 3.58 3.00E-01 0.642 0.695
20 0.2 0.70 1.76E-04 3.75 2.89E-04 3.54 6.00E-01 0.574 0.647
21 0.4 0.40 1.76E-04 3.75 3.18E-04 3.50 1.20E+00 0.509 0.602
22 0.8 0.10 1.76E-04 3.75 3.50E-04 3.46 2.40E+00 0.450 0.564
23 1 0.00 1.76E-04 3.75 3.60E-04 3.44 3.00E+00 0.433 0.553
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39 Spreadsheet Documentation
40 B7=-LOG(A7) F7=-LOG(E7)
41 C7=SQRT(A$4) G7=0.5((A7)*(2)*(2)+(2*A7)*(1)*(1)+(C7)*(1)*(1)+(C7)*(-1)*(-1))
42 D7=-LOG(C7) H7=10^(-0.51*(1)*(1)*SQRT(G7)/(1+(3.3*0.25*SQRT(G7)))
43 E7=SQRT(A$4/(H7*I7)) I7=10^(-0.51*(-1)*(-1)*SQRT(G7)/(1+(3.3*0.425*SQRT(G7)))
(Ag (IO3
-
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 1010
 A  A B B C C D D E E F F G G H H II
1 1 ProbProb lem lem 10-1610-16(c)(c)
22
33 KKspsp(BaSO(BaSO44))
44 1.10E-101.10E-10
55
66 cc(Ba(NO(Ba(NO33))22)) ppcc(Ba(NO(Ba(NO33))22)) ss((ccoonncc)) ppss(conc)(conc) ss(act)(act) ppss(act)(act)
2+2+))
77 00..00000011 44..000 0 11..1100EE--0066 5.965.96 11..3322EE--0066 55..888 8 44..0099EE--0044 00..991122 00..991122
88 00..00000022 33..770 0 55..5500EE--0077 6.266.26 77..1100EE--0077 66..115 5 88..0044EE--0044 00..888800 00..887799
99 00..00000044 33..440 0 22..7755EE--0077 6.566.56 33..9922EE--0077 66..441 1 11..6600EE--0033 00..883388 00..883366
1010 00..00000088 33..110 0 11..3388EE--0077 6.866.86 22..2255EE--0077 66..665 5 33..2200EE--0033 00..778844 00..778811
1111 00..000011 33..000 0 11..1100EE--0077 6.966.96 11..8899EE--0077 66..772 2 44..0000EE--0033 00..776644 00..776600
1212 00..000022 22..770 0 55..5500EE--0088 7.267.26 11..1155EE--0077 66..994 4 88..0000EE--0033 00..669933 00..668877
1313 00..000044 22..440 0 22..7755EE--0088 7.567.56 77..4488EE--0088 77..113 3 11..6600EE--0022 00..661122 00..660011
1414 00..000088 22..110 0 11..3388EE--0088 7.867.86 55..1199EE--0088 77..228 8 33..2200EE--0022 00..552233 00..550077
1515 00..0011 22..000 0 11..1100EE--0088 7.967.96 44..6699EE--0088 77..333 3 44..0000EE--0022 00..449933 00..447766
1616 00..0022 11..770 0 55..5500EE--0099 8.268.26 33..5588EE--0088 77..445 5 88..0000EE--0022 00..440044 00..338800
1717 00..0044 11..440 0 22..7755EE--0099 8.568.56 22..9922EE--0088 77..554 4 11..6600EE--0011 00..332222 00..229922
1818 00..0088 11..110 0 11..3388EE--0099 8.868.86 22..4499EE--0088 77..660 0 33..2200EE--0011 00..225533 00..221188
1919 00..11 11..000 0 11..1100EE--0099 8.968.96 22..3388EE--0088 77..662 2 44..0000EE--0011 00..223344 00..119988
2020 00..22 00..770 0 55..5500EE--1100 9.269.26 22..0066EE--0088 77..669 9 88..0000EE--0011 00..118833 00..114466
2121 00..44 00..440 0 22..7755EE--1100 9.569.56 11..7744EE--0088 77..776 16 1..6600EE++0000 00..114466 00..110088
2222 00..88 00..110 0 11..3388EE--1100 9.869.86 11..4400EE--0088 77..885 35 3..2200EE++0000 00..111199 00..008822
2323 11 00..000 0 11..1100EE--1100 9.969.96 11..2299EE--0088 77..889 49 4..0000EE++0000 00..111133 00..007766
2424
2525
2626
2727
2828
2929
3030
3131
3232
3333
3434
3535
3636
3737
3838
39 39 Spreadsheet Spreadsheet DocumentDocumentationation
40 40 B7=-LB7=-LOG(A7) OG(A7) F7=-LOG(E7)F7=-LOG(E7)
41 41 C7=(A$4/(AC7=(A$4/(A7)) 7)) G7=0.5((A7)*(2)*(2)+(2*AG7=0.5((A7)*(2)*(2)+(2*A7)*(1)*(1))7)*(1)*(1))
42 42 D7=-LOG(C7) D7=-LOG(C7) H7=10^H7=10^(-0.51*(2)*(2)*SQRT(G7)/(1+(3.3*0.50*SQRT(G7)))(-0.51*(2)*(2)*SQRT(G7)/(1+(3.3*0.50*SQRT(G7)))
43 43 E7=(A$4/(AE7=(A$4/(A7*H7*I7)) 7*H7*I7)) I7=10^I7=10^(-0.51*(-2)*(-2)*SQRT(G7)/(1+(3.3*0.40*SQRT(G7)))(-0.51*(-2)*(-2)*SQRT(G7)/(1+(3.3*0.40*SQRT(G7)))
(Ba(Ba (SO(SO44
2-2-))
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 1010
 A  A B B C C D D E E F F G G H H II
1 1 ProbProb lem lem 10-1610-16(d)(d)
22
33 KKspsp(La(IO(La(IO33))33
44 1.00E-111.00E-11
55
66 cc(Ba(NO(Ba(NO33))22)) ppcc(Ba(NO(Ba(NO33))22)) ss((ccoonncc)) ppss(conc)(conc) ss(act)(act) ppss(act)(act)
3+3+))
77 00..00000011 44..000 0 77..8800EE--0044 3.113.11 99..6633EE--0044 33..002 2 44..9988EE--0033 00..554400 00..992277
88 00..00000022 33..770 0 77..8800EE--0044 3.113.11 99..6699EE--0044 33..001 1 55..2288EE--0033 00..553322 00..992255
99 00..00000044 33..440 0 77..8800EE--0044 3.113.11 99..7788EE--0044 33..001 1 55..8888EE--0033 00..551177 00..992211
1010 00..00000088 33..110 0 77..8800EE--0044 3.113.11 99..9966EE--0044 33..000 0 77..0088EE--0033 00..449911 00..991155
1111 00..000011 33..000 70 7..8800EE--0044 3.113.11 11..0000EE--0033 33..000 0 77..6688EE--0033 00..448800 00..991122
1212 00..000022 22..770 70 7..8800EE--0044 3.113.11 11..0044EE--0033 22..998 8 11..0077EE--0022 00..443344 00..889999
1313 00..000044 22..440 70 7..8800EE--0044 3.113.11 11..1100EE--0033 22..996 6 11..6677EE--0022 00..337733 00..887788
1414 00..000088 22..110 70 7..8800EE--0044 3.113.11 11..1199EE--0033 22..993 3 22..8877EE--0022 00..330044 00..885500
1515 00..0011 22..000 0 77..8800EE--0044 3.113.11 11..2222EE--0033 22..991 1 33..4477EE--0022 00..228822 00..883399
161600..0022 11..770 0 77..8800EE--0044 3.113.11 11..3355EE--0033 22..887 7 66..4477EE--0022 00..221166 00..880000
1717 00..0044 11..440 0 77..8800EE--0044 3.113.11 11..5522EE--0033 22..882 2 11..2255EE--0011 00..116622 00..775544
1818 00..0088 11..110 0 77..8800EE--0044 3.113.11 11..7722EE--0033 22..776 6 22..4455EE--0011 00..112200 00..770044
1919 00..11 11..000 0 77..8800EE--0044 3.113.11 11..8800EE--0033 22..775 5 33..0055EE--0011 00..111100 00..668877
2020 00..22 00..770 0 77..8800EE--0044 3.113.11 22..0033EE--0033 22..669 9 66..0055EE--0011 00..008833 00..663377
2121 00..44 00..440 0 77..8800EE--0044 3.113.11 22..2299EE--0033 22..664 4 11..2200EE++0000 00..006666 00..559911
2222 00..88 00..110 0 77..8800EE--0044 3.113.11 22..5544EE--0033 22..660 0 22..4400EE++0000 00..005544 00..555500
2323 11 00..000 0 77..8800EE--0044 3.113.11 22..6611EE--0033 22..558 8 33..0000EE++0000 00..005511 00..553388
2424
2525
2626
2727
2828
2929
3030
3131
3232
3333
3434
3535
3636
3737
3838
39 39 Spreadsheet Spreadsheet DocumentDocument ationation
40 40 B7=-LOG(A7) B7=-LOG(A7) F7=-LOG(F7=-LOG(E7)E7)
41 41 C7=(A$4/(AC7=(A$4/(A7)) 7)) G7=0.5((A7)*(2)*(2)+(2*AG7=0.5((A7)*(2)*(2)+(2*A7)*(-1)*(-1)+(C7)*(3)*(3)+(3*C7)*(-1)*(-1))7)*(-1)*(-1)+(C7)*(3)*(3)+(3*C7)*(-1)*(-1))
42 42 D7=-LOG(C7) D7=-LOG(C7) H7=10^H7=10^(-0.51*(3)*(3)*SQRT(G7)/(1+(3.3*0.90*SQRT(G7)))(-0.51*(3)*(3)*SQRT(G7)/(1+(3.3*0.90*SQRT(G7)))
43 43 E7=(A$4/(27*H7*I7^E7=(A$4/(27*H7*I7^ 3))^3))^ (1/4) (1/4) I7=10^I7=10^(-0.51*(-1)*(-1)*SQRT(G7)/(1+(3.3*0.40*SQRT(G7)))(-0.51*(-1)*(-1)*SQRT(G7)/(1+(3.3*0.40*SQRT(G7)))
(La(La (IO(IO33
--))
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 1010
10-1710-17
 A  A B B C C D D E E F F GG
1 1 ProbProblem lem 10-1710-17
22
33
44 xx Z Z 0.001 0.001 0.005 0.005 0.01 0.05 0.01 0.05 0.10.1
55 0.9 0.9 1 1 0.967 0.967 0.934 0.934 0.913 0.913 0.854 0.854 0.8260.826
66 0.6 0.6 1 1 0.966 0.966 0.930 0.930 0.907 0.907 0.834 0.834 0.7960.796
77 0.425 0.425 1 1 0.965 0.965 0.927 0.927 0.902 0.902 0.819 0.819 0.7730.773
88 0.35 0.35 1 1 0.965 0.965 0.926 0.926 0.900 0.900 0.812 0.812 0.7620.762
99 0.3 0.3 1 1 0.965 0.965 0.925 0.925 0.899 0.899 0.807 0.807 0.7540.754
1010 0.25 0.25 1 1 0.964 0.964 0.925 0.925 0.897 0.897 0.801 0.801 0.7450.745
1111 0.8 0.8 2 2 0.872 0.872 0.756 0.756 0.690 0.690 0.517 0.517 0.4450.445
1212 0.6 0.6 2 2 0.870 0.870 0.747 0.747 0.676 0.676 0.483 0.483 0.4010.401
1313 0.5 0.5 2 2 0.868 0.868 0.743 0.743 0.668 0.668 0.464 0.464 0.3770.377
1414 0.45 0.45 2 2 0.868 0.868 0.740 0.740 0.664 0.664 0.455 0.455 0.3640.364
1515 0.4 0.4 2 2 0.867 0.867 0.738 0.738 0.660 0.660 0.444 0.444 0.3510.351
1616 0.9 0.9 3 3 0.737 0.737 0.539 0.539 0.443 0.443 0.242 0.242 0.1780.178
1717 0.4 0.4 3 3 0.726 0.726 0.505 0.505 0.393 0.393 0.161 0.161 0.0950.095
1818 1.1 1.1 4 4 0.587 0.587 0.347 0.347 0.252 0.252 0.098 0.098 0.0630.063
1919 0.5 0.5 4 4 0.569 0.569 0.304 0.304 0.199 0.199 0.046 0.046 0.0200.020
2020
21 21 Spreadsheet Spreadsheet DocumentDocumentationation
22 C5=10^(-0.51*$B5^2*SQRT(C$4)/(1+3.3*$A5*SQRT(C$4)))22 C5=10^(-0.51*$B5^2*SQRT(C$4)/(1+3.3*$A5*SQRT(C$4)))
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 1111
Chapter 11Chapter 11
11-111-1 The overall dissociation constant for HThe overall dissociation constant for H22S isS is
2211
22
--2222
33
S]S]HH[[
]]SS[[]]OOHH[[
 K  K  K  K 

In a solution saturated with the gas, [HIn a solution saturated with the gas, [H22S] is constant, thereforeS] is constant, therefore
221122
33
22--22
]]OOHH[[
S]S]HH[[
]]SS[[ K K  K  K 

 oror
22
33
--22
]]OOHH[[
]]SS[[


K K 
11-211-2 The simplifications in equilibrium calculations involve assuming that the concentrationsThe simplifications in equilibrium calculations involve assuming that the concentrations
of one or more speciof one or more species can be approximated as 0.00 M. es can be approximated as 0.00 M. Adding or subtracting aAdding or subtracting a
concentration that can be approximated as concentration that can be approximated as 0.00 leads to a meaningful res0.00 leads to a meaningful result. ult. In contrast,In contrast,
multiplying or dividing by 0.00 in the equilibrium multiplying or dividing by 0.00 in the equilibrium constant expression causes theconstant expression causes the
constant to become equal to zero or constant to become equal to zero or infinity. infinity. Thus, the expression is meaninglThus, the expression is meaningless.ess.
11-311-3 A charge-balance equation is derived by relating the concentration of cations and anionsA charge-balance equation is derived by relating the concentration of cations and anions
such thatsuch that
no. mol/L positive charge = no. mol/L negative chargeno. mol/L positive charge = no. mol/L negative charge
For a doubly charged ion, such as BaFor a doubly charged ion, such as Ba
2+2+
, the concentration electrons for each , the concentration electrons for each mole is twicemole is twice
the molarthe molar concentrationconcentration of the Ba of the Ba2+2+. . That That is,is,
mol/L positive charge = 2[Bamol/L positive charge = 2[Ba
2+2+
]]
Thus, the molar concentration of all multiply charged species is aThus, the molar concentration of all multiply charged species is a lways multiplied by thelways multiplied by the
charge in a charge-balance equation.charge in a charge-balance equation.
11-411-4 (a)(a)  0.20   0.20 = = [H[H33AsOAsO44] + [H] + [H22AsOAsO44
--] + [HAsO] + [HAsO44
2-2-] + [AsO] + [AsO44
3-3-]]
(b)(b)  0.10   0.10 = = [H[H33AsOAsO44] + [H] + [H22AsOAsO44
--] + [HAsO] + [HAsO44
2-2-] + [AsO] + [AsO44
3-3-]]
2(0.10) 2(0.10) = = [Na[Na
++
] ] = = 0.200.20
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 1111
(c)(c) 0.0500 0.0500 + + 0.100 0.100 = = [ClO[ClO--] + [HClO]] + [HClO]
0.100 0.100 = = [Na[Na
++
]]
(d)(d) [F[F--] ] + + [HF] [HF] = = 0.25 0.25 + + 2[Ca2[Ca2+2+]]
[Na[Na
++
] ] = = 0.250.25
(e)(e) 0.100 0.100 = = [Na[Na++] ] = = [OH[OH--] + 2[Zn(OH)] + 2[Zn(OH)44
2-2-]]
(f)(f) [Ba[Ba2+2+] ] = = [C[C22OO44
2-2-] + [HC] + [HC22OO44
--] + [H] + [H22CC22OO44]]
(g)(g) [Ca[Ca
2+2+
] ] = = ½([F½([F
--
] + [HF])] + [HF])
11-511-5 (a)(a) [H[H33OO
++] ] = = [OH[OH--] + [H] + [H22AsOAsO44
--] + 2[HAsO] + 2[HAsO44
2-2-] + 3[AsO] + 3[AsO44
3-3-]]
(b)(b) [Na[Na
++
] + [H] + [H33OO
++
] ] = = [OH[OH
--
] + [H] + [H22AsOAsO44
--
] + 2[HAsO] + 2[HAsO44
2-2-
] + 3[AsO] + 3[AsO44
3-3-
]]
(c)(c) [Na[Na++] + [H] + [H33OO
++] ] = = [OH[OH--] + [ClO] + [ClO--]]
(d)(d) [Na[Na
++
] + [H] + [H33OO
++
]+ 2[Ca]+ 2[Ca
2+2+
] ] = = [F[F
--
] + [OH] + [OH
--
]]
(e)(e)  2[Zn  2[Zn2+2+] + [Na] + [Na++] + [H] + [H33OO
++] ] = = [OH[OH--] + 2[Zn(OH)] + 2[Zn(OH)44
2-2-]]
(f)(f)  2[Ba  2[Ba2+2+] + [H] + [H33OO
++] ] = = [OH[OH--] + 2[C] + 2[C22OO44
2-2-] + [HC] + [HC22OO44
--]]
(g)(g)  2[Ca  2[Ca
2+2+
] + [H] + [H33OO
++
] ] = = [OH[OH
--
] + [F] + [F
--
]]
11-611-6 Step 1 AgStep 1 Ag22CC22OO44  2Ag2Ag
++ + + CC22OO44
2-2-
HH22CC22OO44 + + HH22OO  HH33OO
++
+ + HCHC22OO44
--
HCHC22OO44
-- + + HH22OO  HH33OO
++ + + CC22OO44
2-2-
Step Step 2 S 2 S = solubility = solubility = [Ag= [Ag++]/2]/2
Step 3 [AgStep 3 [Ag++]]22[C[C22OO44
2-2-] ] ==  K  K spsp  = 3.5  = 3.51010
-11-11  (1)  (1)
  
22
11
442222
--
442233 10106060..55
OOCCHH
]]OOHCHC][][OOHH[[ 

  K  K  (2)(2)
55
22--
4422
--22
442233 10104242..55
]]OOHCHC[[
]]OOCC][][OOHH[[ 

 K  K  (3)(3)
Step 4 [AgStep 4 [Ag
++
] = 2([C] = 2([C22OO44
2-2-
] + [HC] + [HC22OO44
--
] + [H] + [H22CC22OO44]) (4)]) (4)
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth ed. ed. Chapter Chapter 1111
[H[H33OO
++] = 1.0] = 1.0  10 10-6-6 (5)(5)
Step 5 Step 5 No charge-balance equation because a buffer of unknown compositiNo charge-balanceequation because a buffer of unknown composition is present.on is present.
Step 6 Unknowns: [AgStep 6 Unknowns: [Ag++], [C], [C22OO44
2-2-], [HC], [HC22OO44
--], [H], [H22CC22OO44]]
Equations: Equations: (1), ((1), (2), (3) 2), (3) and (4)and (4)
Step 7 Step 7 No approximations No approximations needed.needed.
Step 8 Step 8 Substituting (5) Substituting (5) into (3) and reinto (3) and rearranging givesarranging gives
[HC[HC22OO44
--
] ] ==
  
55
--22
4422
66
10104242..55
OOCC101000..11




  = 0.01845[C  = 0.01845[C22OO44
2-2-
]]
Substituting this relationship and (5) into (2) and rearranging givesSubstituting this relationship and (5) into (2) and rearranging gives
[H[H22CC22OO44] ] == 22
--22
4422
66
10106060..55
OOCC0184501845..00101000..11




  = 3.29  = 3.291010
-7-7
[C[C22OO44
2-2-
]]
Substituting these two relationships into (4) givesSubstituting these two relationships into (4) gives
[Ag[Ag++] = 2[C] = 2[C22OO44
2-2-] ] + + 22(0.01845[C(0.01845[C22OO44
2-2-]) + 2]) + 2(3.29(3.291010-7-7[C[C22OO44
2-2-])])
[Ag[Ag
++
] = 2.037[C] = 2.037[C22OO44
2-2-
] or] or
[C[C22OO44
2-2-] = 0.4909[Ag] = 0.4909[Ag++]]
Substituting this relationship into (1) and rearranging givesSubstituting this relationship into (1) and rearranging gives
[Ag[Ag++] = (3.5] = (3.51010-11-11/0.4909)/0.4909)1/31/3  = 4.15  = 4.151010-4-4 M M
FinallyFinally
S = 4.15S = 4.151010-4-4/2 = 2.1/2 = 2.11010-4-4
Substituting other values for [HSubstituting other values for [H33OO
++
] yields the following solubility data:] yields the following solubility data:
Fundamentals of Analytical Chemistry: 8th ed. Chapter 11
[H3O
+], M Solubility, M
(a) 1.010
-6
2.110
-4
(b) 1.010
-7
2.110
-4
(c) 1.010-9 2.110-4
(d) 1.010
-11
2.110
-4
11-7 Proceeding as in Problem 11-6, we write
BaSO4  Ba
2+
+ SO4
2-
 K sp  = 1.110
-10
HSO4
- + H2O  H3O
+ + SO4
2-  K 2  = 1.0210
-2
S = [Ba
2+
]
[Ba2+][SO4
2-] = 1.110-10  (1)



]HSO[
]SO][OH[
4
2
43   1.0210-2  (2)
Mass balance requires that
[Ba2+] = [SO4
2-] + [HSO4
-] (3)
[H3O
+
] = 2.5 (4)
Substituting (4) into (2) gives upon rearranging
[HSO4
-
] = (2.5)[SO4
2-
]/(1.0210
-2
) = 245.1[SO4
2-
] (5)
Substituting (5) into (3) gives
[Ba
2+
] = [SO4
2-
] + 245.1[SO4
2-
] = 246.1[SO4
2-
] (6)
Substituting (6) into (1) gives
246.1[SO4
2-
]
2
  = 1.110
-10
[SO4
2-
] = 6.710
-7
 M
From (6) [Ba2+] = 1.610-4  M = S
Fundamentals of Analytical Chemistry: 8th ed. Chapter 11
Proceeding in a similar manner for the other [H3O
+] gives
[H3O
+
] S, mol/L
(a) 2.5 1.6510-4
(b) 1.5 1.2810-4
(c) 0.060 2.7510
-5
(d) 0.200 4.7610-5
11-8 The following derivation applies to this and the following two problems.
MS( s)  M
2+
+ S
2-
H2S + H2O  H3O
+ + HS-  K 1  = 9.610
-8
HS- + H2O
-
 H3O
+ + S2-  K 2  = 1.310
-14
H2S + 2H2O  2H3O
+
+ S
2-
 K 1 K 2  = 9.610
-8
 1.310
-14
  = 1.2510
-21
S = solubility = [M2+] = [S2-] + [HS-] + [H2S]
[M
2+
][S
2-
] =  K sp  (1)
 


SH
]HS][OH[
2
3  K 2  = 1.310
-14  (2)
 


SH
]S[]OH[
2
22
3  K 1 K 2  = 1.2510
-21
  (3)
From mass-balance consideration
[M2+] = [S2-] + [HS-] + [H2S] (4)
Substituting (2) and (3) into (4) gives
[M2+] = [S2-] + 
 
 

 
 




21
2
3
2
32
21
22
3
2
2
3 ]OH[]OH[1]S[
]S[]OH[]S][OH[
 K  K  K  K  K  K 
  (5)
Substituting (1) into (5)
Fundamentals of Analytical Chemistry: 8th ed. Chapter 11
[M
2+
] = 
 
 

 
 



21
2
3
2
3
2
sp ]OH[]OH[
1
]M[ K  K  K 
 K 
[M
2+
] = 
 
 

 
 


21
2
3
2
3
sp
]OH[]OH[
1
 K  K  K 
 K 
[M2+] = 
 
 

 
 








21
2
3
14
3
sp
1025.1
]OH[
103.1
]OH[
1 K    (6)
(a)  Substituting  K sp = 810
-37
 and [H3O
+
] = 2.010
-1
 into (6) gives
[M2+] = solubility =
 


 
 


 
 






21
2
14
37
1025.1
20.0
103.1
20.0
1108
= M101.5 9
(b)  Substituting  K sp = 810
-37 and [H3O
+] = 2.010-4 into (6) gives
solubility = M101.5 12
11-9 (a)  Substituting  K sp = 110
-27 and [H3O
+] = 2.010-1 into (6) gives
solubility = M102 4
(b)  Substituting  K sp = 110
-27 and [H3O
+] = 2.010-4 into (6) gives
solubility = 210
-7
 M
11-10 (a)  Substituting  K sp = 310
-14 and [H3O
+] = 2.010-5 into (6) gives
solubility = 0.1 M
(b)  Substituting  K sp = 310
-14 and [H3O
+] = 2.010-7 into (6) gives
solubility = 110
-3
 M
Fundamentals of Analytical Chemistry: 8th ed. Chapter 11
11-11 PbCO3  Pb
2+ + CO3
2-  K sp  = [Pb
2+][CO3
2-] = 7.410-14
H2CO3 + H2O  H3O
+  + HCO3
-  K 1 = 

]COH[
]HCO][OH[
32
-
33   4.4510-7
HCO3
- + H2O  H3O
+ + CO3
2-  K 2 = 

]HCO[
]CO][OH[
3
-2
33   4.6910-11
[Pb2+] = [CO3
2-] + [HCO3
-] + [H2CO3]
[H3O
+
] = 110
-7
Proceeding as in problem 11-8
[M2+] = 
 
 

 
 


21
2
3
2
3
sp
]OH[]OH[
1
 K  K  K 
 K 
[Pb
2+
] =
 


 
 


 
 











117
27
11
7
14
1069.41045.4
100.1
1069.4
100.1
1104.7
solubility = [Pb
2+
] = 1.410
-5
 M
11-12 Ag2SO3  2Ag
+
+ SO3
2-
 K sp  = [Ag
+
]
2
[SO3
2-
] = 1.510
-14
  (1)
H2SO3 + H2O  H3O
+
  + HSO3
-
 K 1 = 

]SOH[
]HSO][OH[
32
-
33   1.2310
-2
  (2)
HSO3
-
+ H2O  H3O
+
+ SO3
2-
 K 2 = 

]HSO[
]SO][OH[
3
-2
33 6.610
-8
  (3)
H2SO3 + 2H2O  2H3O
+ + SO3
2-  K 1 K 2 = 

]SOH[
]SO[]OH[
32
-2
3
2
3 8.110-10  (4)
The mass balances are
½ [Ag+] = [SO3
2-] + [HSO3
-] + [H2SO3] (5)
[H3O
+
] = 110
-8
  (6)
Fundamentals of Analytical Chemistry: 8th ed. Chapter 11
Substituting (2), (4) and (6) into (5) gives
½ [Ag+] = [SO3
2-] +
 
8-
-2
3
8
106.6
]SO[101

 
+
 
10-
-2
3
28
108.1
]SO[101

 
½ [Ag+] = [SO3
2-]
 


 
 


 
 








10
28
8-
8
101.8
101
106.6
101
1
Substituting (1) into this expression gives
½ [Ag+] =
 
 


 
 


 
 








 10
28
8-
8
2
sp
101.8
101
106.6
101
1
Ag
 K 
[Ag+]3 = 21.510-14
 


 
 


 
 








10
28
8-
8
101.8
101
106.6
101
1
[Ag+] = 4.210-5
Solubility = 2.110
-5
 M
11-13 [Cu2+][OH-]2  = 4.810-20 [Mn2+][OH-]2 = 210-13
(a)  Cu(OH)2 precipitates first
(b) Cu2+ begins to precipitate when
[OH
-
] = 050.0/108.4 20   = 9.810-10 M
(c) Mn2+ begins to precipitate when
[OH-] = 040.0/102 13   = 2.210-6
[Cu2+] = 4.810-20/(2.210-6)2  = 9.610-9 M
11-14 Ba(IO3)2  Ba
2+  + 2IO3
-  K sp  = 1.5710
-9
BaSO4  Ba
2+ + SO4
2-  K sp  = 1.310
-10
To initiate precipitation of Ba(IO3)2
Fundamentals of Analytical Chemistry: 8th ed. Chapter 11
[Ba2+] = 1.5710-9/(0.050)2  = 6.310-7
To initiate precipitation of BaSO4
[Ba
2+
] = 1.310
-10
/(0.040) = 3.210
-9
 M
(a)  BaSO4 precipitates first
(b) [Ba
2+
] = 3.210
-9
 M
(c) When [Ba2+] = 6.310-7 M
[SO4
2-] = 1.110-10/(6.310-7) = 1.710-4 M
11-15 (a) [Ag+] =  K sp/[I
-] = 8.310-17/(1.010-6) = 8.310-11 M
(b) [Ag
+
] =  K sp/[SCN
-
] = 1.110
-12
/(0.070) = 1.610
-11
 M
(c) [I
-
] when [Ag
+
] = 1.610
-11
 M
[I-] = 8.310-17/(1.610-11) = 5.210-6 M
[SCN
-
]/[I
-
] = 0.070/(5.210
-6
) = 1.310
4
(d) [I
-
] = 8.310
-17
/(1.010
-3
) = 8.310
-14
 M
[SCN-] = 1.110-12/(1.010-3) = 1.110-9 M
[SCN-]/[I-] = 1.110-9/(8.310-14) = 1.3104
 Note that this ratio is independent of [Ag
+
] as long as some AgSCN( s) is present.
11-16 (a) [Ba2+][SO4
2-] = 1.110-10 [Sr 2+][SO4
2-] = 3.210-7
BaSO4 precipitation is complete when
[SO4
2-
] = 1.110
-10
/(1.010
-6
) = 1.110
-4
 M
SrSO4 begins to precipitate when
[SO4
2-
] = 3.210
-7
/(0.050) = 6.410
-6
 M
SrSO4 begins to precipitate before the Ba
2+ concentration is reduced to 1.010-6 M.
Fundamentals of Analytical Chemistry: 8th ed. Chapter 11
(b) [Ba2+][SO4
2-] = 1.110-10 [Ag+]2[SO4
2-] = 1.610-5
BaSO4 precipitation is completewhen [SO4
2-] = 1.110-4 M
Ag2SO4 begins to precipitate when
[SO4
2-] = 1.610-5/(0.020)2  = 0.040 M
Ag2SO4 does not precipitate before the Ba
2+
 concentration is reduced to 1.010
-6
 M.
(c) Be2+ precipitates when [OH-] = (7.010-22/(0.020))1/2  = 1.910-10 M
Hf 4+ precipitates when [OH-] = (4.010-26/(0.010))1/4  = 1.410-6 M
Be precipitation complete when [OH
-
] = (7.010
-22
/1.010
-6
)
1/2
  = 2.610
-8
 M
Hf(OH)4 does not precipitate before the Be
2+ concentration is reduced to 1.010-6 M.
(d) In3+ precipitates when [IO3
-] = (3.310-11/(0.20))1/3  = 5.510-4 M
Tl
+
 precipitates when [IO3
-
] = 3.110
-6
/(0.090) = 3.410
-5
 M
Tl+ precipitation complete when [IO3
-] = 3.110-6/1.010-6  = 3.1 M
In(IO3)3 begins to precipitate before the Tl
+ concentration is reduced to 1.010-6 M.
11-17 AgBr  Ag+ + Br - 5.010-13  = [Ag+][Br -] (1)
Ag+  + 2CN-   Ag(CN)2
- 1.31021 =
2
2
]CN][Ag[
])CN(Ag[


  (2)
It is readily shown that the reaction
CN
-
+ H2O   HCN + OH
-
 proceeds to such a small extent that it can be neglected in formulating a solution to this
 problem. That is, [HCN] << [CN-], and only the equilibria shown need to be taken into
account.
Solubility = [Br -]
Fundamentals of Analytical Chemistry: 8th ed. Chapter 11
Mass balance requires that
[Br 
-
] = [Ag
+
] + [Ag(CN)2
-
] (3)
0.100 = [CN-] + 2[Ag(CN)2
-] (4)
We now have 4 equations and 4 unknowns.
Based upon the large size of 2 let us assume that
[CN
-
] << 2[Ag(CN)2
-
] and [Ag
+
] << [Ag(CN)2
-
]
(4) becomes [Ag(CN)2
-] = 0.100/2 = 0.0500
and (3) becomes [Br 
-
] = [Ag(CN)2
-
] = 0.0500
To check the assumptions, we calculate [Ag+] by substituting into (1)
[Ag
+
] = 5.010
-13
/0.0500  110
-11
( 110
-11
 << 0.0500)
To obtain [CN-] we substitute into (2) and rearrange
[CN
-
] =
  2111 103.1101
0500.0
 
  = 2.010
-6
( 210
-6
 << 0.100)
Thus, the two assumptions are valid and
Solubility = [Br -] = 0.0500 M
mass AgBr/200 mL = 0.0500
AgBr mmol
AgBr g0.1877
mL200
mL
mmol

= 1.877 g
11-18 CuCl( s)  Cu
+
+ Cl
-
 K sp  = [Cu
+
][Cl
-
] = 1.910
-7
  (1)
Cu+  + 2Cl-   CuCl2
- 2 = 2-
-
2
]Cl][Cu[
]CuCl[

  = 7.9104  (2)
It is convenient to multiply (1) by (2) to give
]Cl[
]CuCl[
-
-
2   = 1.910-77.9104  = 1.510-2  (3)
Fundamentals of Analytical Chemistry: 8th ed. Chapter 11
From a charge balance consideration, we can write (if we assume [H3O
+] = [OH-])
[Cu
+
] + [Na
+
] = [Cl
-
] + [CuCl2
-
] (4)
By rearranging (1) and (3) and substituting into (4) we obtain
[Cl
-
] = [Na
+
] +
][Cl
109.1
-
7
  - 1.510
-2
[Cl
-
]
which rearranges to the quadratic
0 = 1.015[Cl-]2  - [Na+][Cl-] - 1.910-7  (5)
By using [Na+] = the NaCl analytical concentration, (5) can be solved to give the
following [Cl-]
(a) 2.0 M (c) 0.020 M (e) 5.410
-4
 M
(b) 0.20 M (d) 0.0021 M
 Note that the equilibrium [Cl
-
] concentration is larger  than the NaCl analytical
concentration for parts (d) and (e). The reason for this apparent anomaly is that the
dissolution of CuCl to give Cu
+
 and Cl
-
 contributes significantly to the equilibrium [Cl
-
]
at the lower NaCl analytical concentrations.
The solubility of CuCl can be obtained from the calculated [Cl
-
] and the expression
S = [Cu+] + [CuCl2
-] =
]Cl[
109.1 7


  + 1.510-2[Cl-]
Solution of this equation for each of the [Cl-] gives
(a) 0.030 M (c) 3.110
-4
 M (e) 3.610
-4
 M
(b) 3.010-3 M (d) 1.210-4 M
Fundamentals of Analytical Chemistry: 8th ed. Chapter 11
11-19 (a) CaSO4( s)  Ca
2+ + SO4
2-  K sp = [Ca
2+][SO4
2-] = 2.610-5  (1)
CaSO4(aq)  Ca
2+ + SO4
2-  K d =
)](CaSO[
]SO][Ca[
4
-2
4
2
aq

  = 5.210-3  (2)
CaSO4( s)   CaSO4(aq) (3)
The mass balance gives
[Ca2+] = [SO4
2-] (4)
We have 3 equations and 3 unknowns ([Ca2+], [SO4
2-], and [CaSO4(aq)]
To solve we divide (1) by (2) to give
[CaSO4(aq)] =  K sp/ K d  = (2.610
-5
)/(5.210
-3
) = 5.010
-3
 Note that this is the equilibrium constant expression for (3) and indicates that the
concentration of un-ionized CaSO4 is always the same in a saturated solution of CaSO4.
Substituting (4) into (1) gives
[Ca2+] = (2.610-5)1/2  = 5.110-3
and since S = [CaSO4(aq)] + [Ca
2+] we obtain
S = 5.010
-3
  + 5.110
-3
  = 0.0101 M
%CaSO4(aq) = (5.010
-3/1.0110-2)100% = 49%
(b) (Note: In the first printing of the text, the answer in the back of the book was in error.)
Here [CaSO4(aq)] is again equal to 5.010
-3 and the mass balance gives
[SO4
2-
] = 0.0100 + [Ca
2+
] (5)
Substituting (1) into (5) and rearranging gives
0 = [SO4
2-
]
2
  - 0.0100[SO4
2-
] -  K sp
which may be solved using the quadratic equation to give
[SO4
2-
] = 0.0121 and [Ca
2+
] = 2.1410
-3
Fundamentals of Analytical Chemistry: 8th ed. Chapter 11
S = 5.010-3  + 2.1410-3  = 7.1410-3
%CaSO4(aq) = (5.010
-3/7.1410-3)100% = 70%
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
14-33 In each of the parts of this problem, we are dealing with a weak base B and its conjugate
acid BHCl or (BH)2SO4. The pH determining equilibrium can then be written as
BH+ + H2O  H3O
+ + B
The equilibrium concentration of BH
+
 and B are given by
[BH+] = cBHCl + [OH
-] –  [H3O
+] (1)
[B] = cB - [OH
-] + [H3O
+] (2)
In many cases [OH
-
] and [H3O
+
] will be much smaller than cB and cBHCl and [BH
+
] ≈
cBHCl and [B] ≈ cB so that
[H3O
+] =
B
BHCl
a
c
c
 K     (3)
(a) Amount NH4
+ = 3.30 g (NH4)2SO4 
424
4
424
424
SO)(NHmmol
 NHmmol2
SO)(NHg13214.0
SO)(NHmmol1 
  =
49.95 mmol
Amount NaOH = 125.0 mL0.1011 mmol/mL = 12.64 mmol
mL0.500
1
 NaOHmmol
 NHmmol1
 NaOHmmol64.12 3 NH3 c  = 2.52810
-2 M
mL0.500
1
 NHmmol)64.1295.49( 4 NH4
 c  = 7.46210
-2 M
Substituting these relationships in equation (3) gives
[H3O
+
] =
B
BHCl
a
c
c
 K    = 5.7010-10  7.46210-2 / (2.52810-2) = 1.68210-9 M
 pH = -log 1.68210
-9
  = 8.77
(b) Substituting into equation (3) gives
[H3O
+
] = 7.510
-12
 0.080 / 0.120 = 5.0010
-12
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
 pH = -log 5.0010-12  = 11.30
(c) cB = 0.050 and cBHCl  = 0.167
[H3O
+
] = 2.3110
-11
 0.167 / 0.050 = 7.71510
-11
 pH = -log 7.71510-11  = 10.11
(d) Original amount B = 2.32 g B
Bg0.09313
Bmmol1
  = 24.91 mmol
Amoung HCl = 100 mL  0.0200 mmol/mL = 2.00 mmol
cB  = (24.91 –   2.00)/250.0 = 9.16410
-2
 M
cBH+  = 2.00/250.0 = 8.0010
-3 M
[H3O
+] = 2.5110-5  8.0010-3 / 9.16410-2  = 2.19110-6 M
 pH = -log 2.19110
-6
  = 5.66
14-34 (a)  pH = 0.00
(b) [H3O
+] changes to 0.00500 M from 0.0500 M
 pH = -log 0.00500 –   (-log0.0500) = 2.301 –   1.301 = 1.000
(c) pH diluted solution = 14.000 – (-log 0.00500) = 11.699
 pH undiluted solution = 14.000 – (-log 0.0500) = 12.699
 pH = -1.000
(d) In order to get a better picture of the pH change with dilution, we will dispense with
the usual approximations and write
5
-
3
a 1075.1
HOAc][
]][OAcOH[ 

 K 
[H3O
+]2 + 1.7510-5[H3O
+] –  0.0500  1.7510-5 = 0
Solving by the quadratic formula or by successive approximations gives
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
[H3O
+] = 9.26710-4  and pH = -log 9.26710-4  = 3.033
For diluted solution, the quadratic becomes
[H3O
+
]
2
 + 1.7510
-5
 –  0.005001.7510
-5
[H3O
+] = 2.87210-4  and pH = 3.542
 pH = 3.033 –   3.542 = -0.509
(e) OAc
- + H2O  HOAc + OH
-
5
14
-
-
1075.1
1000.1
]OAc[
]HOAc][OH[




   = 5.7110-10 =  K  b
Here we can use an approximation solution because K  b is so very small. For the
undiluted sample
0500.0
]OH[ 2-
  = 5.7110-10
[OH-] = (5.7110-10  0.0500)1/2  = 5.34310-6 M
 pH = 14.00 –  (-log 5.34310
-6
) = 8.728
For the diluted sample
[OH
-
] = (5.7110
-10
 0.00500)
1/2
  = 1.69010
-6
 M
 pH = 14.00 –  (-log 1.69010
-6
) = 8.228
 pH = 8.228 –   8.728 = -0.500
(f) Here we must avoid the approximate solution because it will not reveal the small pHchange resulting from dilution. Thus, we write
[HOAc] = cHOAc + [OH
-
] –  [H3O
+
] ≈ cHOAc –  [H3O
+
]
[OAc-] = c NaOAc –  [OH
-] + [H3O
+
] ≈ c NaOAc + [H3O
+]
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
 K a  = 1.7510
-5
=
 
]OH[0500.0
]OH[0500.0]OH[
3
33




Rearranging gives
[H3O
+]2 + 5.001810-2[H3O
+] –  8.7510-7 = 0
[H3O
+
] = 1.74910
-5
  and pH = 4.757
Proceeding in the same way we obtain for the diluted sample
1.7510-5 =
 
]OH[00500.0
]OH[00500.0]OH[
3
33




[H3O
+]2 + 5.017510-3[H3O
+] –  8.7510-8 = 0
[H3O
+
] = 1.73810
-5
  and pH = 4.760
 pH = 4.760 –   4.757 = 0.003
(g) Proceeding as in part (f) a 10-fold dilution of this solution results in a pH change that
is less than 1 in the third decimal place. Thus for all practical purposes,
 pH = 0.000
14-35 (a) After addition of acid, [H3O
+] = 1 mmol/100 mL = 0.0100 M and pH = 2.00
Since original pH = 7.00
 pH = 2.00 –   7.00 = -5.00
(b) After addition of acid
cHCl  = (1000.0500 + 1.00)/100 = 0.0600 M
 pH = -log 0.0600 – (-log 0.0500) = 1.222 –   1.301 = -0.079
(c) After addition of acid,
c NaOH  = (1000.0500 – 1.00)/100 = 0.0400 M
[OH-] = 0.0400 M and pH = 14.00 – (-log 0.0400) = 12.602
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
From Problem 14-34 (c), original pH = 12.699
 pH = -0.097
(d) From Solution 14-34 (d), original pH = 3.033
Upon adding 1 mmol HCl to the 0.0500 M HOAc, we produce a mixture that is
0.0500 M in HOAc and 1.00/100 = 0.0100 M in HCl. The pH of this solution is
approximately that of a 0.0100 M HCl solution, or 2.00. Thus,
 pH = 2.000 –   3.033 = -1.033
(If the contribution of dissociation of HOAc to the pH is taken into account, a pH
of 1.996 is obtained and  pH = -1.037 is obtained.)
(e) From Solution 14-34 (e), original pH = 8.728
Upon adding 1.00 mmol HCl we form a buffer having the composition
cHOAc  = 1.00/100 = 0.0100
c NaOAc  = (0.0500  100 –   1.00)/100 = 0.0400
Applying Equation 14-xx gives
[H3O
+] = 1.7510-5   0.0100/0.0400 = 4.57510-6 M
 pH = -log 4.57510-6  = 5.359
 pH = 5.359 –   8.728 = -3.369
(f) From Solution 14-34 (f), original pH = 4.757
With the addition of 1.00 mmol of HCl we have a buffer whose concentrations are
cHOAc  = 0.0500 + 1.00/100 = 0.0600 M
c NaOAc  = 0.0500 – 1.00/100 = 0.0400 M
Proceeding as in part (e), we obtain
[H3O
+] = 2.62510-5 M and pH = 4.581
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
 pH = 4.581 –   4.757 = -0.176
(g) For the original solution
[H3O
+
] = 1.7510
-5
  0.500/0.500 = 1.7510
-5
 M
 pH = -log 1.7510-5  = 4.757
After addition of 1.00 mmol HCl
cHOAc  = 0.500 + 1.00/100 = 0.510 M
c NaOAc  = 0.500 – 1.00/100 = 0.490 M
Proceeding as in part (e), we obtain
[H3O
+
] = 1.7510
-5
  0.510/0.490 = 1.82110
-5
 M
 pH = -log 1.82110-5  = 4.740
 pH = 4.740 –   4.757 = -0.017
14-36 (a) c NaOH  = 1.00/100 = 0.0100 = [OH
-]
 pH = 14.00 – (-log 0.0100) = 12.00
Original pH = 7.00 and  pH = 12.00 –   7.00 = 5.00
(b) Original pH = 1.301 [see Problem 14-34 (b)]
After addition of base, cHCl  = (100  0.0500 – 1.00)/100 = 0.0400 M
 pH = -log 0.0400 –   1.301 = 1.398 – 1.301 = 0.097
(c) Original pH = 12.699 [see Problem 14.34 (c)]
After addition of base, c NaOH  = (100  0.0500 + 1.00)/100 = 0.0600 M
 pH = 14.00 – (-log 0.0600) = 12.778
 pH = 12.778 – 12.699 = 0.079
(d) Original pH = 3.033 [see Problem 14-34 (d)]
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
Addition of strong base gives a buffer of HOAc and NaOAc.
c NaOAc  = 1.00 mmol/100 = 0.0100 M
cHOAc  = 0.0500 – 1.00/100 = 0.0400 M
Proceeding as in Solution 14-35 (e) we obtain
[H3O
+] = 1.7510-5   0.0400/0.0100 = 7.0010-5 M
 pH = -log 7.0010
-5
  = 4.155
 pH = 4.155 –   3.033 = 1.122
(e) Original pH = 8.728 [see Problem 14.34 (e)]
Here, we have a mixture of NaOAc and NaOH and the pH is determined by the
excess NaOH.
c NaOH  = 1.00 mmol/100 = 0.0100 M
 pH = 14.00 – (-log 0.0100) = 12.00
 pH = 12.00 –   8.728 = 3.272
(f) Original pH = 4.757 [see Problem 14-34 (f)]
c NaOAc  = 0.0500 + 1.00/100 = 0.0600 M
cHOAc  = 0.0500 – 1.00/100 = 0.0400 M
Proceeding as in Solution 14.35 (e) we obtain
[H3O
+
] = 1.16710
-5
M and pH = 4.933
 pH = 4.933 –   4.757 = 0.176
(g) Original pH = 4.757 [see Problem 14-34 (f)]
cHOAc  = 0.500 – 1.00/100 = 0.490 M
c NaOAc  = 0.500 + 1.00/100 = 0.510 M
Substituting into Equation 9-29 gives
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
[H3O
+] = 1.7510-5   0.400/0.510 = 1.68110-5 M
 pH = -log 1.68110-5  = 4.774
 pH = 4.774 –   4.757 = 0.017
14-37 For lactic acid,  K a  = 1.3810
-4
= [H3O
+
][A
-
]/[HA]
Throughout this problem we will base calculations on Equations 9-25 and 9-26.
[A
-
] = c NaA + [H3O
+
] –  [OH
-
]
[HA] = cHA –  [H3O
+] –  [OH-]
 
]OH[
]OH[]OH[
3HA
3 NaA3




c
c
  = 1.3810-4
This equation rearranges to
[H3O
+]2 + (1.3810-4 + 0.0800)[H3O
+] –  1.3810-4 cHA = 0
(a) Before addition of acid
[H3O
+
]
2
 + (1.3810
-4
 + 0.0800)[H3O
+
] –  1.3810
-4
  0.0200 = 0
[H3O
+] = 3.44310-5  and pH = 4.463
Upon adding 0.500 mmol of strong acid
cHA  = (100  0.0200 + 0.500)/100 = 0.0250 M
c NaA  = (100  0.0800 – 0.500)/100 = 0.0750 M
[H3O
+
]
2
 + (1.3810
-4
 + 0.0750)[H3O
+
] –  1.3810
-4
  0.0250 = 0
[H3O
+] = 4.58910-5  and pH = 4.338
 pH = 4.338 –   4.463 = -0.125
(b) Before addition of acid
[H3O
+]2 + (1.3810-4 + 0.0200)[H3O
+] –  1.3810-4   0.0800 = 0
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
[H3O
+] = 5.34110-5  and pH = 3.272
After adding acid
cHA  = (100  0.0800 + 0.500)/100 = 0.0850 M
c NaA  = (100  0.0200 – 0.500)/100 = 0.0150 M
[H3O
+
]
2
 + (1.3810
-4
 + 0.0150)[H3O
+
] –  1.3810
-4
  0.0850 = 0
[H3O
+] = 7.38810-4  and pH = 3.131
 pH = 3.131 –   3.272 = -0.141
(c) Before addition of acid
[H3O
+]2 + (1.3810-4 + 0.0500)[H3O
+] –  1.3810-4   0.0500 = 0
[H3O
+] = 1.37210-4  and pH = 3.863
After adding acid
cHA  = (100  0.0500 + 0.500)/100 = 0.0550 M
c NaA  = (100  0.0500 – 0.500)/100 = 0.0450 M
[H3O
+
]
2
 + (1.3810
-4
 + 0.0450)[H3O
+
] –  1.3810
-4
  0.0550 = 0
[H3O
+] = 1.67510-4  and pH = 3.776
 pH = 3.776 –   3.863 = -0.087
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
14-38
A B C D E F G H I
1 V i , NaOH 50.00
2 c i , NaOH 0.1000 M
3 c , HCl 0.1000 M
4 V eq. pt. 50.00
5 K w 1.00E-14
6
7 Vol. HCl, mL [H3O
+
] pH
8 0.00 1.00E-13 13.000
9 10.00 1.50E-13 12.824
10 25.00 3.00E-13 12.523
11 40.00 9.00E-13 12.046
12 45.00 1.90E-12 11.721
13 49.00 9.90E-12 11.004
14 50.00 1.00E-07 7.000
15 51.00 9.90E-04 3.004
16 55.00 4.76E-03 2.322
17 60.00 9.09E-03 2.041
18
19 Spreadsheet Documentation
20 B4 = B2*B1/B3
21 B8 = $B$5/(($B$2*$B$1-A8*$B$3)/($B$1+A8))
22 B14 = SQRT(B5)
23 B15 = (A15*$B$3-$B$1*$B$2)/(A15+$B$1)
24 C8 = -LOG(B8)
14-39 Let us calculate pH when 24.95 and 25.05 mL of reagent have been added.
24.95 mL reagent
cA- 
95.74
495.2
solnmL95.74
KOHmmol1000.095.24
solnvolumetotal
addedKOHamount


   = 0.03329 M
cHA  [HA] =
solnvolumetotal
addedKOHamount-HAamountoriginal
=
solnmL74.95
HAmmol0.1000)24.95-0.0500(50.00 
=
95.74
005.0
95.74
495.2500.2


  = 6.6710
-5
 M
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
Substituting into Equation 9-29
[H3O
+
] =  K a cHA / cA-  = 1.80  10
-4
 6.6710
-5
/ 0.03329 =3.60710
-7
 M
 pH = -log 3.60710
-7
  = 6.44
25.05 mL KOH
cKOH =
solnvolumetotal
HAamountinitial-addedKOHamount
=
solnmL75.05
0.0500050.00-0.100025.05 
  = 6.6610
-5
  = [OH
-
]
 pH = 14.00 –  (-log 6.6610
-5
) = 9.82
Thus, the indicator should change color in the range of pH 6.5 to 9.8. Cresol purple
(range 7.6 to 9.2, Table 14-1) would be quite suitable.
14-40 (See Solution 14-39) Let us calculate the pH when 49.95 and 50.05 mL of HClO4 have
 been added.
49.95 mL HClO4
B = C2H5 NH2 BH
+ = C2H5 NH3
+
95.99
995.495.99
10000.095.49
solnvolumetotal
HClOmmolno. 4
BH


c = 0.04998 M ≈ [BH
+
]
cB =
 
95.99
00500.0
95.99
1000.095.491000.000.50


 = 5.0010
-5
M ≈  [B]
[H3O
+
] = 2.31  10
-11
 0.04998 / 5.0010
-5
  =2.30910
-8
 M
 pH = -log 2.30910-8  = 7.64
50.05 mL HClO4
 
05.100
1000.000.501000.005.50
4HClO

c   = 4.99810-5 = [H3O+]
 pH = -log 4.99810-5  = 4.30
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
Indicator should change color in the pH range of 7.64 to 4.30. Bromocresol purple would
 be suitable.
For Problems 14-41 through 14-43 we set up spreadsheets that will solve a quadratic equation
to determine [H3O
+] or [OH-], as needed. While approximate solutions are appropriate for many
of the calculations, the approach taken represents a more general solution and is somewhat easier
to incorporate in a spreadsheet. As an example consider the titration of a weak acid with a strong
 base.
Before the equivalence point: [HA] =
 
  NaOHHAi
 NaOH NaOHiHAiHAi
V V 
V cV c


 - [H3O
+]
and [A
-
] =
 
  NaOHHAi
 NaOH NaOHi
V V 
V c

 + [H3O
+
]
Substituting these expressions into the equilibrium expression for HA and rearranging gives
0 = [H3O
+
]
2
+
 
  

 
 

 
 

 a NaOHHAi
 NaOH NaOHi
 K 
V V 
V c
[H3O
+
] -
 
  NaOHHAi
 NaOH NaOHiHAiHAia
V V 
V cV c K 


From which [H3O
+] is directly determined.
At and after the equivalence point : [A
-
] =
 
  NaOHHAi
HAHAi
V V 
V c

 - [HA]
and [OH-] =
 
  NaOHHAi
HAiHAi NaOH NaOHi
V V 
V cV c


 + [HA]
Substituting these expressions into the equilibrium expression for A
-
 and rearranging gives
0 = [HA]2 +
 
  

 
 

 
 



a
w
 NaOHHAi
HAiHAi NaOH NaOHi
 K 
 K 
V V 
V cV c
[HA] -
 
  NaOHHAia
HAHAiw
V V  K 
V c K 

From which [HA] can be determined and [OH-] and [H3O
+] subsequently calculated. A similar
approach is taken for the titration of a weak base with a strong acid.
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
14-41
A B C D E F
1 Part (a)
2 V i , HNO2 50.00
3 c i , HNO2 0.1000
4 K a, HNO2 7.10E-04
5 K w, H2O 1.00E-14
6
7 c , NaOH 0.1000
8 V eq. pt. 50.00
9
10 Vol. NaOH, mL b c [OH-] [H3O
+
] pH
11 0.00 7.1000E-04 -7.1000E-05 8.0786E-03 2.0927
12 5.00 9.8009E-03 -5.8091E-05 4.1607E-03 2.3808
13 15.00 2.3787E-02 -3.8231E-05 1.5112E-03 2.8207
14 25.00 3.4043E-02 -2.3667E-05 6.8155E-04 3.1665
15 40.00 4.5154E-02 -7.8889E-06 1.7404E-04 3.7594
16 45.00 4.8078E-02 -3.7368E-06 7.7599E-05 4.1101
17 49.00 5.0205E-02 -7.1717E-07 1.4281E-05 4.8452
18 50.00 1.4085E-11 -7.0423E-13 8.3917E-07 1.1916E-08 7.9239
19 51.00 9.9010E-04 -6.9725E-13 9.9010E-04 1.0100E-11 10.9957
20 55.00 4.7619E-03 -6.7069E-13 4.7619E-03 2.1000E-12 11.6778
21 60.00 9.0909E-03 -6.4020E-13 9.0909E-03 1.1000E-12 11.9586
22
23 Spreadsheet Documentation
24 C8 = C2*C3/C7
25 B11 = $C$7*A11/($C$2+A11)+$C$4
26 C11 = -$C$4*($C$3*$C$2-$C$7*A11)/($C$2+A11)
27 E11 = (-B11+SQRT(B11^2-4*C11))/2
28 F11 = -LOG(E11)
29 B18 = ($C$7*A18-$C$3*$C$2)/($C$2+A18)+$C$5/$C$4
30 C18 = -($C$5/$C$4)*($C$2*$C$3/($C$2+A18))
31 D18 = (-B18+SQRT(B18^2-4*C18))/2+($C$7*A18-$C$2*$C$3)/($C$2+A18)
32 E18 = $C$5/D18
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
A B C D E F
1 Part (b)
2 V i , Lactic Acid 50.00
3 c i , Lactic Acid 0.1000
4 K a, Lactic Acid 1.38E-04
5 K w, H2O 1.00E-14
6
7 c , NaOH 0.1000
8 V eq. pt. 50.00
9
10 Vol. NaOH, mL b c [OH-] [H3O
+
] pH
11 0.00 1.3800E-04 -1.3800E-05 3.6465E-03 2.4381
12 5.00 9.2289E-03 -1.1291E-05 1.0938E-03 2.9611
13 15.00 2.3215E-02 -7.4308E-06 3.1579E-04 3.5006
14 25.00 3.3471E-02 -4.6000E-06 1.3687E-04 3.8637
15 40.00 4.4582E-02 -1.5333E-06 3.4367E-05 4.4639
16 45.00 4.7506E-02 -7.2632E-07 1.5284E-05 4.8158
17 49.00 4.9633E-02 -1.3939E-07 2.8083E-06 5.5516
18 50.00 7.2464E-11 -3.6232E-12 1.9034E-06 5.2537E-09 8.2795
19 51.00 9.9010E-04 -3.5873E-12 9.9010E-04 1.0100E-11 10.9957
20 55.00 4.7619E-03 -3.4507E-12 4.7619E-03 2.1000E-12 11.6778
21 60.00 9.0909E-03 -3.2938E-12 9.0909E-03 1.1000E-12 11.9586
A B C D E F
1 Part (c)
2 V i , C5H5NH
+
50.00
3 c i , C5H5NH
+
0.1000
4 K a, C5H5NH
+
5.90E-06
5 K w, H2O 1.00E-14
6
7 c , NaOH 0.1000
8 V eq. pt. 50.00
9
10 Vol. NaOH, mL b c [OH-] [H3O
+
] pH
11 0.00 5.9000E-06 -5.9000E-07 7.6517E-04 3.1162
12 5.00 9.0968E-03 -4.8273E-07 5.2760E-05 4.2777
13 15.00 2.3083E-02 -3.1769E-07 1.3755E-05 4.8615
14 25.00 3.3339E-02 -1.9667E-07 5.8979E-06 5.2293
15 40.00 4.4450E-02 -6.5556E-08 1.4748E-06 5.8313
16 45.00 4.7374E-02 -3.1053E-08 6.5546E-07 6.1835
17 49.00 4.9501E-02 -5.9596E-09 1.2039E-07 6.9194
18 50.00 1.6949E-09 -8.4746E-11 9.2049E-06 1.0864E-09 8.9640
19 51.00 9.9010E-04 -8.3907E-11 9.9018E-04 1.0099E-11 10.9957
20 55.00 4.7619E-03 -8.0710E-11 4.7619E-03 2.1000E-12 11.6778
21 60.00 9.0909E-03 -7.7042E-11 9.0909E-03 1.1000E-12 11.9586
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
14-42
A B C D E F
1 Part (a)
2 V i , NH3 50.00
3 c i , NH3 0.1000
4 K a, NH4
+
5.70E-10
5 K w, H2O 1.00E-14
6
7 c , HCl 0.1000
8 V eq. pt. 50.00
9
10 Vol. HCl, mL b c [OH-] [H3O
+
] pH
11 0.00 1.7544E-05 -1.7544E-06 1.3158E-03 7.6000E-12 11.1192
12 5.00 9.1085E-03 -1.4354E-06 1.5495E-04 6.4535E-11 10.1902
13 15.00 2.3094E-02 -9.4467E-07 4.0832E-05 2.4490E-10 9.6110
14 25.00 3.3351E-02 -5.8480E-07 1.7525E-05 5.7060E-10 9.2437
15 40.00 4.4462E-02 -1.9493E-07 4.3838E-06 2.2811E-09 8.6419
16 45.00 4.7386E-02 -9.2336E-08 1.9485E-06 5.1321E-09 8.2897
17 49.00 4.9512E-02 -1.7721E-08 3.5791E-07 2.7940E-08 7.5538
18 50.00 5.7000E-10 -2.8500E-11 5.3383E-06 5.2726
19 51.00 9.9010E-04 -2.8218E-11 9.9013E-04 3.0043
20 55.00 4.7619E-03 -2.7143E-11 4.7619E-03 2.3222
21 60.00 9.0909E-03 -2.5909E-11 9.0909E-03 2.0414
22
23 Spreadsheet Documentation
24 C8 = C2*C3/C7
25 B11 = $C$7*A11/($C$2+A11)+$C$5/$C$4
26 C11 = -$C$5/$C$4*($C$3*$C$2-$C$7*A11)/($C$2+A11)
27 D11 = (-B11+SQRT(B11^2-4*C11))/2
28 E11 = $C$5/D11
29 F11 = -LOG(E11)
30 B18 = ($C$7*A18-$C$3*$C$2)/($C$2+A18)+$C$4
31 C18 = -($C$4)*($C$2*$C$3/($C$2+A18))
32 E18 = (-B18+SQRT(B18^2-4*C18))/2+($C$7*A18-$C$2*$C$3)/($C$2+A18)
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
A B C D E F
1 Part (b)
2 V i , H2NNH2 50.00
3 c i , H2NNH2 0.1000
4 K a, H2NNH3
+
1.05E-08
5 K w, H2O 1.00E-14
6
7 c , HCl 0.1000
8 V eq. pt. 50.00
9
10 Vol. HCl, mL b c [OH-] [H3O
+
] pH
11 0.00 9.5238E-07 -9.5238E-08 3.0813E-04 3.2454E-11 10.4887
12 5.00 9.0919E-03 -7.7922E-08 8.5625E-06 1.1679E-09 8.9326
13 15.00 2.3078E-02 -5.1282E-08 2.2219E-06 4.5006E-09 8.3467
14 25.00 3.3334E-02 -3.1746E-08 9.5233E-07 1.0501E-08 7.9788
15 40.00 4.4445E-02 -1.0582E-08 2.3809E-07 4.2001E-08 7.3767
16 45.00 4.7369E-02 -5.0125E-09 1.0582E-07 9.4502E-08 7.0246
17 49.00 4.9496E-02 -9.6200E-10 1.9436E-08 5.1451E-07 6.2886
18 50.00 1.0500E-08 -5.2500E-10 2.2908E-05 4.6400
19 51.00 9.9011E-04 -5.1980E-10 9.9062E-04 3.0041
20 55.00 4.7619E-03 -5.0000E-10 4.7620E-03 2.3222
21 60.00 9.0909E-03 -4.7727E-10 9.0910E-03 2.0414
A B C D E F
1 Part (c)
2 V i , NaCN 50.00
3 c i , NaCN 0.1000
4 K a, HCN 6.20E-10
5 K w, H2O 1.00E-14
6
7 c , HCl 0.1000
8 V eq. pt. 50.00
9
10 Vol. HCl, mL b c [OH-] [H3O
+
] pH
11 0.00 1.6129E-05 -1.6129E-06 1.2620E-03 7.9242E-12 11.1010
12 5.00 9.1070E-03 -1.3196E-06 1.4267E-04 7.0092E-11 10.1543
13 15.00 2.3093E-02 -8.6849E-07 3.7547E-05 2.6633E-10 9.5746
14 25.00 3.3349E-02 -5.3763E-07 1.6113E-05 6.2060E-10 9.2072
15 40.00 4.4461E-02 -1.7921E-07 4.0304E-06 2.4811E-09 8.6054
16 45.00 4.7385E-02 -8.4890E-08 1.7914E-06 5.5821E-09 8.2532
17 49.00 4.9511E-02 -1.6292E-08 3.2905E-07 3.0390E-08 7.5173
18 50.00 6.2000E-10 -3.1000E-11 5.5675E-06 5.2543
19 51.00 9.9010E-04 -3.0693E-11 9.9013E-04 3.0043
20 55.00 4.7619E-03 -2.9524E-11 4.7619E-03 2.3222
21 60.00 9.0909E-03 -2.8182E-11 9.0909E-03 2.0414
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
A B C D E F
1 Part (c)2 V i , HOCl 50.00
3 c i , HOCl 0.1000
4 K a, HOCl 3.00E-08
5 K w, H2O 1.00E-14
6
7 c , NaOH 0.1000
8 V eq. pt. 50.00
9
10 Vol. NaOH, mL b c [OH-] [H3O
+
] pH
11 0.00 3.0000E-08 -3.0000E-09 5.4757E-05 4.2616
12 5.00 9.0909E-03 -2.4545E-09 2.6999E-07 6.5687
13 15.00 2.3077E-02 -1.6154E-09 7.0000E-08 7.1549
14 25.00 3.3333E-02 -1.0000E-09 3.0000E-08 7.5229
15 40.00 4.4444E-02 -3.3333E-10 7.5000E-09 8.1249
16 45.00 4.7368E-02 -1.5789E-10 3.3333E-09 8.4771
17 49.00 4.9495E-02 -3.0303E-11 6.1224E-10 9.2131
18 50.00 3.3333E-07 -1.6667E-08 1.2893E-04 7.7560E-11 10.1104
19 51.00 9.9043E-04 -1.6502E-08 1.0065E-03 9.9355E-12 11.0028
20 55.00 4.7622E-03 -1.5873E-08 4.7652E-03 2.0985E-12 11.6781
21 60.00 9.0912E-03 -1.5152E-08 9.0926E-03 1.0998E-12 11.9587
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
A B C D E F
1 Part (d)
2 V i , HONH3
+
50.00
3 c i , HONH3
+
0.1000
4 K a, HONH3
+
1.10E-06
5 K w, H2O 1.00E-14
6
7 c , HCl 0.1000
8 V eq. pt. 50.00
9
10 Vol. HCl, mL b c [OH-] [H3O
+
] pH
11 0.00 9.0909E-09 -9.0909E-10 3.0147E-05 3.3171E-10 9.4792
12 5.00 9.0909E-03 -7.4380E-10 8.1817E-08 1.2222E-07 6.9128
13 15.00 2.3077E-02 -4.8951E-10 2.1212E-08 4.7143E-07 6.3266
14 25.00 3.3333E-02 -3.0303E-10 9.0909E-09 1.1000E-06 5.9586
15 40.00 4.4444E-02 -1.0101E-10 2.2727E-09 4.4000E-06 5.3565
16 45.00 4.7368E-02 -4.7847E-11 1.0101E-09 9.9000E-06 5.0044
17 49.00 4.9495E-02 -9.1827E-12 1.8553E-10 5.3900E-05 4.2684
18 50.00 1.1000E-06 -5.5000E-08 2.3397E-04 3.6308
19 51.00 9.9120E-04 -5.4455E-08 1.0423E-03 2.9820
20 55.00 4.7630E-03 -5.2381E-08 4.7729E-03 2.3212
21 60.00 9.0920E-03 -5.0000E-08 9.0964E-03 2.0411
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
14-44
A B C D E F G
1 Species pH [H3O
+
] K a 0 1
2 (a) Acetic Acid 5.320 4.7863E-06 1.75E-05 0.215 0.785
3 (b) Picric Acid 1.250 5.6234E-02 4.3E-01 0.116 0.884
4 (c) HOCl 7.000 1.0000E-07 3.0E-08 0.769 0.231
5 (d) HONH3
+
5.120 7.5858E-06 1.10E-06 0.873 0.127
6 (e) Piperidine 10.080 8.3176E-11 7.50E-12 0.917 0.083
7
8 Spreadsheet Documentation
9 D2 = 10^(-C2)
10 F2 = D2/(D2+E2)
11 G2 = E2/(D2+E2)
14-45 [H3O
+] = 6.31010-4 M. Substituting into Equation 9-35 gives,
0 = 44
4
1080.110310.6
10310.6




  = 0.778
   
0850.0
HCOOHHCOOH
T

c
= 0
[HCOOH] = 0.778   0.0850 = 6.6110-2 M
14-46 [H3O
+] = 3.3810-12 M. For CH3 NH3
+, Equation 9-36 takes the form,
1 = 1112
11
a3
a
T
23
103.21038.3
103.2
]OH[
] NHCH[


 




 K 
 K 
c
= 0.872 =
120.0
] NHCH[ 23
[CH3 NH2] = 0.872  0.120 = 0.105 M
14-47 For lactic acid,  K a  = 1.38  10
-4
0 =
]OH[1038.1
]OH[
]OH[
]OH[
3
4
3
3a
3






 K 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14
= 0.640 =
   
120.0
HAHA
T

c
[HA] = 0.640  0.120 = 0.0768 M
1 = 1.000 –   0.640 = 0.360
[A
-
] = 1   0.120 = (1.000 –  0.640)  0.120 = 0.0432 M
[H3O
+] =  K a cHA / cA-  = 1.38  10
-4
 0.640 / (1 –   0.640) = 2.45310-4 M
 pH = -log 2.45310-4  = 3.61
The remaining data are obtained in the same way.
Acid cT  pH [HA] [A
-
] 0 1
Lactic 0.120 3.61 0.0768 0.0432 0.640 0.360
Iodic 0.200 1.28 0.0470 0.153 0.235 0.765
Butanoic 0.162 5.00 0.0644 0.0979 0.397 0.604
 Nitrous 0.179 3.30 0.0739 0.105 0.413 0.587
HCN 0.366 9.39 0.145 0.221 0.396 0.604
Sulfamic 0.250 1.20 0.095 0.155 0.380 0.620
Fundamentals of Analytical Chemistry: 8th ed. Chapter 15
Chapter 15
15-1 The HPO4
2- is such a weak acid ( K a3 = 4.510
-13) that the change in pH in the vicinity of
the third equivalence point is too small to be observable.
15-2 (a) NH4
+
 + H2O NH3 + H3O
+
 K a = 5.7010
-10
OAc
-
 + H2O HOAc + OH
-
 K  b =
10
5
14
a
1071.5
1075.1
1000.1 






 K 
 K 
w
Since the  K ’s are essentially identical, the solution should be approximately neutral.
(b) NO2
- + H2O HNO2 + OH
- Solution will be basic
(c) Neither K 
+ nor NO3
- reacts with H2O. Solution will be neutral
(d) HC2O4
-
 + H2O C2O4
2-
 + H3O
+
 K a2 = 5.4210
-5
HC2O4
-
 + H2O H2C2O4 + OH
-
 K  b2 =
13
2
14
1079.1
1060.5
1000.1 





Solution will be acidic
(e) C2O4
2- + H2O HC2O4
- + OH-  K  b =
10
5
14
1084.1
1042.5
1000.1 





Solution will be basic
(f)  HAsO4
2- + H2O AsO4
3- + H3O
+
 K a3 = 3.210
-12
HAsO4
2- + H2O H2AsO4
- + OH-  K  b2 =
8
7
14
101.9
101.1
1000.1 





Solution will be basic
(g) H2AsO4
- + H2O HAsO4
2- + H3O
+
 K a2 = 1.110
-7
H2AsO4
- + H2O H3AsO4 + OH
-
 K  b3 =
12
3
14
107.1
108.5
1000.1 





Solution will be acidic
Fundamentals of Analytical Chemistry: 8th ed. Chapter 15
(h) AsO4
3-
 + H2O HAsO4
2-
 + OH
-
 K  b1 =
3
12
14
101.3
102.3
1000.1 





Solution will basic
15-3 H3PO4 + H2O H2PO4
- + H3O
+
 K a1 = 7.1110
-3
H2PO4
-
 + H2O HPO4
2-
 + H3O
+
 K a2 = 6.3210
-8
Here both  K a2 and  K w /K a1 are small and we may assume that [H2PO4
-] >> [H3PO4] and
[HPO4
2-
] at the first equivalence point. If we further assume that K w <<  K a2[H2PO4
-
] and
[H2PO4
-]/ K a1 >> 1 we may use equation 15-16 to give
[H3O
+
] = (7.1110
-3
6.3210
-8
)
1/2
 = 2.1210
-5
 pH = -log(2.1210
-5
) = 4.674
Bromocresol green would be satisfactory
15-4 (Note: In the first printing of the text, the answer in the back of the book was in error.)
H2PO4
-
 + H2O HPO4
2-
 + H3O
+
 K a2 = 6.3210
-8
HPO4
2- + H2O PO4
3- + H3O
+
 K a3 = 4.510
-13
Here again both  K a3 and  K w /K a2 are small and we may assume that [HPO4
2-
] >> [H2PO4
-
]
and [PO4
3-] at the first equivalence point. However, the further approximation K w <<
 K a3[HPO4
2-
] is likely not valid and we should use equation 15-15. If we assume, for
example, [HPO4
2-] = 0.01 M we obtain
[H3O
+] =
 8
1413
1032.6/01.01
100.1105.401.0




 = 3.010-10
 pH = -log(3.010-10) = 9.52
Phenolphthalein or Thymolphthalein would work.
Fundamentals of Analytical Chemistry: 8th ed. Chapter 15
15-5 Curve  A in figure 15-4 is the titration curve for H3PO4. Note that one end point occurs at
about pH 4.5 and a second at about pH 9.5. Thus, H3PO4 would be determined by titration
with bromocresol green as an indicator (pH 3.8 to 5.4). A titration to the second end point
with phenolphthalein would give the number of millimoes of NaH2PO4 plus twice the
number of millimoles of H3PO4. Thus, the concentration of NaH2PO4 is obtained from the
difference in volume for the two titrations.
15-6 (a) For this system we should use equation 15-15 since the approximation
[H2AsO4
-
]/5.810
-3
 >> 1 is not valid.
[H3O
+
] =
)108.5/(05.01
100.1101.105.0
3
147




 = 2.410
-5
 pH = -log(2.410
-5
) = 4.62
Bromocresol green (3.8 to 5.4)
(b) P
2- + H2O HP
- + OH-  K  b1 =
9
6
14
-2
--
1056.2
1091.3
1000.1
]P[
]][HPOH[ 






[OH-] = [HP-] and we assume [P2-] = 0.05 –  [OH-]  0.05
[OH
-
] = (0.052.5610
-9
)
1/2
 = 1.1310
-5
 pH = 14.00 –  (-log(1.1310
-5)) = 9.05
Phenolphthalein (8.3 to 10)
(c) As in part (b)
[OH-] = (0.051.0010-14/4.3110-5)1/2 = 3.4110-6
 pH = 14.00 –  (-log(3.4110
-6
)) = 8.53
Cresol purple (7.6 to9.2)
Fundamentals of Analytical Chemistry: 8th ed. Chapter 15
(d) Here we are able to use equation 15-16
[H3O
+
] = (1.4210
-7
1.1810
-10
)
1/2
 = 4.0910
-9
 pH = -log(4.0910
-9
) = 8.39
Cresol purple (7.6 to 9.2)
(e) NH3C2H4 NH3
2+
 + H2O NH3C2H4 NH2
+
 + H3O
+
 K a1 = 1.4210
-7
[H3O
+] = (0.051.4210
-7)1/2 = 8.4310
-5
 pH = -log(8.4310-5) = 4.07
Bromocresol green (3.8 to 5.4)
(f) Proceeding as in part (a)
[H3O
+
] =
)1023.1/(05.01
100.1106.605.0
2
148




 = 2.5510
-5
 pH = -log(2.5510
-5
) = 4.59
Bromocresol green (3.8 to 5.4)
(g) Proceeding as in part (b) we obtain pH = 9.94
Phenolphthalein(8.5 to 10.0)
15-7 (a) H3AsO4 + H2O H2AsO4
- + H3O
+
 K a1 = 5.810
-3
3
3
2
3
43
-
423 108.5
]OH[0400.0
]OH[
]AsOH[
]AsO][HOH[ 





0 = [H3O
+]2 + 5.810-3[H3O
+] –  5.810-30.0400
Solving using the quadratic formula gives
[H3O
+] = 1.2610-2 and pH = 1.90
Proceeding in the same way we obtain
(b)  2.20
Fundamentals of Analytical Chemistry: 8th ed. Chapter 15
(c)  1.64
(d)  1.77
(e) 4.21
(f) NH2C2H4 NH2 + H2O NH3C2H4 NH2
+
 + OH
-
 K  b1 =
5
10
14
1047.8
1018.1
1000.1 






0400.0
]OH[ 2-
8.4710
-5
[OH
-
] = (0.04008.4710
-5
)
1/2
 = 1.8410
-3
 pH = 14.00 –  (-log(1.8410
-3
)) = 11.26
15-8 (a) [H3O
+
] =
)108.5/(0400.01
100.1101.10400.0
3
147




 = 2.3610
-5
 pH = -log(2.3610
-5
) = 4.63
Proceeding as in part (a) we obtain
(b) pH = 2.95
(c) pH = 4.28
(d) pH = 4.60
(e) pH = 9.80
(f)  pH = 8.39
15-9 (a) AsO4
3-
 + H2O HAsO4
2-
 + OH
-
 K  b1 =
3
12
14
1012.3
102.3
1000.1 





3
-
2-
-3
4
-2
4
-
1012.3
]OH[0400.0
]OH[
]AsO[
]][HAsOOH[ 


0 = [OH-]2 + 3.1210
-3[OH-] –  3.1210
-3
0.0400
Solving using the quadratic formula gives
Fundamentals of Analytical Chemistry: 8th ed. Chapter 15
[OH-] = 9.7210
-3 and pH = 14.00 –  (-log(9.7210
-3)) = 11.99
(b) C2O4
2- + H2O HC2O4
- + OH-  K  b1 =
10
5
14
1084.1
1042.5
1000.1 





[OH-] = (0.04001.8410-10)1/2 = 2.7210-6  and pH = 8.43
Proceeding as in part (b) we obtain
(c) pH = 9.70
(d) pH = 9.89
(e) Proceeding as in part (a) gives pH = 12.58
(f) NH3C2H4 NH3
2+ + H2O NH3C2H4 NH2
+ + H3O
+
 K a1 = 1.4210
-7
[H3O
+] = (0.04001.4210-7)1/2 = 7.5410-5 and pH = 4.12
15-10 (a) H3PO4 + H2O H2PO4
- + H3O
+
 K a1 = 7.1110
-3
7.1110-3 =
]OH[0500.0
]OH)[]OH[0200.0(
]POH[
]PO][HOH[
3
33
43
-
423





Rearranging gives 0 = [H3O
+]2 +(0.0200+7.1110-3)[H3O
+] –  (7.1110-3)(0.0500)
and solving the quadratic gives [H3O
+
] = 9.6710
-3
and pH = 2.01
(b) H2AsO4
- + H2O HAsO4
2- + H3O
+
 K a2 = 1.1110
-7
1.1110
-7 =
0300.0
]OH)[0500.0(
]OH[0300.0
]OH)[]OH[0500.0(
]AsOH[
]][HAsOOH[ 3
3
33
-
42
-2
43







[H3O
+] = 6.6610
-8 and pH = 7.18
(c) HCO3
- + H2O CO3
2- + H3O
+
 K a2 = 4.6910
-11
Proceeding as in part (b) we obtain [H3O
+
] = 2.3410
-11
and pH = 10.63
(d) H3PO4 + HPO4
2-
 2H2PO4
-
For each milliliter of solution, 0.0200 mmol Na2HPO4 reacts with 0.0200 mmol
Fundamentals of Analytical Chemistry: 8th ed. Chapter 15
H3PO4 to give 0.0400 mmol NaH2PO4 and to leave 0.0200 mmol H3PO4. Thus,
we have a buffer that is 0.0200 M in H3PO4 and 0.0400 M in NaH2PO4.
Proceeding as in part (a) we obtain [H3O
+] = 2.8210
-3 and pH = 2.55
(e) HSO4
- + H2O SO4
2- + H3O
+
 K a2 = 1.0210
-2
Proceeding as in part (a) we obtain [H3O
+
] = 8.6610
-3
and pH = 2.06
15-11 (a) Proceeding as in 15-10(a) we obtain [H3O
+
] = 3.4810
-3
and pH = 2.46
(b) Proceeding as in 15-10(b) we obtain [H3O
+
] = 3.1010
-8
and pH = 7.51
(c) HOC2H4 NH3
+ + H2O HOC2H4 NH2 + H3O
+
 K a = 3.1810
-10
Proceeding as in 15-10(b) we obtain [H3O
+
] = 3.7310
-10
and pH = 9.43
(d) H2C2O4 + C2O4
2-
 2HC2O4
-
For each milliliter of solution, 0.0240 mmol H2HPO4 reacts with 0.0240 mmol
C2O4
2-
 to give 0.0480 mmol HC2O4
-
 and to leave 0.0120 mmol C2O4
2-
. Thus,
we have a buffer that is 0.0480 M in HC2O4
- and 0.0120 M in C2O4
2-.
Proceeding as in 15-10(a) we obtain [H3O
+
] = 2.1710
-4
and pH = 3.66
(e) Proceeding as in 15-10(b) we obtain [H3O
+] = 2.1710-4 and pH = 3.66
15-12 (a) (NO2)3C6H2OH + H2O (NO2)3C6H2O
- + H3O
+
 K a = 0.43
0.43 =
 x
 x x




0200.0
)0100.0(
OH]HC)(NO[
]OHC)][(NOOH[
2632
-
26323
Rearranging gives 0 =  x
2
 +(0.0100+0.43) x –  (0.43)(0.0200)
and solving the quadratic gives  x = 1.8710
-2
the total [H3O
+
] = 0.0100 +  x = 0.0287 and pH = 1.54
(b) Proceeding as in part (a) we obtain [H3O
+] = 1.01310
-2 and pH = 1.99
Fundamentals of Analytical Chemistry: 8th ed. Chapter 16
A B C D E
1 Problem 16-34
2
3 Conc. HCl, M 0.1750
4 Vol. HCl, mL 100.0
5 Conc. NaOH, M 0.1080
6 Vol. NaOH, mL 11.37
7 No. Tablets 4
8
9 mg CH5N3/tablet 80.0991169
10
11 Patient Wt, lb Proper dose No. tablets
12 (a) 100 5.67546831 6
13 (b) 150 8.51320247 9
14 (c) 275 15.6075379 16
15
16 Spreadsheet Documentation
17 B9 = (((B3*B4)-(B5*B6))*(1/3)*59.07)/B7
18 C12 = 10*B12*0.4546*(1/$B$9)
19 D12 = ROUND(C12,0)
16-35
 N%92.3%100
sampleg992.0
mmol1000
 Ng007.14
HClmmol
 Nmmol1
HClmL66.22
mL
HClmmol1224.0
 N% 

 
 

 
 


16-36 Multiplication factor for meat is 6.25 protein/N
can/ proteing0.45
tunag100
 proteing48.24
oz
g3.28
can
tunaoz50.6
 protein%48.24
 N
 protein25.6
 N%916.3


Fundamentals of Analytical Chemistry: 8th ed. Chapter 16
16-37
A B C D E F
1 16-37 Nitrogen in a plant food preparation
2 Weight sample, g 0.5843
3 Volume of HCl, mL 50.00
4 Conc. HCl, M 0.1062
5 Vol. NaOH, mL 11.89
6 Conc. NaoH, M 0.0925 In Cell B8, the mmol of HCl/g sample is calculated by subtracting
7 the mmol of NaOH used from the total mmol of HCl added and
8 mmol HCl/g sample 7.20550 dividing by the sample weight.
9 Molar masses Percentages
10 (a) N 14.007 10.09 %N
11 (b) urea 60.06 21.64 % urea
12 (c) (NH4)2SO4 132.141 47.61 % (NH4)2SO4
13 (d) (NH4)3PO4 149.09 35.81 % (NH4)3PO4
14 Spreadsheet Documentation
15 B8=($B$3*$B$4-$B$5*$B$6)/$B$2
16 C10=$B$8*1*B10/1000*100 The percentages are calculated in Cells C10:C13 from the no. of
17 C11=$B$8*1/2*B11/1000*100 mmol of HCl/g sample times the no. of mmol compound/mmol HCl
18 C12=$B$8*1/2*B12/1000*100 times the molar mass of the compound divided by 1000 (mmolar
19 C13=$B$8*1/3*B13/1000*100 mass).
16-38
 protein%0.197.5%335.3 protein%
 N%335.3%100
g9092.0
mmol1000
g007.14
HClmmol1
 Nmmol1
HClmmol165.2
 N%
mmol165.2mL46.7
mL
mmol04917.0
mL00.50
mL
mmol05063.0
consumedHClmmol





 
 

 
 

 
 

 
 

Fundamentals of Analytical Chemistry: 8th ed. Chapter 16
16-39 In the first titration,
HClmmol63114.1mL17.10
mL
 NaOHmmol08802.0
mL00.30
mL
HClmmol08421.0
consumedHClmmol

 
 

 
 


 
 

 
 

and
 42434 SO) NH(mmol2 NO NHmmolHClmmol63114.1 
The amounts of the two species in the original sample are
    mmol5246.6
mL50
mL200
mmol63114.1SO) NH(mmol2 NO NHmmol 42434   (1)
In the second titration,
HClmmol27994.1mL16.14
mL
 NaOHmmol08802.0
mL00.30
mL
HClmmol08421.0
consumedHClmmol

 
 

 
 


 
 

 
 

and
 42434 SO) NH(mmol2) NO NHmmol2(HClmmol27994.1 
The amounts of the two species in the original sample are
 
mmol2395.10
mL25
mL200
mmol27994.1SO) NH(mmol2) NO NHmmol2( 42434


  (2)
Fundamentals of Analytical Chemistry: 8th ed. Chapter 16
Subtracting equation (1) from equation (2) gives
424
424
424
424
34
34
34
34
424424
3434
SO) NH(%23.15
%100
sampleg219.1
mmol1000
SO) NH(g14.132
SO) NH(mmol4048.1
SO) NH(%
 NO NH%39.24%100
sampleg219.1
mmol1000
 NO NHg04.80
 NO NHmmol7149.3
 NO NH%
SO) NH(mmol4048.1
2
)mmol7149.32(mmol2395.10
SO) NH(mmol
 NO NHmmol7149.3mmol52455.6mmol2395.10 NO NHmmol











16-40 For the first aliquot,
mmol8854.1 NaOHmL74.4
mL
 NaOHmmol04983.0
HClmL00.40
mL
HCLmmol05304.0
)COK mmol2(KOHmmol
)COK mmol2(KOHmmol NaOHmmolconsumedHClmmol
32
32

 
 

 
 


 
 

 
 


For the second aliquot,
3232 COK mmol1853.0
2
5148.1mmol8854.1
COK mmol
)KOH(HClmmol5148.1HClmL56.28
mL
HClmmol05304.0
KOHmmolHClmmol




OH%12.9%)04.21%84.69(%100
COK %04.21%100
mL500
mL50
g217.1
mmol1000
COK g21.138
COK mmol18530.0
COK %
KOH%84.69%100
mL500
mL50
g217.1
mmol1000
KOHg11.56
KOHmmol5148.1
KOH%
2
32
32
32
32



 
 

 
 





 
 

 
 



Fundamentals of Analytical Chemistry: 8th ed. Chapter 16
16-41 For the firstaliquot,
mmol6026.0 NaOHmL34.2
mL
 NaOHmmol01063.0
HClmL00.50
mL
HClmmol01255.0
)CO Nammol2( NaHCOmmol
)CO Nammol2( NaHCOmmol NaOHmmolHClmmol
323
323

 
 

 
 


 
 

 
 


For the second aliquot,
3
3
 NaHCOmmol1700.0
HClmL63.7
mL
HClmmol01255.0
 NaOHmL00.25
mL
 NaOHmmol01063.0
HClmmol NaOHmmol NaHCOmmol


 
 

 
 

 
 

 
 


3232 CO Nammol2163.0
2
mmol1700.0mmol6026.0
CO Nammol 


OH%59.25%)85.45%56.28(%100
CO Na%85.45%100
mL0.250
mL00.25
g5000.0
mmol1000
CO Nag99.105
CO Nammol2163.0
CO Na%
 NaHCO%56.28%100
g0.250
g00.25
g5000.0
mmol1000
 NaHCOg01.84
 NaHCOmmol1700.0
 NaHCO%
2
32
32
32
32
3
3
3
3



 
 

 
 





 
 

 
 



Fundamentals of Analytical Chemistry: 8th ed. Chapter 16
16-42
A B C D E
1 16-42 Titrations with 0.06122 M HCl
2 M HCl 0.06122
3 M Na3PO4  0.05555
4 (a) mL Na3PO4 mmol base mL HCl
5  Add one proton 10.00 0.55550 9.07
6 to thymolphthalein 15.00 0.83325 13.61
7 endpoint 25.00 1.38875 22.68
8 40.00 2.22200 36.30
9 (b) mL Na3PO4 mmol base mL HCl
10  Add two protons 10.00 1.11100 18.15
11 to bromocresol 15.00 1.66650 27.22
12 green endpoint 20.00 2.22200 36.30
13 25.00 2.77750 45.37
14
15 (c) mL solution mmol base mL HCl
16 M Na3PO4 0.02102 20.00 1.17180 19.14
17 M Na2HPO4 0.01655 25.00 1.46475 23.93
18  Add two protons 30.00 1.75770 28.71
19 to phosphate and 40.00 2.34360 38.28
20 one to mono-
21 hydrogen phosphate
22 (d) 0.01655mL Na3PO4 mmol base mL HCl
23 M NaOH 15.00 0.56355 9.21
24  Add one proton 20.00 0.75140 12.27
25 35.00 1.31495 21.48
26 40.00 1.50280 24.55
27 Spreadsheet Documentation
28 D5=C5*$B$3
29 D10=C10*$B$3*2
30 D16=$B$15*2*C16+$B$16*C16
31 D23=C23*$B$16+C23*$B$22
32 E5=D5/$B$2
33 E10=D10/$B$2
34 E16=D16/$B$2
35 E23=D23/$B$2
Fundamentals of Analytical Chemistry: 8th ed. Chapter 16
16-43
A B C D E
1 16-43 Titrations with 0.07731 M NaOH
2 M NaOH 0.07731
3 (a) and (b)
4 M HCl 0.03000
5 M H3PO4  0.01000
6 (a) React with one mL solution mmol acid mL NaOH
7 proton to bromo- 25.00 1.0000 12.93
8 cresol green
9 endpoint
10 (b) React with two mL solution mmol acid mL NaOH
11 protons to thymol- 25.00 1.2500 16.17
12 phthalein end point
13
14 (c) NaH2PO4 mL solution mmol acid mL NaOH
15 M NaH2PO4 0.06407 10.00 0.64070 8.29
16 One proton reacts 20.00 1.28140 16.57
17 to thymol-phthalein 30.00 1.92210 24.86
18 endpoint 40.00 2.56280 33.15
19 (d) Mixture mL solution mmol acid mL NaOH
20 M H3PO4 0.02000 20.00 1.40000 18.11
21 M NaH2PO4 0.03000 25.00 1.75000 22.64
22 30.00 2.10000 27.16
23 (d) React with two
24 protons from H3PO4
25 and one from
26 NaH2PO4
27 Spreadsheet Documentation
28 D7=C7*$B$4+C7*$B$5
29 E6=D6/$B$2
30 D11=C6*$B$4+2*C6*$B$5
31 E11=D11/$B$2
32 D15=$B$15*C15
33 E15=D15/$B$2
34 D20=2*$B$20*C20+$B$21*C20
35 E20=D20/$B$2
Fundamentals of Analytical Chemistry: 8th ed. Chapter 16
16-44
A B C D E F
1 16-44 Titrations of carbonate mixtures ol. to phenol., mL ol. to BCG, mL
2 M HCl 0.1202 (a) 22.42 22.44
3 Volume, mL 25.00 (b) 15.67 42.13
4 M NaOH 40.00 (c) 29.64 36.42
5 M Na2CO3 105.99 (d) 16.12 32.23
6 M NaHCO3 84.01 (e) 0.00 33.33
7 Table 14-2 gives the volume relationships in titrations of these mixtures.
8
9 (a) Since essentially the same volume is used for each endpoint, there is only NaOH
10 present. We use the average volume to calculate the no. of mg NaOH/mL
11 mmol NaOH mg NaOH/mL
12 2.6961 4.314
13 (b) Since V phth< ½V bcg, only carbonate and bicarbonate are present.
14 mmol carbonate mmol total mmol bicarbonate mg Na2CO3 /ml mg NaHCO3 /ml
15 1.8835 5.0640 1.2970 7.985 4.358
16 (c) Now V phth > ½V bcg, so we have a mixture of NaOH and Na2CO3
17 mmol carbonate plus NaOH mmol carbonate mmol NaOH mg Na2CO3 /ml mg NaOH/ml
18 3.5627 0.8150 2.7478 3.455 4.396
19 (d)  Since V phth = ½V bcg, we have only Na2CO3 present
20 mmol carbonate mg Na2CO3 /ml
21 1.9376 8.215
22 (e)  Since V phth = 0, we have only NaHCO3 present which gains one proton.
23 mmol NaHCO3 mg NaHCO3 /ml
24 4.0063 13.46
25 Spreadsheet Documentation
26 B12=((D2+E2)/2)*$B$2 D18=B18-C18
27 C12=B12*1*$B$4/$B$3 E18=C18*$B$5/$B$3
28 B15=D3*$B$2 F18=D18*$B$4/$B$3
29 C15=E3*$B$2 B21=$D$5*$B$2
30 D15=C15-2*B15 C21=B21*1*$B$5/$B$3
31 E15=B15*$B$5/$B$3 B24=E6*$B$2
32 F15=D15*$B$6/$B$3 C24=B24*1*$B$6/$B$3
33 B18=D4*$B$2
34 C18=($E$4-$D$4)*$B$2
Fundamentals of Analytical Chemistry: 8th ed. Chapter 16
16-45
A B C D E F
1 16-45 Titrations of Mixtures ol. to phenol., mL ol. to BCG, mL
2 M HCl 0.08601 (a) 0.00 18.15
3 Volume, mL 25.00 (b) 21.00 28.15
4 M NaOH 40.00 (c) 19.80 39.61
5 M Na3 AsO4 207.89 (d) 18.04 18.03
6 M Na2HAsO4 185.91 (e) 16.00 37.37
7 We use the method of Problem 14-44. Table 14-2 gives the volume relationships in titrations of similar mixtures.
8 (a) Since V phth = 0, we have only Na2HAsO4 present which gains one proton.
9 mmol NaHCO3 mg Na2HAsO4 /ml
10 1.5611 11.61
11 (b) Now V phth > ½V bcg, so we have a mixture of NaOH and Na3 AsO4
12 mmol Na3AsO4 plus NaOH mmol Na3AsO4 mmol NaOH mg Na3AsO4 /ml mg NaOH/ml
13 1.8062 0.6150 1.1912 5.114 1.906
14 (c) Since V phth = ½V bcg, we have only Na3 AsO4 present
15 mmol Na3AsO4 mg Na3AsO4 /ml
16 1.7030 14.16
17 (d) Since essentially the same volume is used for each endpoint, there is only NaOH
18 present. We use the average volume to calculate the no. of mg NaOH/mL
19 mmol NaOH mg NaOH/mL
20 1.5512 2.482
21 (e) Since V phth< ½V bcg, only Na3 AsO4 and Na2HAsO4 are present.
22 mmol Na3AsO4 mmol total mmol Na2HAsO4 mg Na3AsO4 /ml mg Na2HAsO4 /ml
23 1.3762 3.2142 0.4619 11.44 3.435
24 Spreadsheet Documentation
25 B10=$E2*$B$2 B16=$D$4*$B$2
26 C10=B10*1*$B$6/$B$3 C16=$E$6*$B$2
27 B13=D3*$B$2 B23=$D$6*$B$2
28 C13=($E$3-$D$3)*$B$2 C23=$E$6*$B$2
29 D13=B13-C13 D23=C23-2*B23
30 E13=C13*$B$5/$B$3 E23=B23*$B$5/$B$3
31 F13=D13*$B$4/$B$3 F23=D23*$B$6/$B$3
Fundamentals of Analytical Chemistry: 8th ed. Chapter 16
16-46 (a) The equivalent weight of an acid is that weight of the pure material that contains one
mole of titratable protons in a specified reaction.
(b) The equivalent weight of a base is that weight of the pure material that consumes one
mole of protons in a specified reaction.
16-47 (a) With bromocresol green, only one of the two protons in the oxalic acid will react.
Therefore, the equivalent mass is the molar mass, or 126.1 g.
(b) When phenolphthalein is the indicator, two of the protons are consumed. Therefore,
the equivalent mass of oxalic acid is one-half the molar mass, or 63.0 g.
16-48 (a)
COOHCHM4598.0
mL00.10
 NaOHmmol
COOHCHmmol1
 NaOHmL62.45
mL
 NaOHmmol1008.0
3
3


(b)
COOHCH%75.2
%100
g004.1
COOHCHmL1
mg1000
g1
mmol
COOHCHmg03.60
mL
COOHCHmmol4598.0
3
333


Fundamentals of Analytical Chemistry: 8th ed. Chapter 17
Chapter 17
17-1 (a) A  chelate is a cyclic complex consisting of metal ion and a reagent that contains two
or more electron donor groups located in such a position that they can bond with the
metal ion to form a Heterocyclic ring structure.
(b) A  tetradentate chelating agent  is a molecule that contains four pairs of donor electron
located in such positions that they all can bond to a metal ion, thus forming two rings.
(c) A  ligand  is a species that contains one or more electron pair donor groups that tend to
form bonds with metal ions.
(d) The  coordination number  is the number of covalent bonds that a cation tends to form
with electron donor groups.
(e) A  conditional formation constant  is an equilibrium constant for the reaction between a
metal ion and a complexing agent that applies only when the pH and/or the concentration
of other complexing ions are carefully specified.
(f) NTA is the acronym for nitrilotriacetic acid, a tetradentate complexing agent that
contains three carboxylate groups and one tertiary amine. As an electron donor, NTA has
found applications in the titrationof a variety of cations.
(g) Water hardness is the concentration of calcium carbonate that is equivalent to the
total concentration of all of the multivalent metal carbonates in the water.
(h) In an EDTA displacement titration, an unmeasured excess of a solution containing the
magnesium or zinc complex of EDTA is introduced into the solution of an analyte that
forms a more stable complex that that of magnesium or zinc. The liberated magnesium
or zinc ions are then titrated with a standard solution of EDTA. Displacement titrations
are used for the determination of cations for which no good indicator exists.
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth eedd. . CChhaapptteer r 1177
17-3117-31
A A B B C C D D E E F F G G H H II
1 1 17-317-31 1 ConConditditionional al conconstastants nts for for BaBa2+2+- EDTA complex- EDTA complex
22 Note: Note: The The conditional conditional constantconstant K  K ''MYMY is the product of  is the product of   44 and and  K  K MYMY (Equation 17-25). (Equation 17-25).
33 The The value value of of  K  K MYMY is found in  is found in TablTable 17-3e 17-3
44 K K MYMY   5.80E+07  5.80E+07 pH pH DD 44 K K ''MYMY
55 K K 11 11..0022EE--002 2 77..0 0 11..7733EE--118 8 44..8800EE--004 4 22..88EE++0044
66 K K 22 22..1144EE--003 3 99..0 0 11..6600EE--220 0 55..2211EE--002 2 33..00EE++0066
77 K K 33 66..9922EE--007 7 1111..0 0 99..8822EE--222 2 88..4466EE--001 1 44..99EE++0077
88 K K 44   5.50E-11  5.50E-11
9 9 SprSpreadeadshesheet et DocDocumumententatioationn
1010   D5=(10^-C5)^4+$B$5*(10^-C5)^3+$B$5*$B$6*(10^-C5)^2+$B$5*$B$6*$B$7*(10^-C5)+$B$5*$B$6*$B$7*$B$8  D5=(10^-C5)^4+$B$5*(10^-C5)^3+$B$5*$B$6*(10^-C5)^2+$B$5*$B$6*$B$7*(10^-C5)+$B$5*$B$6*$B$7*$B$8
1111   E5=$B$5*$B$6*$B$7*$B$8/D5  E5=$B$5*$B$6*$B$7*$B$8/D5
1212   F5==E5*$B$4  F5==E5*$B$4
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth eedd. . CChhaapptteer r 1177
17-3217-32
A A B B C C D D E E F F G G HH
1 1 1717-3-32 2 TiTitrtratatioion n of of 5050.0.00 0 mL mL of of 0.00.0101000 00 M M Sr Sr 
2+2+
with 0.02000 M EDTAwith 0.02000 M EDTA
22 Note: Note: The The conditiconditional onal constconstantant K K ''MYMY is the product of  is the product of   44 andand  K  K MYMY  (Equation 17-25). (Equation 17-25).
33 The The value value of of K K MYMY is found in Table 17-3. is found in Table 17-3.
44 K K MYMY   4.30E+08  4.30E+08 pH pH DD 44 K K ''MYMY
55   EDTA  EDTA K K 11 11..0022EE--002 2 1111. 0 . 0 99. 8. 822EE--222 2 88. 4. 466EE--001 1 33..6644EE++0088
66 K K 22   2.14E-03  2.14E-03
77 K K 33   6.92E-07  6.92E-07
88 K K 44   5.50E-11  5.50E-11
99 Initial Initial conc. conc. Sr Sr 
2+2+
00..001100000 0 IInniittiiaal l VVooll. . 5500..0000
1010 InInititiaial l coconcnc. . EDEDTA TA 0.0.0202000000
111 1 VVooll. . EEDDTTAA, , mmLL c c Sr2+Sr2+ c c SrY2-SrY2- c c TT [Sr [Sr 
2+2+] ] [[SSrrYY2-2-] ] ppSSr  r  
1212 00..000 0 00..001100000 0 0 0 00..001100000 0 22..0000
1313 1100..000 0 00..000055000 0 00..000033333 3 00..000055000 0 22..3300
1414 2244..000 0 00..000000227 7 00..000066449 9 00..000000227 7 33..5577
1515 2244..990 0 00..000000003 3 00..000066665 5 00..000000003 3 44..5577
1616 2255..000 0 00..000000000 0 00..000066667 7 00..000066667 7 44..2288EE--006 6 00..000066667 7 55..3377
1717 2255..110 0 00..000066666 26 2..6666EE--005 5 66..8877EE--007 7 00..000066666 6 66..1166
1818 2266..000 0 00..000066558 28 2..6633EE--004 4 66..8877EE--008 8 00..000066558 8 77..1166
1919 3300..000 0 00..000066225 15 1..2255EE--003 3 11..3377EE--008 8 00..000066225 5 77..8866
20 20 SprSpreadeadshesheet et DocDocumeumentantationtion
21 21 B12=(B12=($B$9*$B$9*$D$9-$D$9-$B$10$B$10*A12)*A12)/($D$/($D$9+A129+A12))
22 22 C12C12=($=($B$1B$10*A0*A12)12)/($/($D$9D$9+A+A12)12)
23 23 C16C16=($=($B$1B$10*$A0*$A$16$16)/()/($D$$D$9+A9+A16)16)
24 24 D17=(D17=($B$10$B$10*A17-*A17-$D$9*$D$9*$B$9)$B$9)/($D$/($D$9+A179+A17))
25 25 D16D16=($=($B$1B$10*$A0*$A$16$16)/()/($D$$D$9+A9+A16)16)
226 6 EE1122==BB1122
27 27 E1E16=S6=SQRQRT(T(C1C16/6/$F$$F$5)5)
28 28 E1E17=7=C1C17/7/(D(D1717*$*$F$F$5)5)
229 9 FF1166==CC1166
30 30 H1H12=2=-L-LOGOG1010(E(E1212))
3131
3232
3333
3434
3535
3636
3737
3838
3939
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth eedd. . CChhaapptteer r 1177
17-3317-33
A A B B C C D D E E F F G G HH
1 1 1717-3-33 3 TiTitratratition on of of 5050.0.00 0 mL mL of of 0.00.015150 0 M M FeFe
2+2+
with 0.0300 M EDTAwith 0.0300 M EDTA
22 Note: Note: The The conditional conditional constantconstant K K ''MYMY is the product of  is the product of   44 andand  K  K MYMY  (Equation 17-25). (Equation 17-25).
33 The The value value of of  K  K MYMY is found in Table 17-3. is found in Table 17-3.
44 K K MYMY   2.10E+14  2.10E+14 pH pH DD 44 K K ''MYMY
55   EDTA  EDTA K K 11 11..0022EE--002 2 77..0 0 11..7733EE--118 8 44..8800EE--004 1 .4 1 .0011EE++1111
66 K K 22   2.14E-03  2.14E-03
77 K K 33   6.92E-07  6.92E-07
88 K K 44   5.50E-11  5.50E-11
99 Initial Initial conc. conc. FeFe
2+2+
00..0011550 0 IInniittiiaal l VVooll. . 5500..0000
1010 InInititiaial l coconnc. c. EEDDTA TA 0.0.03030000
111 1 VVooll. . EEDDTTAA, , mmLL c c Fe2+Fe2+ c c FeY2-FeY2- c c TT [Fe[Fe
2+2+
] ] [[FFeeYY
2-2-
] ] ppFFee
1212 00..000 0 00..001155000 0 0 0 00..001155000 0 11..8822
1313 1100..000 0 00..000077550 0 00..000055000 0 00..000077550 0 22..1122
1414 2244..000 0 00..000000441 1 00..000099773 3 00..000000441 1 33..3399
1515 2244..990 0 00..000000004 4 00..000099997 7 00..000000004 4 44..4400
1616 2255..000 0 00..000000000 0 00..001100000 0 00..001100000 0 33..1155EE--007 7 00..001100000 0 66..5500
1717 2255..110 0 00..000099999 39 3..9999EE--005 5 22..4488EE--009 9 00..000099999 9 88..6611
1818 2266..000 0 00..000099887 37 3..9955EE--004 4 22..4488EE--110 0 00..000099887 7 99..6611
1919 3300..000 0 00..00009933775 5 11..8888EE--003 3 44..9966EE--111 1 00..000099338 8 1100..3300
20 20 SpSpreareadsdsheheet et DoDocumcumenentatatiotionn
21 21 B12=(B12=($B$9*$$B$9*$D$9-D$9-$B$10$B$10*A12)/*A12)/($D$($D$9+A129+A12))
22 22 C1C12=(2=($B$$B$10*A10*A12)12)/($/($D$9D$9+A1+A12)2) Note: Note: The The method method is is identiidentical cal to to ProbleProblemm
23 23 C1C16=(6=($B$$B$10*10*$A$A$16$16)/()/($D$D$9+A$9+A16)16)   17-32.  17-32.
24 24 D17=(D17=($B$10*A$B$10*A17-$D$17-$D$9*$B$9*$B$9)/($9)/($D$9+AD$9+A17)17)
25 25 D1D16=(6=($B$$B$10*10*$A$A$16$16)/()/($D$D$9+A$9+A16)16)
226 6 EE1122==BB1122
27 27 E1E16=6=SQSQRTRT(C(C1616/$/$F$F$5)5)
28 28 E1E17=7=C1C17/7/(D(D1717*$*$F$F$5)5)
229 9 FF1166==CC1166
30 30 H1H12=2=-L-LOGOG1010(E(E1212))
3131
3232
3333
3434
3535
3636
3737
3838
3939
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8thth eedd. . CChhaapptteer r 1177
17-3417-34
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2222
22
22
22
2222
MgMgmmolmmol10991099..0017511751..0028502850..00MgMgmmolmmol
CaCammolmmol17511751..00
EDTAEDTAmmolmmol
CaCammolmmol11
EDTAEDTAmLmL5353..1414
mLmL
EDTAEDTAmmolmmol0120501205..00
CaCammolmmol
mmolmmol28502850..00EDTAEDTAmLmL6565..2323
mLmL
EDTAEDTAmmolmmol0120501205..00
MgMgmmolmmolCaCammolmmol
(a)(a)
See discussSee discussion of ion of water hardneswater hardness in s in 17D-9.17D-9.
33
33
2222
33
CaCOCaCO ppm ppm55..570570
mLmL10001000
LL
mLmL0000..5050
mmolmmol
CaCOCaCOmgmg087087..100100
mmolmmol28502850..00
MgMgCaCa ppm ppmCaCOCaCO ppm ppmhardnesshardnessWater Water 
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33
33
22
3322
CaCOCaCO ppm ppm55..350350
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33
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MgCOMgCO ppm ppm33..185185
mLmL10001000
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Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8 thth ed. ed. Chapter Chapter 1818
Chapter 18Chapter 18
18-118-1 (a)(a) OxidationOxidation is a process in  is a process in which a species loses one or more electrons.which a species loses one or more electrons.
(b)(b) An An oxidizing agent oxidizing agent  is an electron acceptor. is an electron acceptor.
(c)(c) A A salt bridge salt bridge is  is a device that provides electrical contact but prevents mixing of dissimilar solutions ina device that provides electrical contact but prevents mixing of dissimilar solutions in
an electrochemical cell.an electrochemical cell.
(d)(d) A A liquid junctionliquid junction is the  is the interface between dissimilar liquids. A potential developinterface between dissimilar liquids. A potential develops across s across the interface.the interface.
(e)(e) The The Nernst equati Nernst equationon relates the  relates the potential to the concentrations (strictlypotential to the concentrations (strictly, activities) of , activities) of the participants inthe participants in
an electrochemical reaction.an electrochemical reaction.
18-218-2 (a)(a) The The electrode potential electrode potential  is  is the potential of an electrochemical cell in which a the potential of an electrochemical cell in which a standard hydrogen electrodestandard hydrogen electrode
acts as the reference electrode on the left and the half-cell acts as the reference electrode on the left and the half-cell of interest is on the right as written in cell of interest is on the right as written in cell notation.notation.
(b)(b) The The formal poten formal potential tial  of a half-reaction if the  of a half-reaction if the potential of the system (measured against the standardpotential of the system (measured against the standard
hydrogen electrode) when the concentration of each solute participating in the half-hydrogen electrode) when the concentration of each solute participating in the half- reaction has areaction has a
concentration of exactly one molar and the concentrations of all concentration of exactly one molar and the concentrations of all other constituents of the other constituents of the solution are carefullysolution are carefully
specified.specified.
(c)(c) The The standard ele standard electrode potentiactrode potential l  for a  for a half-reaction is the potential of a cell consisting of the half-reactionhalf-reaction is the potential of a cell consisting of the half-reaction
of interest on the right and a standard hyof interest on the right and a standard hydrogen electrode on the left as written in cell notadrogen electrode on the left as written in cell notation. tion. The activitiesThe activities
of all of the participants in the half-reaction are specof all of the participants in the half-reaction are specified as having a value of unityified as having a value of unity. . The additionalThe additional
specification that the standard hydrogen electrode is the reference electrode implies that the specification that the standard hydrogen electrode is the reference electrode implies that the standard potentialstandard potential
for the half-reaction is for the half-reaction is always a reduction potential.always a reduction potential.
(d)(d) A A liquid-junction potential  liquid-junction potential  is the  is the potential that develops across the interface between two dissimilarpotential that develops across the interface between two dissimilar
solutions.solutions.
(e)(e) An An oxidation potential oxidation potential  is the potential of an electrochemical cell in which the cathode is a standard is the potential of an electrochemical cell in which the cathode is a standard
hydrogen electrode and the half-cell of interest acts as anode.hydrogen electrode and the half-cell of interest acts as anode.
18-318-3 (a)(a) Reduction Reduction is the process whereby a substance acquires electrons; a is the process whereby a substance acquires electrons; a reducing agent reducing agent  is a supplier of is a supplier of
electrons.electrons.
Fundamentals of Analytical Chemistry: 8Fundamentals of Analytical Chemistry: 8 thth ed. ed. Chapter Chapter 1818
(b)(b) A A galvanic cell  galvanic cell  is one in which a  is one in which a spontaneous electspontaneous electrochemical reaction occurs and is thus a source ofrochemical reaction occurs and is thus a source of
energy. energy. The reaction The reaction in anin an electrolytic cell electrolytic cell  is  is forced in a forced in a nonspontannonspontaneous direction through application ofeous direction through application of
an external source of an external source of electrical energy.electrical energy.
(c)(c) The The anodeanode of an electrochemical ce of an electrochemical cell is the electrode at which oxidation oll is the electrode at which oxidation occurs. ccurs. TheThe cathode cathode is the is the
electrode at which reduction occurs.electrode at which reduction occurs.
(d)(d) In a In a reversible cell reversible cell , alteration of the direction of the current simply causes a reversal in the, alteration of the direction of the current simply causes a reversal in the
electrochemielectrochemical procal process. cess. In anIn an irreversible cell irreversible cell , reversal of the current results in a , reversal of the current results in a different reaction at onedifferent reaction at one
or both of the electrodes.or both of the electrodes.
(e)(e) The The standard ele standard electrode potentiactrode potential l  is  is the potential of an electrochemical cell in which the the potential of an electrochemical cell in which the standardstandard
hydrogen electrode acts as the reference hydrogen electrode acts as the reference electrode on the left and electrode on the left and all participants in the all participants in the right-hand electroderight-hand electrode
 process have unit a process have unit activity. ctivity. TheThe formal pot formal potential ential  differs in that the molar concentrations of the reactants and differs in that the molar concentrations of the reactants and
 products are unity  products are unity and the concentand the concentrations of other sperations of other species in the soluticies in the solution are carefully on are carefully specified.specified.
18-418-4 The first standard potential is for a solution that is saturated with IThe first standard potential is for a solution that is saturated with I22 and has an I and has an I22 ( (aqaq) ) activity significantlyactivity significantly
less than one. less than one. The second potentiaThe second potential if for a hypothetical half-cell in which the Il if for a hypothetical half-cell in which the I22 ( (aqaq) activity is unity.) activity is unity.
Such a half-cell, if it existed, would have a Such a half-cell, if it existed, would have a greater potential becausgreater potential because the driving force for the re the driving force for the reductioneduction
would be greater at the higher Iwould be greater at the higher I22 concentration.  concentration. The second half-cell potentThe second half-cell potential, although hypial, although hypothetical, isothetical, is
nevertheless useful for calculating electrode potentials for solutions that are undersaturated in nevertheless useful for calculating electrode potentials for solutions that are undersaturated in II22..
18-518-5 It is necessary to It is necessary to bubble hydrogen through the electrolyte in a hydrogen electrode in order bubble hydrogen through the electrolyte in a hydrogen electrode in order to keep theto keep the
solution saturated with the gas. solution saturated with the gas. Only underthese conditions is the hydrogen activity cOnly under these conditions is the hydrogen activity constant so that theonstant so that the
electrode potential is constant and electrode potential is constant and reproducible.reproducible.
18-618-6 The potential in the presence of The potential in the presence of base would be more negative because the base would be more negative because the nickel ion activity in thisnickel ion activity in this
solution would be far less than 1 M. solution would be far less than 1 M. ConsequentConsequently the driving force for the reduction if Ni (II) to thely the driving force for the reduction if Ni (II) to the
metallic state woumetallic state would also be far less, and the electrode potential would also be far less, and the electrode potential would be significantly mld be significantly more negative. ore negative. (In(In
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 18
fact the standard electrode potential for the reaction 

  OH2)( Nie2)OH( Ni 2 s  has a value of – 0.72
V, whereas the standard electrode potential for )( Nie2 Ni2  s

   is – 0.250 V.)
18-7 (a)
  4223 SnFe2SnFe2
(b) )(Ag3Cr Ag3)(Cr  3  s s  
(c)
  2223 CuOH2)( NO2H4)(Cu NO2 g  s
(d) OH3H4SO5Mn2SOH5MnO2 2
2
4
2
324 

(e)
  H2)CN(FeTiOOH)CN(FeTi 46
2
2
3
6
3
(f)
  H2Ce2)(OCe2OH 32
4
22 g 
(g)
  24 Sn)(AgI2SnI2)(Ag2 s s
(h) OH2ZnUH4)(ZnUO 2
242
2 
  s
(i) OH3Mn2 NO5HMnO2HNO5 2
2
342 

(j) OH3ICl)( NCl2H2IO NNHH 222322 
  g 
18-8 (a) Oxidizing agent Fe3+;


  23 FeeFe
Reducing agent Sn2+;


  e2SnSn 42
(b) Oxidizing agent Ag+; )(AgeAg s

 
Reducing agent Cr;


  e3Cr )(Cr  3 s
(c) Oxidizing agent NO3
-; OH)( NOeH2 NO 223  
  g 
Reducing agent Cu;


  e2Cu)(Cu 2 s
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 18
(d) Oxidizing agent MnO4
-; OH4Mne5H8MnO 2
2
4 



Reducing agent H2SO3;


  e2H4SOOHSOH 24232
(e) Oxidizing agent Fe(CN)6
3-;


  46
3
6 )CN(Fee)CN(Fe
Reducing agent Ti3+; 

  eH2TiOOHTi 22
3
(f) Oxidizing agent Ce4+;


  34 CeeCe
Reducing agent H2O2;


  e2H2)(OOH 222 g 
(g) Oxidizing agent Sn4+;


  24 Sne2Sn
Reducing agent Ag;


  e)(AgII)(Ag s s
(h) Oxidizing agent UO2
2+; OH2Ue2H4UO 2
42
2 



Reducing agent Zn;


  e2Zn)(Zn 2 s
(i) Oxidizing agent MnO4
-; OH4Mne5H8MnO 2
2
4 



Reducing agent HNO2;


  e2H3 NOOHHNO 322
(j) Oxidizing agent IO3
-; OH3ICle4Cl2H6IO 223 



Reducing agent H2 NNH2;


  e4H4)( N NNHH 222 g 
18-9 (a) 
  H2)OH(V5MnOH11VO5MnO 4
2
2
2
4
(b)
  H2)(SI2)(SHI 22 s g 
(c) OHUO3Cr 2H2U3OCr  2
2
2
342
72 

Fundamentals of Analytical Chemistry: 8 th ed. Chapter 18
(d) OH2Mn)(ClH4)(MnOCl2 2
2
22 
  g  s
(e) OHI3I5H6IO 223 

(f) OH3ICl3Cl6H6I2IO 223 

(g) OH2MnO2POOH3MnO2HPO 2
2
4
3
44
2
3 

(h)
  HBr HCNSOOHBrOSCN 2423
(i) OH5VO3H2)OH(V2V 2
2
4
2  
(j) OH2)(MnO5OH4Mn3MnO2 22
2
4 
  s
18-10 (a) Oxidizing agent MnO4
-; OH4Mne5H8MnO 2
2
4 



Reducing agent VO2+;


  eH2)OH(VOH3VO 42
2
(b) Oxidizing agent I2;


 I2e2)(I2 ag 
Reducing agent H2S;


  e2H2)(S)(SH2 s g 
(c) Oxidizing agent Cr 2O7
2-; OH7Cr 2e6H14OCr  2
32
72 



Reducing agent U4+;


  e2H4UOOH2U 222
4
(d) Oxidizing agent MnO2; OH2Mne2H4)(MnO 2
2
2 


 s
Reducing agent Cl-;


  e2)(ClCl2 2 g 
(e) Oxidizing agent

3IO ; OH3I
2
1
e5H6IO 223  

Reducing agent I-;


  eI
2
1
I 2
(f) Oxidizing agent

3IO ; OH3ICle4Cl2H6IO 223 



Fundamentals of Analytical Chemistry: 8 th ed. Chapter 18
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2Zn
2


 
 

 
 




 
 

 
 

 
 

 
 



c K 
 E 
18-15
)(He2H2 2 g 
 
    

 
 


 
 




 
 


 
 


 2
H
22
H
Ho
]H[
00.1
log
2
0592.0
00.0
 p
log
2
0592.0
2
a
 E  E 
The ionic strength of the solution  is given by
       0100.010100.010100.0
2
1 22 
From Table 10-2
   
V121.0121.000.0
913.00100.0
00.1
log
2
0592.0
00.0
913.0
22
H

 
 

 
 


 
 E 
18-16 V73.0Cl4)(Pte2PtCl o
2
4 


  E  s
(a)
 
V78.0)051.0(73.0
0263.0
1492.0
log
2
0592.0
73.0
4
Pt 
 
 

 
 
 E 
(b)
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 18
V198.0)044.0(154.0
1050.7
1050.2
log
2
0592.0
154.0
2
3
Pt 
 
 

 
 


 

 E 
(c)
 
V355.0
1000.1
00.1
log
2
0592.0
000.0
26Pt


 
 


 
 



 E 
(d) V359.0OHVeH2VO o2
32  

  E 
 
   
V210.0149.0359.0
100.00353.0
20586.0
log
2
0592.0
359.0
2Pt

 
 

 
 


 E 
(e)


  4223 SnFe2SnFe2


  2
2
2
22 Snmmol30.2L00.25
SnClmmol
Snmmol1
mL
SnClmmol0918.0
consumedSnmmol











22
4
3
4
34
3
3
3
33
Snmmol340.0960.130.2remainingSnmmol
Snmmol960.1
Femmol2
Snmmol1
Femmol920.3formedSnmmol
Femmol920.3L00.25
FeClmmol
Femmol1
mL
FeClmmol1568.0
consumedFemmol
 
 
V177.0)023.0(154.0
0.50/960.1
0.50/340.0
log
2
0592.0
154.0Pt 
 
 

 
 
 E 
(f) OH2VO2V)OH(V 2
23
4 





  4
4
4 )OH(Vmmol08.2L00.25
mL
)OH(Vmmol0832.0
consumed)OH(Vmmol
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 18











44
2
3
2
32
3
342
3
3423
)OH(Vmmol993.0087.108.2remaining)OH(Vmmol
VOmmol174.2
Vmmol
VOmmol2
Vmmol087.1formedVOmmol
Vmmol087.1L00.50
)SO(Vmmol
Vmmol2
mL
)SO(Vmmol01087.0
consumedVmmol
 
  
V86.0139.000.1
1000.000.75/993.0
00.75/174.2
log0592.000.1
2Pt

 
 

 
 
 E 
18-17 (a)
V29.0068.036.0
00566.0
0813.0
log0592.036.0Pt 
 
 

 
  E 
(b)
 
V749.0022.0771.0
200845.0
0400.0
log0592.0771.0Pt 
 
 

 
 

 E 
(c) 63 1082.2]OH[55.5 pH
 
 
V329.0329.0000.0
1082.2
00.1
log
2
0592.0
000.0
26Pt


 
 


 
 



 E 
(d)
  
V894.0106.000.1
0800.01996.0
0789.0
log0592.000.1
2Pt

 
 

 
 E 
(e)





4
24
4
244
Cemmol04.3
L00.50
)SO(Cemmol
Cemmol1
mL
)SO(Cemmol0607.0
consumedCemmol








22
343
2
2
2
22
Femmol196.004.300.5remainingFemmol
Femmol04.3consumedCemmolformedFemmol
Femmol00.5L00.50
FeClmmol
Femmol1
mL
FeClmmol100.0
consumedFemmol
Fundamentals of Analytical Chemistry: 8 th ed. Chapter 18
V69.0)011.0(68.0
0.100/04.3
0.100/965.1
log0592.068.0Pt 
 
 

 
  E 
(f)
OH2VO2V)OH(V 2
23
4 





  4
4
4 )OH(Vmmol314.0L00.50
mL
)OH(Vmmol0628.0
consumed)OH(Vmmol












44
2
3
2
32
3
342
3
3423
)OH(Vmmol85.3314.016.4remaining)OH(Vmmol
VOmmol628.0
Vmmol
VOmmol2
Vmmol314.0formedVOmmol
Vmmol16.4
L00.25
)SO(Vmmol
Vmmol2
mL
)SO(Vmmol0832.0
consumedVmmol
  
V194.0165.0359.0
100.000.75/628.0
00.75/85.3
log0592.0359.0
2Pt

 
 

 
 
 E 
18-18 (a)
anodeV280.0030.0250.0
0943.0
00.1
log
2
0592.0
250.0 Ni 
 
 

 
  E 
(b)
    anodeV090.0)061.0(151.00922.0log0592.0151.0Ag  E 
(c)
  
  cathodeV003.1226.0229.1
1050.1760/780
00.1
log
4
0592.0
229.1
44O2


 
 


 
 



 E 
(d)
cathodeV171.0)017.0(154.0
350.0
0944.0
log
2
0592.0
154.0Pt 
 
 

 
  E 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 20
20-4 Amalgamation of the zinc prevents loss of reagent by reaction of the zinc with hydronium
ions.
20-5 OH)(AgCl2UCl2H4)(Ag2UO 2
42
2 



 s s
20-6 OH2ZnTi2H4)(ZnTiO2 2
232  


 s
20-7 Standard solutions of reductants find somewhat limited use because of their susceptibility
to air oxidation.
20-8 Standard KMnO4 solutions are seldom used to titrate solutions containing HCl because of
the tendency of MnO4
-
 to oxidize Cl
-
 to Cl2, thus causing overconsumption of MnO4
-
.
20-9 Cerium (IV) precipitates as a basic oxide in alkaline solution.
20-10 

 H4)(MnO5OH2Mn3MnO2 22
2
4 s
20-11 Freshly prepared solutions of permanganate are inevitably contaminated with small
amounts of solid manganese dioxide, which catalyzes the further decompositions of
 permanganate ion. By removing the dioxide at the outset, a much more stable standard
reagent is produced.
20-12 Standard permanganate and thiosulfate solutions are generally stored in the dark because
their decomposition reactions are catalyzed by light.
20-13 

 OH4)(O3)(MnO4OH2MnO4 2
 brown
224 g  s
Fundamentals of Analytical Chemistry: 8th ed. Chapter 20
20-14 Solutions of K 2Cr 2O7 are used extensively for back-titrating solutions of Fe
2+ when the
latter is being used as a standard reductant for the determination of oxidizing agents.
20-15 Iodine is not sufficiently soluble in water to produce a useful standard reagent. It is quite
soluble in solutions containing excess I- because of formation of triiodide.
20-16 The solution concentration of I3
-
 becomes stronger because of air oxidation of the excess
I-. The reaction is
OH2I2H4)(OI6 232 

 g 
20-17 )(SHSOHOS 3
2
32 s

20-18 When a measured volume of a standard solution of KIO3 is introduced into an acidic
solution containing an excess of iodide ion, a known amount of iodine is produced as a
consequence of a reaction.
OH3I3H6I5IO 223 

20-19




2
64
2
322
22
excess
3
OSI2OS2I
OH3I3Br H6I6BrO
20-20




2
64
2
322
22
3
excess
2
72
OSI2OS2I
OH7I3Cr 2H14I6OCr 
20-21   I4H4)( NH NI2 2422 g 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 20
20-22 Starch is decomposed in the presence of high concentrations of iodine to give products
that do not behave satisfactorily as indicators. This reaction is prevented by delaying the
addition of the starch until the iodine concentration is very small.
20-23


 2
2
Femmol03961.4
g847.55
Femmol1000
sampleg2256.0
(a) 


 4
2
42
CeM1142.0
Femmol
Cemmol1
mL37.35
Femmol03961.4
(b)




2
722
2
72
2
OCr M01904.0
Femmol6
OCr mmol1
mL37.35
Femmol03961.4
(c)



 42
4
2
MnOM02284.0
Femmol5
MnOmmol1
mL37.35
Femmol03961.4
(d)



 42
4
2
)OH(VM1142.0
Femmol
)OH(Vmmol1
mL37.35
Femmol03961.4
(e)



 32
3
2
IOM02855.0
Femmol4
IOmmol1
mL37.35
Femmol03961.4
20-24
722
722722 OCr K g677.3
mmol1000
OCr K g185.294
mL0.500
mL
OCr K mmol02500.0

Dissolve 3.677 g K 2Cr 2O7 in water and dilute to 500.0 mL.
20-25
3
33 KBrOg350.8
mmol1000
KBrOg001.167
L
mL1000
L000.2
mL
KBrOmmol02500.0

Dissolve 8.350 g KBrO3 in water and dilute to 2.000 L.
Fundamentals of Analytical Chemistry: 8th ed. Chapter 20
A B C D E F G H
1 20-54 (a) Titration of 25.00 mL of 0.025 M SnCl2 with 0.050 M FeCl3
2 Reaction: Sn
2+
 + 2Fe
3+
 Sn
4+
+ 2Fe
2+
3 For Fe3+/Fe2+, Eo  0.771
4 For Sn4+/Sn2+, Eo  0.154Equivalence point will be at 25.00 x 0.025 x 2/0.050=25.00 mL
5 Initial concentration Sn2+  0.025
6 Concentration Fe3+  0.05
7 Volume SnCl2 solution 25.00
8 Equivalence point volume 25.00
9 Percentages Vol. Fe3+, mL [Sn4+] [Sn2+] [Fe3+] [Fe2+] E, V
10 10 2.50 0.0023 0.0205 0.126
11 20 5.00 0.0042 0.0167 0.136
12 30 7.50 0.0058 0.0135 0.143
13 40 10.00 0.0071 0.0107 0.149
14 50 12.50 0.0083 0.0083 0.154
15 60 15.00 0.0094 0.0063 0.159
16 70 17.50 0.0103 0.0044 0.165
17 80 20.00 0.0111 0.0028 0.172
18 90 22.50 0.0118 0.0013 0.182
19 95 23.75 0.0122 0.000641 0.192
20 99 24.75 0.0124 0.000126 0.213
21 99.9 24.98 0.0125 0.000013 0.243
22 100 25.00 0.360
23 101 25.25 0.0002 0.0249 0.653
24 105 26.25 0.0012 0.0244 0.694
25 110 27.50 0.0024 0.0238 0.712
26 120 30.00 0.0045 0.0227 0.730
27 Spreadsheet Documentation
28 B10=(A10/11)*$B$8 G22=($B$3+2*$B$4)/3
29 C10=($B$6*B10/2)/($B$7+B10) E23=($B$6*B23-$B$5*$B$7*2)/($B$7+B23)
30 D10=($B$5*$B$7-$B$6*B10.2)/($B$7+B10) F23=($B$5*$B$7*2)/($B$7+B23)
31 G10=$B$4-(0.0592/2)*LOG10(D10/C10) G23=$B$3-0.0592*LOG10(F23/E23)
32
33
34
35
36
37
38
39
40
41
42
43
44
45
Fundamentals of Analytical Chemistry: 8th ed. Chapter 20
A B C D E F G H
1 20-54 (b) Titration of 25.00 mL of 0.08467 M Na2S2O3 with 0.10235 M I2 (I3
-)
2 Reaction: 2S2O3
2-
 + I3
-
 S4O6
2-
+ 3I
-
3 For S2O3
2-
/S4O6
2-
, E
o
  0.08
4 For I3
-
/I-, E
o
  0.536
Equivalence point will be at 25.00 x 0.08467 / 2 / 0.10235=10.34
mL
5 Initial concentration S2O3
2-
  0.08467
6 Concentration I3
-
  0.10235
7 Volume Na2S2O3 solution 25.00
8 Equivalence point volume 10.34
9 Percentages Vol. I3
-, mL [S4O6
2-] [S2O3
2-] [I3
-] [I-] E, V
10 10 1.03 0.0041 0.0732 0.076
11 20 2.07 0.0078 0.0626 0.089
12 30 3.10 0.0113 0.0527 0.098
13 40 4.14 0.0145 0.0436 0.106
14 50 5.17 0.0175 0.0351 0.114
15 60 6.20 0.0203 0.0271 0.123
16 70 7.24 0.0230 0.0197 0.132
17 80 8.27 0.0254 0.0127 0.145
18 90 9.31 0.0278 0.0062 0.165
19 95 9.82 0.0289 0.0030 0.183
20 99 10.24 0.0297 0.000605 0.225
21 99.9 10.33 0.0299 0.000064 0.283
22 100 10.34 0.384
23 101 10.44 0.000296 0.0896 0.525
24 105 10.86 0.0014736 0.0885 0.546
25 110 11.37 0.0029074 0.0873 0.555
26 120 12.41 0.00565611 0.0849 0.565
27 Spreadsheet Documentation
28 B10=(A10/11)*$B$8 G22=($B$3+2*$B$4)/3
29 C10=($B$6*B10/2)/($B$7+B10) E23=($B$6*B23-$B$5*$B$7*2)/($B$7+B23)
30 D10=($B$5*$B$7-$B$6*B10*2)/($B$7+B10) F23=($B$5*$B$7*2)/($B$7+B23)
31 G10=$B$3-(0.0592/2)*LOG10(D10^2/C10) G23=$B$3-0.0592*LOG10(F23/E23)
32
33
34
35
36
37
38
39
40
41
42
43
44
45
Fundamentals of Analytical Chemistry: 8th ed. Chapter 20
A B C D E F G H
1 20-54 (c) Titration of 0.1250 g Na2S2O3 with 0.01035 M KMnO4
2 Reaction: 2MnO4
-
 + 5H2C2O4 +6H
+
 2Mn
2+
+ 2CO2 +8H2O
3 For oxalate acid, Eo  -0.49
4 For MnO4
-
, E
o
1.51 There are 0.1250 g X 1000 mg/g/133.999 = 0.93284 mmol oxalate
5 Initial mmol oxalate 0.93284 acid initially present. Every 5 mmol of oxalate requires 2 mmol
6 Concentration MnO4
-
0.01035 MnO4
-
. Equivalence point will be at (0.93284 mmol X 2 / 5) / 0.01035
7 Initialvolume solution 25.00 mmol/mL=36.05 mL KMnO4
8 Equivalence point volume 36.05
9 Percentages Vol. MnO4
-, mL pCO2 [H2C2O4] [MnO4
-] [Mn2+] [H+] E, V
10 10 3.61 1.00 0.0294 1.00 -0.44
11 20 7.21 1.00 0.0232 1.00 -0.44
12 30 10.82 1.00 0.0182 1.00 -0.44
13 40 14.42 1.00 0.0142 1.00 -0.44
14 50 18.03 1.00 0.0108 1.00 -0.43
15 60 21.63 1.00 0.0080 1.00 -0.43
16 70 25.24 1.00 0.0056 1.00 -0.42
17 80 28.84 1.00 0.0035 1.00 -0.42
18 90 32.45 1.00 0.0016 1.00 -0.41
19 95 34.25 1.00 0.000788 1.00 -0.40
20 99 35.69 1.00 0.000154 1.00 -0.38
21 99.9 36.01 1.00 0.000016 1.00 -0.35
22 100 36.05 1.00 0.0061 1.00 0.94
23 101 36.41 6.05E-05 0.0061 1.00 1.49
24 105 37.85 2.97E-04 0.0059 1.00 1.49
25 110 39.66 5.77E-04 0.0058 1.00 1.50
26 120 43.26 0.0011 0.0055 1.00 1.50
27 Spreadsheet Documentation
28 B10=(A10/11)*$B$8 F22=$B$5*(2/5)/($B$8+B22)
29 D10=($B$5-$B$6*B10*5/2)/($B$7+B10) E23=($B$6*B23-$B$5*2/5)/($B$7+B23)
30 H10=$B$3-(0.0592/2)*LOG10(D10/(C10^2*G10^2)) H23=$B$4-(0.0592/5)*LOG10(F23/(E23*G23^8))
31 H22=((2*$B$3+5*$B$4)/7)-(0.0592/7)*LOG10(1/(C22*2*G22^10))
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Fundamentals of Analytical Chemistry: 8th ed. Chapter 20
A B C D E F G H
1 20-54 (d) Titration of 20.00 mL 0.1034 M Fe
2+
 with 0.01500 M K2Cr 2O7
2 Reaction: 6Fe
2+
 + Cr 2O7
2-
 +14H
+
 6Fe
3+
+ 2Cr 
3+
 +7H2O
3 For dichromate, Eo  1.33
4 For Fe3+/Fe2+, Eo 0.771 There are 20.00 mL X 0.1034mmol/mL = 2.068 mmol Fe 2+
5 Initial Fe2+ concentration 0.1034 initial present. Every mmol of dichromate requires 6 mmol Fe 2+.
6 Concentration Cr 2O7
2-
0.015 Equivalence point will be at (20.00 mL / 0.93284 mmol / 6) / 0.01500
7 Initial volume solution 20.00 mmol/mL=22.98 mL
8 Equivalence point volume 22.98
9 Percentages Vol. Cr 2O7
2-, mL [Fe3+] [Fe2+] [Cr 2O7
2-] [Cr 3+] [H+] E, V
10 10 2.30 0.0093 0.0835 1.00 0.715
11 20 4.60 0.0168 0.0673 1.00 0.735
12 30 6.89 0.0231 0.0538 1.00 0.749
13 40 9.19 0.0283 0.0425 1.00 0.761
14 50 11.49 0.0328 0.0328 1.00 0.771
15 60 13.79 0.0367 0.0245 1.00 0.781
16 70 16.09 0.0401 0.0172 1.00 0.793
17 80 18.38 0.0431 0.0108 1.00 0.807
18 90 20.68 0.0458 0.0051 1.00 0.828
19 95 21.83 0.0470 0.0025 1.00 0.847
20 99 22.75 0.0479 0.00047911 1.00 0.889
21 99.9 22.96 0.0481 4.349E-05 1.00 0.951
22 100 22.98 0.0160 1.00 1.26
23 101 23.21 8.05E-05 0.0160 1.00 1.33
24 105 24.13 3.91E-04 0.0156 1.00 1.33
25 110 25.28 7.62E-04 0.0152 1.00 1.34
26 120 27.58 0.0014 0.0145 1.00 1.34
27 Spreadsheet Documentation
28 B10=(A10/11)*$B$8 F22=($B$5*$B$7/3)/($B$7+B22)
29 C10=($B$6*B10*6)/($B$7+B10) H22=(($B$4+6*$B$3)/7)-(0.0592/7)*LOG10(2*F22/G22^14)
30 D10=($B$5*$B$7-$B$6*B10*6)/($B$7+B10) E23=($B$6*B23-($B$5*$B$7/6)/($B$7+B23)
31 H10=$B$4-0.0592*LOG10(D10/C10) H23=$B$3-(0.0592/6)*LOG10(F23^2/(E23*G23^14))
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Fundamentals of Analytical Chemistry: 8th ed. Chapter 20
A B C D E F G
1 20-54 (e) Titration of 35.00 mL 0.0578 M IO3
-
 with 0.05362 M Na2S2O3
2 Reactions: IO3
-
 + 5I
-
 + 6H
+
 3I2 + 3H2O; I2 + 2S2O3
2-
 2I
-
+ S4O6
2-
3 For thiosulfate, Eo  0.08
4 For I2/I-, E
o
  0.615There are 35.00 mL X 0.0578 mmol/mL = 2.023 mmol IO3
-
5 Initial IO3
-
concentration 0.0578initially present. Every mmol of IO3
-
 requires 6 mmol S2O3
2-
.
6 Concentration S2O3
2-
  0.05362Equivalence point will be at ( 2.023 mmol * 6) /
7 Initial volume solution 35.000.05362 mmol/mL=226.37 mL
8 Equivalence point volume 226.37
9 Percentages Vol. S2O3
2-, mL [I-] [I2] [S2O3
2-] [S4O6
2-] E, V
10 10 22.64 0.0211 0.0948 0.684
11 20 45.27 0.0302 0.0605 0.669
12 30 67.91 0.0354 0.0413 0.660
13 40 90.55 0.0387 0.0290 0.653
14 50 113.19 0.0410 0.0205 0.647
15 60 135.82 0.0426 0.0142 0.641
16 70 158.46 0.0439 0.0094 0.635
17 80 181.10 0.0449 0.0056 0.628
18 90 203.73 0.0458 0.0025 0.617
19 95 215.05 0.0461 0.0012 0.608
20 99 224.11 0.0464 2.34E-04 0.587
21 99.9 226.14 0.0464 2.33E-05 0.557
22 100 226.37 0.35
23 101 228.63 4.60E-04 0.0230 0.23
24 105 237.69 2.23E-03 0.0223 0.19
25 110 249.01 4.27E-03 0.0214 0.17
26 120 271.64 7.92E-03 0.0198 0.15
27 Spreadsheet Documentation
28 B10=(A10/100)*$B$8 E23=(($B$6*B23)-($B$5*$B$7*6)/($B$7+B23)
29 C10=($B$6*B10)/($B$7+B10) F23=(($B$5*$B$7*3)/($B$7+B23)
30 D10=(($B$5*$B$7*3)-($B$6*B10/2))/($B$7+B10) H22=($B$3+$B$4)/2
31 H10=$B$4-(0.0592/2)*LOG10(D10^2/C10) H23=$B$3-(0.0592/2)*LOG10(E23^2/F23)
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