SOLUCIONÁRIO LIVRO FUNDAMENTOS DA QUÍMICA ANALÍTICA SKOOG 8ª-EDIÇÃO - CAPÍTULO 4

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Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
Chapter 4 
4-1 (a) The millimole is an amount of a chemical species, such as an atom, an ion, a molecule 
or an electron. A millimole contains 
millimole
particles1002.6
millimole
mole10
mole
particles1002.6 20323 ×=∗× − 
(b) The molar mass is the mass in grams of one mole of a chemical species. 
(c) The millimolar mass is the mass in grams of one millimole of a chemical species. 
(d) Parts per million, cppm, is a term expressing the concentration of very dilute solutions. 
Thus, 
cppm ppm10
solution of mass
solute of mass 6×= 
The units of mass in the numerator and the denominator must be the same. 
4-2 The species molarity of a solution expresses the equilibrium concentration of a chemical 
species in terms of moles per liter. The analytical molarity of a solution gives the total 
number of moles of a solute in one liter. The species molarity takes into account 
chemical reactions that occur in solution. The analytical molarity specifies how the 
solution was prepared, but does not account for any subsequent reactions. 
4-3 33
33
m10
cm100
m
mL
cm1
L1
mL1000
 1L −=





××= 
3333 m10
mole1
m10
L
L
mole1M1
−−
=×= 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
4-4 (a) kHz 320
Hz1000
kHzHz102.3 5 =×× 
(b) ng 6.45
g
ng10g1056.4
9
8
=×× − 
(c) mmol 843
mol10
mmol
mol1043.8 3
5
=
µ
×µ× 
(d) Ms5.6
s10
Ms
s105.6 6
6
=×× 
(e) m 6.89
nm10
m
nm1096.8 3
4 µ=µ×× 
(f) kg 72
g1000
kgg000,72 =× 
4-5 +
+
++
×=
×
××× Na1098.5
Namol
Na1002.6
PONamol
Namol3
g94.163
PONamol1PONag43.5 22
23
43
43
43 
4-6 +
+
++
×=
×
×× K1022.1
Kmol
K1002.6
POKmol
Kmol3POKmol76.6 25
23
43
43 
4-7 (a) 32
32
32
32 OBmol0712.0OBg62.69
OBmolOBg96.4 =× 
(b) 
O10HOBNamol1073.8
381.37g
O10HOBNamol
mg1000
gOH10OBNamg333
2742
4
2742
2742
⋅×=
⋅
×ו
−
 
(c) 43
43
43
43 OMnmol0382.0OMng81.228
OMnmolOMng75.8 =× 
(d) 423
42
42
42 OCaCmol1031.1OCaCg128.10
OCaCmol
mg1000
gOCaCmg2.167 −×=×× 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
4-8 (a) 52
52
52
52 OPmmol40.0
mol
mmol1000
OPg94.141
OPmol
mg1000
gOPmg57 =××× 
(b) 2
2
2
2 COmmol6.293
mol
mmol1000
COg01.44
COmolCOg92.12 =×× 
(c) 3
3
3
3 NaHCOmmol476
mol
mmol1000
NaHCOg01.84
NaHCOmolNaHCOg0.40 =×× 
(d) 
44
44
44
44
POMgNHmmol2.6
mol
mmol1000
POMgNHg137.32
POMgNHmol
mg1000
gPOMgNHmg850
=
×××
 
4-9 (a) 
4
4
3
4
3
KMnOmmol50.6
L00.2
mol
mmol1000
L
KMnOmol1025.3M KMnO1025.3
=
××
×
≡×
−
−
 
(b) 
KSCNmmol6.41
mL750
mL1000
L
mol
mmol1000
L
KSCNmol0555.0
 M KSCN0555.0
=
×××≡
 
(c) 
4
3
4
44
4
CuSOmmol1047.8mL250
mL1000
L
mol
mmol1000
CuSOg61.159
CuSOmol
mg1000
g
L
CuSOmg41.5CuSO ppm 41.5
−×=××
×××≡
 
(d) KClmmol6.1165L50.3
mol
mmol1000
L
KClmol333.0
 M KCl333.0 =××≡ 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
4-10 (a) 
4
4
4
HClOmmol0.56
mL175
mL1000
L
mol
mmol1000
L
HClOmol320.0HClOM320.0
=
×××≡
 
(b) 
42
42
3
42
3
CrOKmmol121
L0.15
mol
mmol1000
L
CrOKmol1005.8CrOKM1005.8
=
××
×
≡×
−
−
 
(c) 
3
3
33
3
AgNOmmol199.0L00.5
mol
mmol1000
AgNOg87.169
AgNOmol
mg1000
g
L
AgNOmg75.6AgNOppm75.6
=
××××≡
 
(d) 
KOHmmol0.17
mL851
mL1000
L
mol
mmol1000
L
KOHmol0200.0KOHM0200.0
=
×××≡
 
4-11 (a) 34
3
3
3 HNOmg1090.4g
mg1000
HNOmol
HNOg01.63HNOmol777.0 ×=×× 
(b) MgOmg10015.2
g
mg1000
MgOmol
MgOg40.30
mmol1000
molMgOmmol500 4×=××× 
(c) 346
34
34
34 NONHmg1080.1g
mg1000
NONHmol
NONHg04.80NONHmol5.22 ×=×× 
(d) 
6324
6
6324
6324
6324
)NO(Ce)NH(mg1037.2
g
mg1000
)NO(Ce)NH(mol
)NO(Ce)NH(g23.548)NO(Ce)NH(mol32.4
×=
××
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
4-12 (a) KBrg840
KBrmol
KBrg0.119KBrmol1.7 =× 
(b) PbOg49.4
PbOmol
PbOg20.223
mmol1000
molPbOmmol1.20 =×× 
(c) 4
4
4
4 MgSOg452MgSOmol
MgSOg37.120MgSOmol76.3 =× 
(d) 
 
OH6)SO()NH(Feg8.3
OH6)SO()NH(Femol
OH6)SO()NH(Feg23.392
mmol1000
molOH6)SO()NH(Femmol6.9
22424
22424
22424
22424
⋅=
⋅
⋅
××⋅
 
4-13 (a) 
sucrosemg1022.2
mL0.26
g
mg1000
sucrosemol
sucroseg342
mL1000
L
L
sucrosemol250.0
sucroseM250.0
3×=
××××≡
 
(b) 
22
22
2222
3
22
3
OHmg8.472
L92.2
g
mg1000
OHmol
OHg02.34
L
OHmol1076.4OHM1076.4
=
×××
×
≡×
−
−
 
(c) 
23
23
23 )NO(Pbmg25.3mL656
mL1000
L
L
)NO(Pbmg96.4)NO(Pbppm96.4 =××≡ 
(d) 
3
33
3
KNOmg2.42
mL75.6
mL1000
L
g
mg1000
mol
KNOg10.101
L
KNOmol0619.0KNOM0619.0
=
××××≡
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
4-14 (a) 
22
22
2222
22
OHg51.2
mL450
OHmol
OHg34.02
mL1000
L
L
OHmol0.164OHM164.0
=
×××≡
 
(b) 
acidbenzoicg1088.2mL0.27
acidbenzoicmol
acidbenzoicg122
mL1000
L
L
acidbenzoicmol1075.8
acidbenzoicM1075.8
3
4
4
−
−
−
×=
×××
×
≡×
 
(c) 222 SnClg0760.0L50.3
mg1000
g
L
SnClmg7.21SnClppm7.21 =××≡ 
(d) 
3
3
33
3
KBrOg0453.0
mL7.21
KBrOmol
KBrOg167
mL1000
L
L
KBrOmol0125.0KBrOM0125.0
=
×××≡
 
4-15 (a) 
077.1)2(923.0
)10log()38.8log(
)M1038.8log()M0838.0log()M0503.0M0335.0log(pNa
2
2
=−−−=
−−=
×−=−=+−=
−
−
 
475.1)2(525.0
)10log()35.3log(
)M1035.3log()M0335.0log(pCl
2
2
=−−−=
−−=
×−=−=
−
−
 
298.1)2(702.0
)10log()03.5log(
)M1003.5log()M0503.0log(pOH
2
2
=−−−=
−−=
×−=−=
−
−
 
 
Fundamentals of Analytical Chemistry: 8th 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: 8th ed. Chapter 4 
 
92.5)6(79.0
)10log()2.1log(
)M102.1log()))M106.5(2())M108.4(2log((pNO
6
678
3
=−−−=
−−=
×−=××+××−=
−
−−−
 
pH = pOH = - log(1.0×10-7M) = 7.00 
4-18 (a) M1014.2)10(antilog)33.0(antilog]OH[ 103 −+ ×=−×= 
as in part (a), 
(b) M733.0]OH[ =− 
(c) M92.0]Br[ =− 
(d) M105.4]Ca[ 132 −+ ×= 
(e) [Li+] = 1.66 M 
(f) M107.1]NO[ 83 −− ×= 
(g) M99.0]Mn[ 2 =+ 
(h) [Cl-] = 0.0955 M 
4-19 (a) 
+
+
+
=××××× NaM0479.0
g99.22
Namol
L
mL1000
mL
g02.1
ppm10
1Nappm1008.1 6
3
 
−
−
−
−
×=××××
2
4
3
3
4
6
2
4 SOM1087.2g96.06
SOmol
L
mL1000
mL
g1.02
ppm10
1SOppm270 
(b) 
320.1)2(680.0
)10log()79.4log()M1079.4log(pNa 22
=−−−=
−−=×−= −−
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
 
543.2)3(458.0
)10log()87.2log()M1087.2log(pSO 334
=−−−=
−−=×−= −−
 
4-20 (a) 
+−
++
×=××× KM106.4
g10.39
Kmol
mg1000
g
L
mL1000
mL100
Kmg18 3
 
−
−−
=××× ClM103.0
g45.35
Clmol
mg1000
g
L
mL1000
mL100
Clmg365
 
(b) 
34.2)3(66.0
)10log()6.4log()M106.4log(pK 33
=−−−=
−−=×−= −−
 
pCl = –log(1.03×10-1M) = –log(1.03)–log(10-1) 
= –0.0133–(–1) = 0.987 
4-21 (a) 
OH6MgClKClM01037.0
g85.277
OH6MgClKClmol
L00.2
OH6MgClKClg76.5
22
2222
⋅⋅=
⋅⋅
×
⋅⋅
(b) 
+
+
=
⋅⋅
×⋅⋅ 2
22
2
22 MgM01037.0OH6MgClKClmol
MgmolOH6MgClKClM01037.0 
(c) −
−
=
⋅⋅
×⋅⋅ ClM0311.0
OH6MgClKClmol
Clmol3OH6MgClKClM01037.0
22
22 
(d) (w/v)%288.0%100
mL1000
L
L00.2
OH6MgClKClg76.5 22
=××
⋅⋅
 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
(e) 
−
−
−
=×××≡ Clmmol777.0mL25.0
mol
mmol1000
mL1000
L
L
Clmol0.0311ClM0311.0 
(f) 
+
+
++
≡=
××
⋅⋅
×⋅⋅
Kppm405
L
mg405
g
mg1000
Kmol
Kg10.39
OH6MgClKClmol1
Kmol1OH6MgClKClM01037.0
22
22
 
(g) 984.1)2(0170.0)10log()04.1log()M1004.1log(pMg 22 =−−−=−−=×−= −− 
(h) 507.1)2(494.0)10log()12.3log()M1012.3log(pCl 22 =−−−=−−=×−= −− 
4-22 (a) 
63
3636363 )CN(FeKM1074.4
g2.329
)CN(FeKmol
L775
)CN(FeKg1210
mL775
)CN(FeKmg1210
−×=×≡
 
(b) +
+
−
=×× KM0142.0)CN(FeKmol
Kmol3)CN(FeKM1074.4
63
63
3
 
(c) −−
−
− ×=××
3
6
3
63
3
6
63
3 )CN(FeM1074.4)CN(FeKmol
)CN(Femol)CN(FeKM1074.4 
(d) (w/v)%156.0%100
mg1000
g
mL775
)CN(FemgK1210 63
=×× 
(e) 
+
+
++−
+−
=
×××
×
≡×
Kmmol710.0
mL50.0
Kmol
mmolK1000
mL1000
L
L
Kmol101.42KM1042.1
2
2
 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
(f) 
−−
−
−
−
−
−
×≡=××
×
×
≡×
3
6
33
63
6
3
6
63
3
663
3
63
3
Fe(CN)ppm1000.1Fe(CN)
L
mg1000
g
mg1000
Fe(CN)mol
Fe(CN)g211.95
Fe(CN)Kmol
Fe(CN)mol
L
Fe(CN)Kmol104.74Fe(CN)KM1074.4
 
(g) 848.1)2(152.0)10log()42.1log()M1042.1log(pK 22 =−−−=−−=×−= −− 
(h) 324.2)3(676.0)10log()74.4log()M1074.4log()CN(pFe 3336 =−−−=−−=×−= −−− 
4-23 (a) 
33
3333
33
)NO(FeM281.0
g86.241
)NO(Femol
L
mL1000
mL
g059.1
100
1
solutiong
)NO(Feg42.6)NO(Fe%42.6
=
××××≡
 
(b) 
−
−
=×≡ 3
33
333
33 NOM844.0)NO(Femol
NOmol3
L
)NO(Femol281.0)NO(FeM281.0 
(c) 
L
)NO(Feg0.68
mol
)NO(Feg86.241
L
)NO(Femol281.0)NO(FeM281.0 33333333 =×≡ 
4-24 (a) 
2
22
2 NiClM11.1g61.129
NiClmol
L
mL1000
mL
g149.1
100
1
solutiong
NiClg5.12NiCl%5.12 =××××≡ 
(b) −
−
=×≡ ClM22.2
NiClmol
Clmol2
L
NiClmol11.1NiClM11.1
2
2
2 
(c) 2223 NiClg144L
mol
NiClg61.129
L
NiClmol11.1NiCl11.1 =××≡ 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
4-25 (a) 
OHHCg8.23solnmL500
100
1
solnmL
OHHCg75.4OHHC)v/w(%75.4 525252 =××≡ 
Weigh 23.8 g ethanol and add enough water to give a final volume of 500 mL. 
(b) 
waterg2.476OHHCg8.23solng500waterg
watergOHHCg23.8solng500
OHHCg8.23solng500
100
1
solng
OHHCg75.4OHHC)w/w(%75.4
52
52
52
52
52
=−=
+=
=××≡
x
x 
Mix 23.8 g ethanol with 476.2 g water. 
(c) 
OHHCmL8.23solnmL500
100
1
solnmL
OHHCmL75.4OHHC)v/v(%75.4 525252 =××≡ 
Dilute 23.8 mL ethanol with enough water to give a final volume of 500 mL. 
4-26 (a) 
383
383
383 OHCg525solnL50.2L
mL1000
100
1
solnmL
OHCg0.21OHC)v/w(%0.21 =×××≡
Weigh 525 g glycerol and add enough water to give a final volume of 2.50 L. 
(b) 
waterkg98.1OHCkg525.0solnkg50.2waterkg
waterkgOHCgk0.525solnkg50.2
OHCg525solnkg50.2
kg
g1000
100
1
solng
OHCg0.21OHC)w/w(%0.21
383
383
383
383
383
=−=
+=
=×××≡
x
x 
Mix 525 g glycerol with 1.98 kg water. 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
(c) 
383
383
383 OHCmL525solnL50.2L
mL1000
100
1
solnmL
OHCmL0.21OHC)v/v(%0.21 =×××≡
Dilute 525 mL glycerol with enough water to give a final volume of 2.50 L. 
4-27 
L300.0
POHmol0.15
LPOHmol50.4requiredPOH)w/w(%86volume
L
POHmol0.15
g0.98
POHmol
L
mL1000
mL
waterg
waterg
reagentg71.1
100
1
reagentg
POHg86POH)w/w(%86
POHmol50.4mL750
mL1000
L
L
POHmol00.6POHM00.6
43
4343
43
4343
43
43
43
43
=×=
=
×××××≡
=××≡
Dilute 300 mL of the concentrated reagent to 750 mL using water. 
4-28 
L170.0
HNOmol9.15
LHNOmol70.2requiredHNO%5.70volume
L
HNOmol9.15
g0.63
HNOmol
L
mL1000
mL
waterg
waterg
reagentg42.1
100
1
reagentg
HNOg5.70HNO)w/w(%5.70
HNOmol70.2mL900
mL1000
L
L
HNOmol00.3HNOM00.3
3
33
3
33
3
3
3
3
=×=
=
×××××≡
=××≡
Dilute 170 mL of the concentrated reagent to 900 mL using water. 
4-29 (a) 
3
33
3
AgNOg37.6
mL500
mL1000
L
mol
AgNOg87.169
L
AgNOmol0750.0AgNOM0750.0
=
×××≡
 
Dissolve 6.37 g AgNO3 in enough water to give a final volume of 500 mL. 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
(b) 
HClL0475.0
HClmol00.6LHClmol285.0
HClmol285.0L1
L
HClmol285.0HClM285.0
=×
=×≡
 
Take 47.5 mL of the 6.00 M HCl and dilute to 1.00 L using water. 
(c) 
64
64642
2
)CN(FeKg98.2
mol
)CN(FeKg43.368
Kmol4
)CN(FeKmolKmol1024.3
Kmol1024.3mL400
mL1000
L
L
Kmol0810.0KM0810.0
=×××
×=××≡
+
+−
+−
+
+
 
Dissolve 2.98 g K4Fe(CN)6 in enough water to give a final volume of 400 mL. 
(d) 
L216.0
BaClmol400.0
L
g23.208
BaClmolBaClg0.18
BaClg0.18mL600
100
1
solnmL
BaClg00.3BaCl)v/w(%00.3
2
2
2
2
2
2
=××
=××≡
 
Take 216 mL of the 0.400 M BaCl2 solution and dilute to 600 mL using water. 
(e) 
L0203.0
HClOmol8.11
LHClOmol240.0requiredHClO)w/w(%71volume
L
HClOmol8.11
g46.100
HClOmol
L
mL1000
mL
waterg
waterg
reagentg67.1
100
1
reagentg
HClOg71HClO)w/w(%71
HClOmol240.0L00.2
L
HClOmol120.0HClOM120.0
4
44
4
44
4
4
4
4
=×=
=
×××××≡
=×≡
Take 20.3 mL of the concentrated reagent and dilute to a final volume of 2.00 L using 
water. 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
 
(f) 
42
4242 SONag67.1
mol
SONag04.142
Namol2
SONamolNamol02348.0
Namol02348.0
g99.22
Namol
mg1000
gNamg540
Namg540L00.9
solnL
Namg60Nappm60
=××
=××
=×≡
+
+
+
+
+
+
+
+
 
Dissolve 1.67 g Na2SO4 in enough water to give a final volume of 9.00 L. 
4-30 (a) 
4
44
4 KMnOg5.39
mol
KMnOg03.158L00.5
L
KMnOmol0500.0KMnOM0500.0 =××≡
Dissolve 39.5 g KMnO4 in enough water to give a final volume of 5.00 L. 
(b) 
reagentL125.0
reagentmol00.8
LHClOmol00.1
HClOmol00.1L00.4
L
HClOmol250.0HClOM250.0
4
4
4
4
=×
=×≡
 
Take 125 mL of the 8.00 M reagent and dilute a final of volume of 4.00 L using water. 
(c) 
2
222 MgIg39.1
mol
MgIg11.278
Imol2
MgImolImol1000.1
Imol0100.0mL400
mL1000
L
L
Imol0250.0IM0250.0
=×××
=××≡
−
−−
−
−
−
 
Dissolve 1.39 g MgI2 in enough water to give a final volume of 400 mL. 
(d) 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
L0343.0
CuSOmol365.0
L
g60.159
CuSOmolCuSOg00.2
CuSOg00.2mL200
100
1
solnmL
CuSOg00.1CuSO)v/w(%00.1
4
4
4
4
4
4
=××
=××≡
 
Take 34.3 mL of the 0.365 M CuSO4 and dilute to a final volume of 200 mL using water. 
(e) 
L0169.0
NaOHmol062.19
LNaOHmol3225.0requiredNaOH)w/w(%50volume
L
NaOHmol062.19
g00.40
NaOHmol
L
mL1000
mL
waterg
waterg
reagentg525.1
100
1
reagentg
NaOHg50NaOH)w/w(%50
NaOHmol3225.0L50.1
L
NaOHmol215.0NaOHM215.0
=×=
=
×××××≡
=×≡
Take 16.9 mL of the concentrated reagent and dilute to a final volume of 1.50 L using 
water. 
(f) 
64
64644
41
1
)CN(FeKg0424.0
mol
)CN(FeKg35.368
Kmol4
)CN(FeKmolKmol1060.4
Kmol1060.4
g10.39
Kmol
mg1000
gKmg108.1
Kmg108.1L50.1
solnL
Kmg12Kppm0.12
=×××
×=×××
×=×≡
+
+−
+−
+
+
+
+
+
 
Dissolve 42.4 mg K4Fe(CN)6 in enough water to give a final volume of 1.50 L. 
4-31 
−
−
−
−
+−
+
+
×=××≡
×=××≡
3
23
3
32
3
3
IOmol1027.2mL0.75
mL1000
L
L
IOmol302.0IOM302.0
Lamol1025.1mL0.50
mL1000
L
L
Lamol250.0LaM250.0
 
Because each mole of La(IO3)3 requires three moles IO3-, IO3- becomes the limiting 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
reagent. Thus, 
formed)IO(Lag01.5
mol
)IO(Lag6.663
IOmol3
)IO(LamolIOmol1027.2 3333
3
33
3
2
=×××
−
−
−
 
4-32 
−−
−
−
+−
+
+
×=××≡
×=××≡
Clmol1000.7mL400
mL1000
L
L
Clmol175.0ClM175.0
Pbmol1050.2mL200
mL1000
L
L
Pbmol125.0PbM125.0
2
22
2
2
 
Because each mole of PbCl2 requires two moles Cl-, Pb2+ becomes the limiting reagent. 
Thus, 
formedPbClg95.6
mol
PbClg10.278
Pbmol
PbClmolPbmol1050.2 222
222
=×××
+
+−
 
4-33 A balanced chemical equation can be written as shown below. 
)(COOHNaCl2HCl2CONa 2232 g++→+ 
(a) 
HClmol1031.7mL0.100
mL1000
L
L
HClmol0731.0HClM0731.0
CONamol10094.2
g99.105
CONamolCONag2220.0
3
32
332
32
−
−
×=××≡
×=×
 
Because one mole of CO2 is evolved for every mole Na2CO3 reacted, Na2CO3 is the 
limiting reagent. Thus, 
evolvedCOg09218.0
mol
COg01.44
CONamol
COmolCONamol10094.2 22
32
2
32
3
=××× − 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
(b) 
HClM0312.0
L
mL1000
mL0.100
HClmol1012.3
mol1012.3))mol10094.2(2(mol1031.7leftHClmol
3
333
=×
×
×=××−×=
−
−−−
 
4-34 A balanced chemical equation can be written as shown below. 
433343 POHgNaNO3HgNO3PONa +→+ 
(a) 
3
3
3
42
343
43
HgNOmol05151.0mL0.100
mL1000
L
L
HgNOmol5151.0HgNOM5151.0
PONamol1039.9mL0.25
mL1000
L
L
PONamol3757.0PONaM3757.0
=××≡
×=××≡ −
The limiting reagent is Na2PO4. Thus, 
formedPOHgg54.6
mol
POHgg74.696
PONamol
POHgmolPONamol1039.9 4343
43
43
42
3
=××× − 
(b) 
3
3
3
32
3
HgNOM187.0
L
mL1000
mL0.125
HgNOmol0233.0
HgNOmol0233.0))mol1039.9(3(mol10151.5unreactedHgNOmol
=×
=××−×= −−
 
4-35 A balanced chemical equation can be written as shown below. 
)(SOOHNaClO2HClO2SONa 224432 g++→+ 
(a)
32
32
32 SONa mol 23490.0mL 00.75
mL 1000
L
L
SONa mol 3132.0SONa M 3132.0 =××≡
4
4
4 HClOmol 06038.0mL 0.150
mL 1000
L
L
 HClOmol 4025.0
 M HClO4025.0 =××≡ 
Because one mole SO2 is evolved per mole Na2SO3, Na2SO3 is the limiting reagent. 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 4 
 
Thus, 
evolved SO g 505.1
moL
SO g 64.06
SONa mol
SO molSONa mol 02349.0 22
32
2
32 =×× 
(b) 
4
4
44
 M HClO0595.0
L
mL 1000
mL 225.0
 HClOmol 0.01340
HClOmol 13400.0mol)) (0.023492-mol (0.06038unreacted HClOmol
=×
=×=
 
4-36 A balanced chemical equation can be written as shown below. 
−++ ++→++ Cl2Na2)(POMgNHNHPONaMgCl 444422 s 
2
22
2
MgClmol02101.0
g21.95
MgClmol
mL0.200
100
1
mL
MgClg000.1MgCl)v/w(%000.1
=
×××≡
 
42
342
42 PONamol1001.7mL0.40
mL1000
L
L
PONamol1753.0PONaM1753.0 −×=××≡
The Na2PO4 is the limiting reagent. Thus, 
2
2
2
3
2
44
443
44
MgClM0583.0
L
mL1000
mL240.0
MgClmol0.0140
MgClmol0140.0)1001.702101.0(unreactedMgClmol
PO MgNHg 9628.0
mol
PO MgNHg 137.351001.7ppt PO MgNHmass
=×
=×−=
=××=
−
−
 
4-37 A balanced chemical equation can be written as shown below. 
−+ ++→+ 33 NOK)(AgIKIAgNO s 
3
3
3
3
AgNOmL2930
L
mL1000
AgNOmol0100.0
L
KImol
AgNOmolKImol02930.0
KImol02930.0
g0.166
KImol
mL0.200
mL
g
ppt10
1KIppt31.24
=×××
=××××
 
2930 mL of 0.0100 M AgNO3 would be required to precipitate all I- as AgI. 
Fundamentals of Analytical Chemistry: 8th 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 44
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(31(1018.6(unreacted)SO(Almol
−
−
−−−
×=×
×
×=××−×=