Solution Cap 13 e 14 Fundamentals of Analytical Chemistry: 8th ed. Skoog

Prévia do material em texto

Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
Chapter 13 
13-1 amount A (mmol) = )mL/Ammol()mL(volume Ac× 
amount A (mole) = )L/Amol()L(volume Ac× 
13-2 (a) The millimole is the amount of an elementary species, such as an atom, an ion, a 
molecule, or an electron. A millimole contains 
mmol
particles1002.6
mmol1000
mole
mole
particles1002.6 2023 ×=×× 
(b) A titration involves measuring the quantity of a reagent of known concentration 
required to react with a measured quantity of sample of an unknown concentration. The 
concentration of the sample is then determined from the quantities of reagent and sample, 
the concentration of the reagent, and the stoichiometry of the reaction. 
(c) The stoichiometric ratio is the molar ratio of two chemical species that appear in a 
balanced chemical equation. 
(d) Titration error is the error encountered in titrimetry that arises from the difference 
between the amount of reagent required to give a detectable end point and the theoretical 
amount for reaching the equivalence point. 
13-3 (a) The equivalence point in a titration is that point at which sufficient titrant has been 
added so that stoichiometrically equivalent amounts of analyte and titrant are present. 
The end point in a titration is the point at which an observable physical change signals the 
equivalence point. 
(b) A primary standard is a highly purified substance that serves as the basis for a 
titrimetric method. It is used either (i) to prepare a standard solution directly by mass or 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
(ii) to standardize a solution to be used in a titration. 
A secondary standard is material or solution whose concentration is determined from the 
stoichiometry of its reaction with a primary standard material. Secondary standards are 
employed when a reagent is not available in primary standard quality. For example, solid 
sodium hydroxide is hygroscopic and cannot be used to prepare a standard solution 
directly. A secondary standard solution of the reagent is readily prepared, however, by 
standardizing a solution of sodium hydroxide against a primary standard reagent such as 
potassium hydrogen phthalate. 
13-4 The Fajans method is a direct titration of the chloride ion, while the Volhard approach 
requires two standard solutions and a filtration step to eliminate AgCl. The Fajans 
method uses a fluorescein dye. At the end point, the fluoresceinate anions are absorbed 
into the counter ion layer that surrounds the colloidal silver particles giving the solid an 
intense red color. In the Volhard method, the silver chloride is more soluble that silver 
thiocyanide such that the reaction ( ) −
←
→− ++ Cl)(AgSCNSCNAgCl ss occurs to a 
significant extent as the end point is approached. The released Cl- ions cause the end 
point color change to fade resulting in an over consumption of SCN- and a low value for 
the chloride analysis. 
13-5 (a) 
2
22
Imoles2
NNHHmole1 
(b) −
4
22
MnOmoles2
OHmoles5 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
(c) +
⋅
Hmoles2
OH10OBNamole1 2742 
(d) 
3KIOmoles3
Smoles2 
13-6 In contrast to Ag2CO3 and AgCN, the solubility of AgI is unaffected by the acidity. In 
addition, AgI is less soluble than AgSCN. The filtration step is thus unnecessary in the 
determination of iodide, whereas it is needed in the determination of carbonate or cyanide. 
13-7 The ions that are preferentially absorbed on the surface of an ionic solid are generally 
lattice ions. Thus, in a titration, one of the lattice ions is in excess and its charge 
determines the sign of the charge of the particles. After the equivalence point, the ion of 
the opposite charge is present in excess and determines the sign of the charge on the 
particle. Thus, in the equivalence-point region, the charge shift from positive to negative, 
or the reverse. 
13-8 (a) 
3
3
3
3
AgNOg37.6
mole
AgNOg87.169mole0375.0
mole0375.0mL500
mL1000
L
L
AgNOmole0750.0AgNOM0750.0
=×
=××≡
 
Dissolve 6.37 g AgNO3 in water and bring to 500 mL total volume. 
(b) 
reagentL108.0
reagentmole00.6
Lmole650.0
mole650.0L00.2
L
HClmole325.0HClM325.0
=×
=×≡
 
Dilute 108 mL of the 6.00 M HCl reagent in enough water to yield 2.00 L total volume. 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
(c) 
64
6464 )CN(FeKg22.6
mole
)CN(FeKg35.368
Kmoles4
)CN(FeKmoleKmole0675.0
Kmole0675.0mL750
mL1000
L
L
Kmole0900.0KM0900.0
=××
=××≡
+
+
+
+
+
 
Dissolve 6.22 g K4Fe(CN)6 in water and bring to 750 mL total volume. 
(d) 
2
2
2
2
22
2
BaClL115.0
BaClmole500.0
LBaClmole0576.0
BaClmole0576.0mL600
g23.208
BaClmole
solutionmL100
BaClg00.2BaCl)v/w(%00.2
=×
=××≡
 
Dilute 115 mL of 0.500M BaCl2 in enough water to yield 600 mL total volume. 
(e) 
reagentL025.0
HClOmole55.9
reagentLHClOmole240.0reagent.vol
reagentL
HClOmole55.9
g5.100
HClOmole
reagentg100
HClOg60
reagentL
reagentg1060.1
HClOmole240.0L00.2
L
HClOmole120.0HClOM120.0
4
4
444
3
4
4
4
=×=
=××
×
=×≡
 
Dilute 25 mL HClO4 reagent in enough water to yield 2.00 L total volume. 
(f) 
42
42422
2
SONag67.1
mole
SONag0.142
Namoles2
SONamole
g99.22
Namole
mg1000
gNamg104.5
Namg1040.5solnL00.9
solnL
Namg60Nappm0.60
=×××××
×=×≡
+
+
+
+
+
+
 
Dissolve 1.67 g Na2SO4 in water and bring to 9.00 L total volume. 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
13-9 (a) 
4
4
4
4
4
4
KMnOg7.23
mole
KMnOg03.158KMnOmole150.0
KMnOmole150.0L00.1
L
KMnOmole150.0KMnOM150.0
=×
=×≡
 
Dissolve 23.7 g KMnO4 in water and bring to 1.00 L total volume. 
(b) 
reagentHClOL139.0
HClOmole00.9
LHClOmole25.1
HClOmole25.1L50.2
L
HClOmole500.0HClOM500.0
4
4
4
4
4
4
=×
=×≡
 
Dilute 139 mL HClO4 reagent in enough water to yield 2.50 L total volume. 
(c) 
2
22 MgIg78.2
mole
MgIg11.278
Imoles2
MgImoleImole0200.0
Imole0200.0mL400
mL1000
L
L
Imole0500.0IM0500.0
=××
=××≡
−
−
−
−
−
 
Dissolve 2.78 g MgI2 in water and bring to 400 mL total volume. 
(d) 
4
4
4
4
44
4
CuSOL0575.0
CuSOmole218.0
LCuSOmole0125.0
CuSOmole0125.0mL200
g61.159
CuSOmole
mL100
CuSOg00.1CuSO)v/w(%00.1
=×
=××≡
 
Dilute 57.5 mL of the 0.218 M CuSO4 solution to yield 200 mL total volume. 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
(e) 
reagentL0169.0
NaOHmole10906.1
reagentLNaOHmole3225.0reagent.vol
reagentL
NaOHmole10906.1
g00.40
NaOHmole
reagentg100
NaOHg50
reagentL
reagentg10525.1
NaOHmole3225.0L50.1
L
NaOHmole215.0NaOHM215.0
1
13
=
×
×=
×
=××
×
=×≡
 
Dilute 16.9 mL of the concentrated reagent to 1.50 L total volume. 
(f) 
64
64641
1
)CN(FeKg0424.0
mole
)CN(FeKg3.368
Kmoles4
)CN(FeKmole
g10.39
Kmole
mg1000
gKmg108.1
Kmg108.1solnL50.1
solnL
Kmg12Kppm12
=
×××××
×=×≡
+
+
+
+
+
+
 
Dissolve 42.4 mg K4Fe(CN)6 in water and bring to 1.50 L total volume. 
13-10 
mole
g59.216HgO =M 
4
44
2
42
HClOM08190.0
mL51.46
mole
HClOmmol1000
OHmole1
HClOmole1
HgOmole
OHmole2
g59.216
HgOmole1HgOg4125.0
OH2HgBrOHBr4)(HgO
=
××××
+++
−
−
−−
←
→−s
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
13-11 
mole
g99.105
32CONa
=M 
42
4242
32
32
32
22
2
3
SOHM1168.0
mL44.36
mole
SOHmmol1000
Hmole2
SOHmole1
CONamole
Hmole2
g99.105
CONamole1CONag4512.0
)(COOHH2CO
=
××××
++
+
+
←
→+− g
 
13-12 
mole
g04.142
42SONa
=M 
2
42
24242
4
2
4
2
BaClM06581.0
mL25.41
mole
mmol1000
SONamole1
BaClmole1
g04.142
SONamole1
sampleg100
SONag4.96sampleg4000.0
)(BaSOSOBa
=
××××
→+ −+ s
 
13-13 (Note: In the first printing of the text, the answer in the back of the book was in error.) 
NaOHmL
HClOmL0972.1
NaOHmL00.25
HClOmL43.27
V
V
44
NaOH
HClO4 == 
The volume of HClO4 required to titrate 0.3125 g Na2CO3 is 
NaOHM2239.0
HClOmole
NaOHmole1
NaOHmL
HClOmL0972.1
L
HClOmole2041.0M2041.0
V
V
cc
and
HClOM2041.0
mole
mmol1000
CONamole1
HClOmole2
g99.105
CONamole1
HClOmL896.28CONag3125.0
,Thus
HClOmL896.28
NaOHmL
HClOmL0972.1NaOHmL12.10HClOmL00.40
4
44
NaOH
HClO
HClONaOH
4
32
432
4
32
4
4
4
4
4
=××≡=
×=
=×××
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×−
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
13-14 OH8)(CO10Mn2H6OCH5MnO2 22
2
4224 ++++
+
←
→+− g 
4
422
4422
422
KMnOM02858.0
mL75.36
mole
mmol1000
OCNamole5
KMnOmole2
mL1000
L
L
OCNamole05251.0OCNamL00.50
=
××××
 
13-15 
mole
g00.214
3KIO
=M 
−−−
+−−
+→+
+
←
→
++
2
64
2
322
223
OSI2OS2I
OH3I3H6I5IO
 
322
2
322
3
23
3
OSNaM09537.0
mL72.30
mole
mmol1000
Imole1
OSNamole2
KIOmole1
Imole3
g00.214
KIOmole1KIOg1045.0
=
××××
 
13-16 
)(AgSCNNHSCNNHAg
,SCNNHwithtitratedisAgunreactedThe
)(AgClHCOOHHOCHOHAgCOOHClCH
44
4
222
s
s
+→+
++→++
++
+
++
 
SCNNHM098368.0
mL98.22
mole
mmol1000
AgNOmole1
SCNNHmole1
mL1000
L
L
AgNOmole04521.0mL00.50
4
3
43
=
××××
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
 
COOHClCHmg7.116
g
mg1000
mole
g50.94
AgClmole1
COOHClCHmole1AgClmole102345.1
AgClmmol2345.1
mmol02598.1mL00.50
mL
mmol04521.0edprecipitat)s(AgClmmol
SCNNHmmol02598.1mL43.10
mL
SCNNHmmol098368.0SCNNHmmol
2
23
4
4
4
=
××××
=
−⎟
⎠
⎞
⎜
⎝
⎛ ×=
=×=
−
 
13-17 
)(AgSCNSCNAg
OH5)(Ag8BOHOH8Ag8BH 2324
s
s
→+
++→++
−+
−−+−
 
mmol excess Ag+ equals mmol KSCN, 
( )
%5.11%100
materialg213.3
KBHg371.0KBHpurity%
KBHg371.0
mole
KBHg941.53
BHmole1
KBHmole1mL500
mL1000
L
L
BHmole0138.0
BHM0138.0
Agmmol8
BHmmol1
mL100
Agmmol1010.1
Agmmol1010.1mmol133.01011.1Agmmolreacted
AgNOmmol1011.1mL00.50
mL
AgNOmmol2221.0AgNOmmol
Agmmol133.0
KSCNmmol1
Agmmol1mL36.3
mL
KSCNmmol0397.0Agexcessmmol
4
4
4
4
4
44
4
4
1
11
3
13
3
=×=
=××××
=×
×
×=−×=
×=×=
=××=
−
−
−
+
−+
++
+
+
+
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
13-18 )(AsOAgH3Ag3AsOH 4343 s+→+
++ 
3
3
3 AgNOmmol4888.2mL00.40mL
AgNOmmol06222.0addedAgNOmmol
Agmmol0760.1mL76.10
KSCNmmol1
Agmmol1
mL
KSCNmmol1000.0Agexcessmmol
,KSCNmmolequalsAgexcessmmol
=×=
=××= +
+
+
+
 
32
32
43
3243
32
OAs%612.4
100
sampleg010.1
mmol1000
OAsg84.197
AsOAgmmol2
OAsmmol1
Agmmol3
AsOAgmmol1Agmmol4128.1
sampleinOAs%
Agmmol4128.1mmol)0760.14888.2(reactedAgmmol
=
×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×××
=
=−=
+
+
++
 
13-19 
mole
g32.373
7510 ClHC
=M 
The stoichiometry of the titration of heptachlor must be one to one (i.e., one chlorine 
reacts with one silver nitrate) for the calculation, 
( )
samplemass
33.37mLmL
heptachlor% SCNSCNAgAg
××−×
=
cc
, to be true. The factor 37.33 (with 
unwritten units of 
mmol
g ) found in the numerator is derived from the equation below, 
100
mmol1000
ClHCg32.373
AgNOmmol.no
ClHCmmol.no
mmol
g33.37 7510
3
7510 ××= 
Thus, 
00.1
100ClHCg32.373
mmol1000
mmol
g33.37
AgNOmmol.no
ClHCmmol.no
75103
7510 =
×
×
= 
confirming that only one of the chlorines in the heptachlor reacts with the AgNO3. 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
13-20 +−+ +→+ H2)(BiPOPOHBi 442
3 s 
eulytite%90.39
%100
sampleg6423.0
mmol1000
SiO3OBi2g1112
Bimmol4
SiO3OBi2mmol1Bimol921758.0
eulytitepurity%
Bimol921758.0
PONaHmmol1
Bimmol1PONaHmol921758.0Bimol
PONaHmol921758.0mL36.27
mL
PONaHmmol03369.0PONaHmol
232
3
2323
3
42
3
42
3
42
42
42
=
×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛ ⋅
×
⋅
×
=
=×=
=×=
+
+
+
+
+
 
13-21 (a) 
2
56
256
56
2
)OH(BaM01190.0
mL42.40
mole
mmol1000
COOHHCmole2
)OH(Bamole1
g12.122
COOHHCmole1COOHHCg1175.0
)OH(Baofmolarity
=
×××
=
 
(b) 
M102.2
42.40
03.0
1175.0
0002.0)M10190.1(s 5
22
2
y
−− ×=⎟
⎠
⎞
⎜
⎝
⎛ ±+⎟
⎠
⎞
⎜
⎝
⎛ ±××= 
molarity of Ba(OH)2 taking into account the uncertainty of the two measurements can be 
written 0.01190(±0.00002) M. 
(c) Weighing error of -0.0003g results in an absolute error, E, in the molarity calculation, 
( )
( ) ( ) M100.3orM10826.2M10190.1M10187.1M10190.1
mL42.40
mole
mmol1000
COOHHCmole2
)OH(Bamole1
g12.122
COOHHCmole1COOHHCg0003.01175.0
E
55222
56
256
56
−−−−− ×−×−=×−×=×
−
⎟
⎟
⎟
⎟
⎠
⎞
⎜
⎜
⎜
⎜
⎝
⎛ ×××−
=
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
The relative error, Er, in the molarity calculation resulting from this weighing error is 
( ) ppt3or100.3
M10190.1
M100.3E 32
5
r −×−=×
×−
= −−
−
 
13-22 
HOAc%529.1
%100
mL00.50
mmol1000
HOAcg05.60
)OH(Bammol1
HOAcmmol2mL17.43
mL
)OH(Bammol1475.0
HOAcpercentagev/w
2
2
=
×
×××
=
 
Similar calculations for samples 2 to 4 yield the results shown in the spreadsheet that 
follows, 
(a) 
HOAc%528.1
4
1134.6
4
x
HOAcpercentagev/wx i === ∑ 
(b) 
HOAc%1071.5
3
4
)1134.6(34351132.9
3
4
)x(
x
s 3
22
i2
i
−×=
−
=
−
=
∑ ∑
 
(c) 
HOAc)%007.0(528.1
2
)1063.5(35.2528.1
4
tsxCI
3
%90 ±=
××
±=±=
−
 
(d) The values of 1.5352 and 1.5213 can be considered for rejection. Applying the Q 
test we find, that both results are less than Qexpt = 0.765, so neither value should be 
rejected. 
(e) 
V
V
HOAc)%v/w(
HOAc)%v/w( ∆
=
∆ 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
001.0
mL00.50
mL05.0
HOAcV
HOAcV,1sampleFor −=−=∆ 
The results for the remaining samples are found in the following spreadsheet. 
00125.0
4
005.0
n
x
errorsystematicrelativemean −=−== ∑ 
HOAc%102or%1091.1528.100125.0
HOAc)%v/w(,HOAcpercent)v/w(meantheFor
33 −− ×−×−=×−
=∆
 
 A B C D E F G 
1 Problem 13-22 
2 
3 Conc. Ba(OH)2 0.1475 
4 MW HOAc 60.05 
5 t 2.35 
6 
7 Sample Sample Vol, mL Ba(OH)2 Vol, mL w/v % HOAc xi xi2 ∆V/V 
8 1 50.00 43.17 1.529 1.52949152 2.33934429 -0.001
9 2 49.50 42.68 1.527 1.52740511 2.33296637 -0.001
10 3 25.00 21.47 1.521 1.52134273 2.31448370 -0.002
11 4 50.00 43.33 1.535 1.53516024 2.35671695 -0.001
12 
13 Σ(xi) 6.11339959 
14 Σ(xi2) 9.34351132 
15 (a) mean xi 1.528 
16 (b) std. dev. % HOAc 5.71E-03 
17 (c) CI90%(t=2.35) 6.70E-03 
18 (d) Q(expt 1.535-1.521) 0.41 
19 Q(expt 1.527-1.521) 0.44 
20 (e) Σ(∆V/V) -0.005 
21 mean relative systematic error -1.25E-03 
22 mean (w/v) % HOAc -1.91E-03 
23 Spreadsheet Documentation 
24 D8 = (($B$3*C8*2*$B$4/1000)/B8)*100 C16 = SQRT((B14-(B13)^2/4)/3) 
25 E8 = D8 C18 = (D11-D8)/(D11-D10) 
26 F8 = E8^2 C19 = (D9-D10)/(D11-D10) 
27 G8 = -0.05/B8 C20 = SUM(G8:G11) 
28 B13 = SUM(E8:E11) C21 = C20/4 
29 B14 = SUM(F8:F11) C22 = C21*C15 
30 C15 = B13/4 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
13-23 
3
3
3
3
AgNOmmol5204.1mL81.2
KSCNmmol1
AgNOmmol1
mL
KSCNmmol04124.0
mL00.20
mL
AgNOmmol08181.0samplebyconsumedAgNOmmol.no
=⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
××
−⎟
⎠
⎞
⎜
⎝
⎛ ×=
 
tablet
saccharinmg60.15
tablets20
g
mg1000
mmol1000
saccharing17.205
AgNOmmol1
saccharinmmol1AgNOmmol5204.1
tablet/saccharinmg
3
3
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×××
=
 
13-24 (a) 
3
3
3
3
100533.2
mL3.502
mole
mmol1000
AgNOmole1
Agmole1
g87.169
AgNOmole1AgNOg1752.0
Agmolarityweight
−
+
+
×=
×××
=
 
(b) 
3
3
3
109386.1
mL171.25
mole
mmol1000mL765.23
mL1000
AgNOmole100533.2
KSCNmolarityweight
−
−
×=
××
×
= 
(c) 
mole
g26.244OH2BaCl 22 =•M 
mmol026653.0
mL543.7
KSCNmmol1
AgNOmmol1
mL
KSCNmmol109386.1
mL102.20
mL
AgNOmmol100533.2consumedAgNOmmol
3
3
3
3
3
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
××
×
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
×
=
−
−
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
 
%4572.0
%100
sampleg7120.0
mmol1000
g26.244
AgNOmmol2
OH2BaClmmol1AgNOmmol026653.0
OH2BaCl% 3
22
3
22
=
×
×
⋅
×
=⋅
 
13-25 (a) 
OH6MgClKClM01821.0
L000.2
g85.277
OH6MgClKClmole1OH6MgClKClg12.10
22
22
22
⋅⋅=
⋅⋅
×⋅⋅
(b) [ ] [ ] ++ =⋅⋅= 2622 MgM01821.0OH6MgClKClMg 
(c) 
[ ] −
−
− =
⋅⋅
×⋅⋅= ClM05463.0
OH6MgClKClmole1
Clmole3OH6MgClKClmole01821.0Cl
22
22 
(d) 
%506.0%100
mL1000
L
L000.2
g12.10OH6MgClKCl)%v/w( 22 =××=⋅⋅ 
(e) 
−
−
=× Clmmol37.1mL0.25
mL
Clmmol05463.0 
(f) 
+
++
=
××
⋅⋅
×
⋅⋅
Kppm0.712
g
mg1000
mole1
Kg10.39
OH6MgClKClmole1
Kmole1
L
OH6MgClKClmole01821.0
22
22
 
Fundamentals of Analytical Chemistry: 8th ed.Chapter 13 
13-26 
mole
g03.30OCH2 =M 
OCH%5.21%100
mL500
mL0.25sampleg00.5
mmol1000
OCHg03.30OCHmmol787.1
OCHmmol787.1mL1.16
mL
SCNNHmmol134.0mL0.40
mL
AgNOmmol100.0
mL0.30
mL
KCNmmol121.0reactedKCNmmolOCHmmol
2
2
2
2
43
2
=×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
=⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛ ×−⎟
⎠
⎞
⎜
⎝
⎛ ×
−⎟
⎠
⎞
⎜
⎝
⎛ ×==
 
13-27 
mole
g34.308
41619 OHC
=M 
41619
41619
41619
41619
3
41619
3
3
341619
3
3
3
OHC%4348.0%100
sampleg96.13
mmol1000
OHCg34.308OHCmmol1968.0
OHCmmol1968.0
CHImmol1
OHCmmol1
AgNOmmol3
CHImmol1AgNOmmol5905.0OHCmmol
AgNOmmol5905.0mL85.2
mL
KSCNmmol05411.0
mL00.25
mL
AgNOmmol02979.0reactedAgNOmmol
=×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
=
××=
=⎟
⎠
⎞
⎜
⎝
⎛ ×
−⎟
⎠
⎞
⎜
⎝
⎛ ×=
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
13-28 
++
−+
++→++
+→+
4322223
32333
NH6)(SeOAg)(SeAg2OH3)(Se3)NH(Ag6
NO)NH(AgNH2AgNO
sss
 
samplemL/Semg94.7
mL00.5
mmol
Semg96.78Semmol503.0
Semmol503.0
)(SeAgmmol2
)(Semmol3
)NH(Agmmol2
)(SeAgmmol1
AgNOmmol1
)NH(Agmmol1AgNOmmol6707.0)(SeAgfromSemmol
AgNOmmol6707.0mL74.16
mL
KSCNmmol01370.0
mL00.25
mL
AgNOmmol0360.0)(SeAgformtoreactedAgNOmmol
223
2
3
23
32
3
3
23
=
⎟
⎠
⎞
⎜
⎝
⎛ ×
=×
××=
=⎟
⎠
⎞
⎜
⎝
⎛ ×
−⎟
⎠
⎞
⎜
⎝
⎛ ×=
+
+
s
ss
s
s
 
13-29 
−
−
−
−
−
−
−
−
−
−
=×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
=
=×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
=
=
×−×=
=××=
4
4
4
4
4
3
43
4
3
3
ClO%65.55%100
mL0.250
mL00.50sampleg998.1
mmol1000
ClOg45.99ClOmmol236.2
ClO%
Cl%60.10%100
mL0.250
mL00.50sampleg998.1
mmol1000
Clg453.35Clmmol195.1
Cl%
ClOmmol236.2
AgNOmmol1
ClOmmol1)mL97.13mL12.40(
mL
AgNOmmol08551.0ClOmmol
Clmmol195.1
AgNOmmol1
Clmmol1g97.13
mL
AgNOmmol08551.0Clmmol
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
13-30 (a) The equivalence point occurs at 50.0 mL, 
mL00.50
SCNNHmmol02500.0
mL1
Agmmol1
SCNNHmmol1Agmmol250.1SCNmL
Agmmol250.1mL00.25
mL
AgNOmmol05000.0Agmmol
4
4
3
=××=
=×=
+
+−
++
 
At 30.00 mL, 
( )
( )
( )
−−−−−−
−
+−
−
+
+
×=××=×=
=×−=
×=
+
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×−
=
SCNM102.11009.9/101.11009.9/K]SCN[
04.21009.9logpAg
AgM1009.9
mL00.30mL00.25
mL00.30
mL
SCNmmol0250.0Agmmol250.1
]Ag[
103123
sp
3
3
 
Proceeding in the same way, we obtain the data for 40.00 mL and 49.00 mL. The results 
are displayed in the spreadsheet at the end of the solution. 
At 50.00 mL, 
98.5)1005.1log(pAg
M1005.1101.1K]SCN[]Ag[
6
612
sp
=×−=
×=×===
−
−−−+
 
At 51.00 mL, 
( )
48.8)103.3log(pAg
M103.31029.3/101.1]Ag[
M1029.3
mL00.25mL00.51
mmol250.1mL00.51
mL
SCNmmol0250.0
]SCN[
9
9412
4
=×−=
×=××=
×=
+
−⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
×
=
−
−−−+
−
−
−
 
At 60.00 mL and 70.00 mL, pAg is calculated in the same way and the results are 
displayed in the spreadsheet below. 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
 
 A B C D E F 
1 Problem 13-30(a) 
2 The equivalence point occurs at 0.05000 mmol/mL X 
3 Conc. AgNO3 0.05000 25.00 mL X (1 mL/0.02500 mmol) = 50.00 mL SCN
- 
4 Vol. AgNO3 25.00 
5 Conc. KSCN 0.02500 
6 Ksp 1.10E-12 
7 Vol. SCN- [Ag+] [SCN-] pAg 
8 30.00 9.09E-03 1.21E-10 2.041 
9 40.00 3.85E-03 2.86E-10 2.415 
10 49.00 3.38E-04 3.26E-09 3.471 
11 50.00 1.05E-06 1.05E-06 5.979 
12 51.00 3.34E-09 3.29E-04 8.48 
13 60.00 3.74E-10 2.94E-03 9.43 
14 70.00 2.09E-10 5.26E-03 9.68 
15 
16 Spreadsheet Documentation 
17 B8=(($B$3*$B$4)-($B$5*A8))/($B$4+A8) C8=$B$6/B8 
18 B11=SQRT($B$6) C12=(($B$5*A12)-($B$3*$B$4))/($B$4+A12)
19 B12=$B$6/C12 D8 = -LOG(B8) 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
(b) Proceeding as in part (a), we obtain the results in the spreadsheet below. 
 A B C D E F 
1 Problem 13-30(b) 
2 The equivalence point occurs at 0.06000 mmol/mL X 
3 Conc. AgNO3 0.06000 20.00 mL X (1 mL/0.03000 mmol) = 40.00 mL I
- 
4 Vol. AgNO3 20.00 
5 Conc. KI 0.03000 
6 Ksp 8.30E-17 
7 Vol. I- [Ag+] [I-] pAg 
8 20.00 1.50E-02 5.53E-15 1.824 
9 30.00 6.00E-03 1.38E-14 2.222 
10 39.00 5.08E-04 1.63E-13 3.294 
11 40.00 9.11E-09 9.11E-09 8.04 
12 41.00 1.69E-13 4.92E-04 12.77 
13 50.00 1.94E-14 4.29E-03 13.71 
14 60.00 1.11E-14 7.50E-03 13.96 
15 
16 Spreadsheet Documentation 
17 B8=(($B$3*$B$4)-($B$5*A8))/($B$4+A8) C8=$B$6/B8 
18 B11=SQRT($B$6) C12=(($B$5*A12)-($B$3*$B$4))/($B$4+A12)
19 B12=$B$6/C12 D8 = -LOG(B8) 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
(c) Proceeding as in part (a), we obtain the results in the spreadsheet below. 
 A B C D E F 
1 Problem 13-30(c) 
2 The equivalence point occurs at 0.07500 mmol/mL X 
3 Conc. AgNO3 0.07500 30.00 mL X (1 mL/0.07500 mmol) = 30.00 mL CI
- 
4 Vol. AgNO3 30.00 
5 Conc. NaCl 0.07500 
6 Ksp 1.82E-10 
7 Vol. CI- [Ag+] [CI-] pAg 
8 10.00 3.75E-02 4.85E-09 1.426 
9 20.00 1.50E-02 1.21E-08 1.824 
10 29.00 1.27E-03 1.43E-07 2.896 
11 30.00 1.35E-05 1.35E-05 4.87 
12 31.00 1.48E-07 1.23E-03 6.83 
13 40.00 1.70E-08 1.07E-02 7.77 
14 50.00 9.71E-09 1.88E-02 8.01 
15 
16 Spreadsheet Documentation 
17 B8=(($B$3*$B$4)-($B$5*A8))/($B$4+A8) C8=$B$6/B8 
18 B11=SQRT($B$6) C12=(($B$5*A12)-($B$3*$B$4))/($B$4+A12)
19 B12=$B$6/C12 D8 = -LOG(B8) 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
(d) The equivalence point occurs at 70.00 mL, 
23
23
2
4
12
2
4
1422
4
)NO(PbmL00.70
)NO(Pbmmol2000.0
mLSOmmol10400.1PbmL
SOmmol10400.1mL00.35
mL
SONammol4000.0SOmmol
=××=
×=×=
−+
−−
 
At 50.00 mL, 
( )
47.6)104.3log(pPb
PbM104.310706.4/106.1]Pb[
SOM10706.4
)mL00.50mL00.35(
mL00.50
mL
)NO(Pbmmol2000.0mmol10400.1
]SO[
7
27282
2
4
2
231
2
4
=×−=
×=××=
×=
+
⎟
⎠
⎞
⎜
⎝
⎛ ×−×
=
−
+−−−+
−−−
 
At 60.00 mL and 69.00 mL, the calculations are made in the same way and the results 
are shown in the following spreadsheet. 
At 70.00 mL, 
90.3)103.1log(pPb
PbM103.1106.1K]SO[]Pb[
4
248
sp
2
4
2
=×−=
×=×===
−
+−−−+
 
At 71.00 mL, 
( )
7243.2)10887.1log(pPb
SOM105.810887.1/106.1]SO[
PbM10887.1
mL00.71mL00.35
SOmmol10400.1mL00.71
mL
)NO(Pbmmol2000.0
]Pb[
3
2
4
6382
4
23
2
4
123
2
=×−=
×=××=
×=
+
×−⎟
⎠
⎞
⎜
⎝
⎛ ×
=
−
−−−−−
+−
−
+
 
At 80.00 mL and 90.00 mL, the calculations are made in the same way and the results 
are shown in spreadsheet below. 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
 
 A B C D E F 
1 Problem 13-30(d) 
2 The equivalence point occurs at 0.4000 mmol/mL X 
3 Conc. Na2SO4 0.4000 35.00 mL X (1 mL/0.2000 mmol) = 70.00 mL Pb
2+ 
4 Vol. Na2SO4 35.00 
5 Conc. Pb(NO3)2 0.2000 
6 Ksp 1.60E-08 
7 Vol. Pb2+ [SO42-] [Pb2+] pPb 
8 50.00 4.71E-02 3.40E-07 6.469 
9 60.00 2.11E-02 7.60E-07 6.119 
10 69.00 1.92E-03 8.32E-06 5.080 
11 70.00 1.26E-04 1.26E-04 3.898 
12 71.00 8.48E-06 1.89E-03 2.724 
13 80.00 9.20E-07 1.74E-02 1.760 
14 90.00 5.00E-07 3.20E-02 1.495 
15 
16 Spreadsheet Documentation 
17 B8=(($B$3*$B$4)-($B$5*A8))/($B$4+A8) C8=$B$6/B8 
18 B11=SQRT($B$6) C12=(($B$5*A12)-($B$3*$B$4))/($B$4+A12) 
19 B12=$B$6/C12 D8 = -LOG(D8) 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
(e) Proceeding as in part (a), we obtain the results in the spreadsheet below. 
 A B C D E F 
1 Problem 13-30(e) 
2 The equivalence point occurs at 0.02500mmol/mL X 
3 Conc. BaCl2 0.0250 40.00 mL X (1 mL/0.05000 mmol) = 20.00 mL SO4
2- 
4 Vol. BaCl2 40.00 
5 Conc. Na2SO4 0.0500 
6 Ksp 1.10E-10 
7 Vol. SO42- [Ba2+] [SO42-] pBa 
8 0.00 2.50E-02 1.602 
9 10.00 1.00E-02 1.10E-08 2.000 
10 19.00 8.47E-04 1.30E-07 3.072 
11 20.00 1.05E-05 1.05E-05 4.979 
12 21.00 1.34E-07 8.20E-04 6.872 
13 30.00 1.54E-08 7.14E-03 7.812 
14 40.00 8.80E-09 1.25E-02 8.056 
15 
16 Spreadsheet Documentation 
17 B8=(($B$3*$B$4)-($B$5*A8))/($B$4+A8) C8=$B$6/B8 
18 B11=SQRT($B$6) C12=(($B$5*A12)-($B$3*$B$4))/($B$4+A12) 
19 B12=$B$6/C12 D8 = -LOG(B8) 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
(f) Proceeding as in part (d), we obtain the results in the spreadsheet below. 
 A B C D E F 
1 Problem 13-30(f) 
2 The equivalence point occurs at 0.2000 mmol/mL X 
3 Conc. NaI 0.2000 50.00 mL X (1 mL/0.4000 mmol) = 25.00 mL Tl
- 
4 Vol. NaI 50.00 
5 Conc. TlNO3 0.4000 
6 Ksp 6.50E-08 
7 Vol. Tl+ [I-] [Tl+] pTl 
8 5.00 1.45E-01 4.47E-07 6.350 
9 15.00 6.15E-02 1.06E-06 5.976 
10 24.00 5.41E-03 1.20E-05 4.920 
11 25.00 2.55E-04 2.55E-04 3.594 
12 26.00 1.24E-05 5.26E-03 2.279 
13 35.00 1.38E-06 4.71E-02 1.327 
14 45.00 7.72E-07 8.42E-02 1.075 
15 
16 Spreadsheet Documentation 
17 B8=(($B$3*$B$4)-($B$5*A8))/($B$4+A8) C8=$B$6/B8 
18 B11=SQRT($B$6) C12=(($B$5*A12)-($B$3*$B$4))/($B$4+A12)
19 B12=$B$6/C12 D8 = -LOG(C8) 
20 
21 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
 
13-31 (Note: In the first printing of the text, the last answer for pAg in the back of the book was 
in error.) 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
KBrmmol00.2mL0.50
mL
KBrmmol0400.0
KBrmmol =×= 
At 5.00 mL, 
( )
80.10)106.1log(pAg
AgM106.11018.3/100.5]Br/[K]Ag[
M1018.3
mL00.5mL0.50
mL00.5
mL
AgNOmmol0500.0mmol00.2
]Br[
11
11213
sp
2
3
=×−=
×=××==
×=
+
⎟
⎠
⎞
⎜
⎝
⎛ ×−
=
−
+−−−−+
−−
 
At 15.00 mL, 25.00 mL, 30.00 mL, 35.00 mL and 39.00 mL, the calculations are 
performed in the same way and the results are shown in the spreadsheet at the end of this 
solution. 
At 40.00 mL, 
15.6)101.7log(pAg
AgM101.7100.5K]Br[]Ag[
7
713
sp
=×−=
×=×===
−
+−−−+
 
At 41.00 mL, 
( )
260.3)1049.5log(pAg
AgM1049.5
mL00.41mL0.50
Brmmol00.2mL00.41
mL
AgNOmmol0500.0
]Ag[
4
4
3
=×−=
×=
+
−⎟
⎠
⎞
⎜
⎝
⎛ ×
=
−
+−
−
+ 
At 45.00 mL and 50.00 mL, the calculations are performed in the same way and the 
results are shown in the spreadsheet that follows. 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
 
 A B C D E F 
1 Problem 13-31 
2 The equivalence point occurs at 0.04000 mmol/mL X 
3 Conc. AgNO3 0.05000 50.00 mL X (1 mL/0.05000 mmol) = 40.00 mL Ag
+ 
4 Vol. KBr 50.00 
5 Conc. KBr 0.04000 
6 Ksp 5.00E-13 
7 Vol. Ag+ [Br-] [Ag+] pAg 
8 5.00 3.18E-02 1.57E-11 10.804 
9 15.00 1.92E-02 2.60E-11 10.585 
10 25.00 1.00E-02 5.00E-11 10.301 
11 30.00 6.25E-03 8.00E-11 10.097 
12 35.00 2.94E-03 1.70E-10 9.770 
13 39.00 5.62E-04 8.90E-10 9.051 
14 40.00 7.07E-07 7.07E-07 6.151 
15 41.00 7.28E+01 5.49E-04 3.260 
16 45.00 1.52E+01 2.63E-03 2.580 
17 50.00 8.00E+00 5.00E-03 2.301 
18 
19 Spreadsheet Documentation 
20 B8=(($B$5*$B$4)-($B$3*A8))/($B$4+A8) C8=$B$6/B8 
21 B14=SQRT($B$6) C15=(($B$3*A15)-($B$4*$B$5))/($B$4+A15)
22 B15=$B$6/C15 D8 = -LOG(C8) 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
Challenge Problem 
]SCN][Fe[
])SCN(Fe[1005.1)SCN(FeSCNFe 3
2
3
f
23
−+
+
+
←
→−+ =×=+ K 
For part (a) we find, 
%81.0%100
Agmol101588.1
SCNmol
Agmol1
)SCN(Femol
SCNmol1)SCN(Femol104030.9
Error%
)SCN(Femol104030.9
L106353.4
mL1000
LmL00.50
L
)SCN(Femol10759.9)SCN(Femol
10759.9
1005.1
101
SCNL106353.4
mol025.0
L
Agmol
SCNmol1Agmol101588.1SCNL
Agmol101588.1
g8682.107
Agmol1Agg125.0Agmol
Agg125.0mL00.50
mL100
g250.0%250.0Agmass
3
2
26
26
2
25
2
5
3
5
)SCN(Fe
23
3
2
==
×
×××
=
×=
⎥
⎦
⎤
⎢
⎣
⎡
×+⎟⎟
⎠
⎞
⎜⎜
⎝
⎛
××
×
=
×=
×
×
=
×=×××=
×=×=
=×≡=
−
−+
−
+−
+−
−
+−
+
−
−
−−
−
−−
−
+c 
Proceeding the same way for parts (b) and (c), we find the results in the following spreadsheet. 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 13 
 
 A B C D E F G 
1 Problem 13-32 
2 
3 mL taken 50 
4 Kf 1.05E+03 
5 conc SCN 0.025 
6 AW Ag 107.8682 
7 min complx 1.00E-05 
8 %Ag g Ag moles Ag L SCN- c SCN cmplx mol SCN cmplx %Error 
9 (a) 0.25 0.125 0.0011588 0.046353 9.759E-05 9.40308E-06 0.811434
10 (b) 0.1 0.05 0.0004635 0.018541 9.759E-05 6.68893E-06 1.443046
11 (c) 0.05 0.025 0.0002318 0.009271 9.759E-05 5.78422E-06 2.495732
12 
13 Spreadsheet Documentation 
14 B9=$B$3*(A9/100) E9=SQRT($B$7/$B$4) 
15 C9=B9/$B$6 F9=E9*(($B$3/1000)+D9) 
16 D9=C9/$B$5 G9=F9/C9*100 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
Chapter 14 
14-1 (a) The initial pH of the NH3 solution will be less than that for the solution containing 
NaOH. With the first addition of titrant, the pH of the NH3 solution will decrease rapidly 
and then level off and become nearly constant throughout the middle part of the titration. 
In contrast, additions of standard acid to the NaOH solution will cause the pH of the 
NaOH solution to decrease gradually and nearly linearly until the equivalence point is 
approached. The equivalence point pH for the NH3 solution will be well below 7, 
whereas for the NaOH solution it will be exactly 7. 
(b) Beyond the equivalence point, the pH is determined b the excess titrant. Thus, the 
curves become identical in this region. 
14-2 Completeness of the reaction between the analyte and the reagent and the concentrations 
of the analyte and reagent. 
14-3 The limited sensitivity of the eye to small color differences requires that there be a 
roughly tenfold excess of one or the other form of the indicator to be present in order for 
the color change to be seen. This change corresponds to a pH range of ± 1 pH unit about 
the pK of the indicator. 
14-4 Temperature, ionic strength, and the presence of organic solvents and colloidal particles. 
14-5 The standard reagents in neutralization titrations are always strong acids or strong bases 
because the reactions with this type of reagent are more complete than with those of their 
weaker counterparts. Sharper end points are the consequence of this difference. 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
14-6 The sharper end point will be observed with the solute having the larger Kb. 
(a) For NaOCl, 78
14
b 103.3100.3
1000.1 −
−
−
×=
×
×
=K 
 For hydroxylamine 96
14
b 101.9101.1
1000.1 −
−
−
×=
×
×
=K Thus, NaOCl 
(b) For NH3, 510
14
b 1075.1107.5
1000.1 −
−
−
×=
×
×
=K 
 For sodium phenolate, 410
14
b 1000.11000.1
1000.1 −
−
−
×=
×
×
=K Thus, sodium phenolate 
(c) For hydroxyl amine Kb = 9.1×10-9 (part a) 
 For methyl amine, 411
14
b 103.4103.2
1000.1 −
−
−
×=
×
×
=K Thus, methyl amine 
(d) For hydrazine 78
14
b 105.91005.1
1000.1 −
−
−
×=
×
×
=K 
 For NaCN, 310
14
b 106.1102.6
1000.1 −
−
−
×=
×
×
=K Thus, NaCN 
14-7 The sharper end point will be observed with the solute having the larger Ka. 
(a) For nitrous acid Ka = 7.1×10-4 
 For iodic acid Ka = 1.7×10-1 Thus, iodic acid 
(b) For anilinium Ka = 2.51×10-5 
 For benzoic acid Ka = 6.28×10-5 Thus, benzoic acid 
(c) For hypochlorous acid Ka = 3.0×10-8 
 For pyruvicacid Ka = 3.2×10-3 Thus, pyruvic acid 
(d) For salicylic acid Ka = 1.06×10-3 
 For acetic acid Ka = 1.75×10-5 Thus, salicylic acid 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
14-8 HIn + H2O H3O+ + In- =
+
 
HIn][
]In][OH[ -3 Ka 
pKa = 7.10 (Table 14-1) 
Ka = antilog(-7.10) = 7.94×10-8 
[HIn]/[In-] = 1.43 
Substituting these values into the equilibrium expression and rearranging gives 
 [H3O+] = 7.94×10-8×1.43 = 1.13×10-7 
 pH = -log(1.13×10-7) = 6.94 
14-9 InH+ + H2O In + H3O+ =+
+
 
]InH[
In]][OH[ 3 Ka 
For methyl orange, pKa = 3.46 (Table 14-1) 
Ka = antilog(-3.46) = 3.47×10-4 
[InH+]/[In] = 1.64 
Substituting these values into the equilibrium expression and rearranging gives 
 [H3O+] = 3.47×10-4×1.64 = 5.69×10-4 
 pH = -log(5.69×10-4) = 3.24 
14-10 [H3O+] = wK and pH = -log(Kw)
1/2 = -½logKw 
At 0oC, pH = -½ log(1.14×10-15) = 7.47 
At 50oC, pH = -½ log(5.47×10-14) = 6.63 
At 100oC, pH = -½ log(4.9×10-13) = 6.16 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
14-11 At 0oC, pKw = -log(1.14×10-15) = 14.94 
At 50oC, pKw = -log(5.47×10-14) = 13.26 
At 100oC, pKw = -log(4.9×10-13) = 12.31 
14-12 pH + pOH = pKw and pOH = -log[OH-] = -log(1.00×10-2) = 2.00 
(a) pH = pKw - pOH = 14.94 - 2.00 = 12.94 
(b) pH = 13.26 - 2.00 = 11.26 
(c) pH = 12.31 - 2.00 10.31 
14-13 
HCl g 0.03646
HCl mmol 1
soln mL
soln g 054.1
soln g 100
HCl g 14.0
×× = 4.047 M 
[H3O+] = 4.047 M and pH = -log4.047 = -0.607 
14-14 
NaOH g 0.04000
NaOH mmol 1
soln mL
soln g 098.1
soln g 100
NaOH g 9.00
×× = 2.471 M 
[OH-] = 2.471 M and pH = 14.00 - (-log2.471) = 14.393 
14-15 The solution is so dilute that we must take into account the contribution of water to [OH-] 
which is equal to [H3O+]. Thus, 
[OH-] = 2.00×10-8 + [H3O+] = 2.00×10-8 + 
]OH[
1000.1
-
14−× 
[OH-]2 – 2.00×10-8[OH-] – 1.00×10-14 = 0 
[OH-] = 1.105×10-7 
pOH = -log 1.105×10-7 = 6.957 and pH = 14.00 – 6.957 = 7.04 
14-16 The solution is so dilute that we must take into account the contribution of water to [H3O+] 
which is equal to [OH-]. Thus, 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
[H3O+] = 2.00×10-8 + [OH-] = 2.00×10-8 + 
]OH[
1000.1
3
14
+
−× 
[H3O+]2 – 2.00×10-8[H3O+] – 1.00×10-14 = 0 
[H3O+] = 1.105×10-7 and pH = -log 1.105×10-7 = 6.96 
14-17 In each part, 
mmol/Mg(OH) g 0.05832
Mg(OH) g 0.102
2
2 = 1.749 mmol Mg(OH)2 taken 
(a) cHCl = (75.0×0.0600 – 1.749×2)/75.0 = 0.01366 M 
 [H3O+] = 0.01366 and pH = -log(0.01366) = 1.87 
(b) 15.0×0.0600 = 0.900 mmol HCl added. Solid Mg(OH)2 remains and 
 [Mg2+] = 0.900 mmol HCl×
solnmL15.0
1
HClmmol2
Mg mmol 1 2
×
+
 = 0.0300 M 
 Ksp = 7.1×10-12 = [Mg2+][OH-]2 
 [OH-] = (7.1×10-12/0.0300)1/2 = 1.54×10-5 
 pH = 14.00 - (-log(1.54×10-5)) = 9.19 
(c) 30.00×0.0600 = 1.80 mmol HCl added, which forms 0.90 mmol Mg2+. 
 [Mg2+] = 0.90/30.0 = 3.00×10-2 
 [OH-] = (7.1×10-12/0.0300)1/2 = 1.54×10-5 
 pH = 14.00 - (-log(1.54×10-5)) = 9.19 
(d) [Mg2+] = 0.0600 M 
 [OH-] = (7.1×10-12/0.0600)1/2 = 1.09×10-5 
 pH = 14.00 - (-log(1.09×10-5)) = 9.04 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
14-18 In each part, (20.0 mL HCl × 0.200 mmol HCl/mL) = 4.00 mmol HCl is taken 
(a) cHCl = [H3O+] = ( ) soln mL0.250.20
HCl mmol 00.4
+
 = 0.0889 M 
 pH = -log 0.0899 = 1.05 
(b) Same as in part (a); pH = 1.05 
(c) cHCl = (4.00 – 25.0 × 0.132)/(20.0 + 25.0) = 1.556×10-2 M 
 [H3O+] = 1.556×10-2 M and pH = -log 1.556×10-2 = 1.81 
(d) As in part (c), cHCl = 1.556×10-2 and pH = 1.81 
 (The presence of NH4+ will not alter the pH significantly.) 
(e) cNaOH = (25.0 × 0.232 – 4.00)/(45.0) = 4.00×10-2 M 
 pOH = -log 4.00×10-2 = 1.398 and pH = 14.00 – 1.398 = 12.60 
14-19 (a) [H3O+] = 0.0500 and pH = -log(0.0500) = 1.30 
(b) µ = ½ {(0.0500)(+1)2 + (0.0500)(-1)2} = 0.0500 
 +γ OH3 = 0.85 (Table 10-2) 
 +OH3a = 0.86×0.0500 = 0.0425 
 pH = -log(0.043) = 1.37 
14-20 (a) [OH-] = 2×0.0167 = 0.0334 M 
 pH = 14 – (-log(0.0334)) = 12.52 
(b) µ = ½ {(0.0167)(+2)2 + (0.0334)(-1)2} = 0.050 
 −γ OH = 0.81 (Table 10-2) 
 −OHa = 0.81×0.0334 = 0.0271 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
 +− × OHOH 3aa = 1.00×10
-14 
 +OH3a = 1.00×10
-14/0.0271 = 3.69×10-13 
 pH = -log(3.69×10-13) = 12.43 
14-21 HOCl + H2O H3O+ + OCl- Ka = 
HOCl][
]][OClOH[ -3
+
 = 3.0×10-8 
[H3O+] = [OCl-] and [HOCl] = cHOCl – [H3O+] 
[H3O+]2/(cHOCl – [H3O+]) = 3.0×10-8 
rearranging gives the quadratic: 0 = [H3O+]2 + 3×10-8[H3O+] - cHOCl×3.0×10-8 
 cHOCl [H3O+] pH 
(a) 0.100 5.476×10-5 4.26 
(b) 0.0100 1.731×10-5 4.76 
(c) 1.00×10-4 1.717×10-6 5.76 
14-22 OCl- + H2O HOCl + OH- Kb = 78
14
-
-
a
w 1033.3
100.3
1000.1
]OCl[
]OH[HOCl][ −
−
−
×=
×
×
==
K
K
 
[HOCl] = [OH-] and [OCl-] = cNaOCl – [OH-] 
[OH-]2/(cNaOCl -[OH-]) = 3.33×10-7 
rearranging gives the quadratic: 0 = [OH-]2 + 3.33×10-7[OH-] - cNaOCl×3.33×10-7 
 cNaOCl [OH-] pOH pH 
(a) 0.100 1.823×10-4 3.74 10.26 
(b) 0.0100 5.754×10-5 4.24 9.76 
(c) 1.00×10-4 5.606×10-6 5.25 8.75 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
14-23 NH3 + H2O NH4+ + OH- Kb = 510
14
1075.1
107.5
1000.1 −
−
−
×=
×
× 
[NH4+] = [OH-] and [NH3] = 
3NH
c – [OH-] 
[OH-]2/(
3NH
c -[OH-]) = 1.75×10-5 
rearranging gives the quadratic: 0 = [OH-]2 + 1.75×10-5[OH-] - 
3NH
c ×1.75×10-5 
 
3NH
c [OH-] pOH pH 
(a) 0.100 1.314×10-3 2.88 11.12 
(b) 0.0100 4.097×10-4 3.39 10.62 
(c) 1.00×10-4 3.399×10-5 4.47 9.53 
14-24 NH4+ + H2O H3O+ + NH3 Ka = 5.7×10-10 
[H3O+] = [NH3] and [NH4+] = +
4NH
c – [H3O+] 
[H3O+]2/( +
4NH
c – [H3O+]) = 5.7×10-10 
rearranging gives the quadratic: 0 = [H3O+]2 + 5.7×10-10[H3O+] - +
4NH
c ×5.7×10-10 
 +
4NH
c [H3O+] pH 
(a) 0.100 7.550×10-6 5.12 
(b) 0.0100 2.387×10-6 5.62 
(c) 1.00×10-4 1.385×10-7 6.62 
14-25 C5H11N + H2O C5H11NH+ + OH- Kb = 312
14
10333.1
105.7
1000.1 −
−
−
×=
×
× 
[C5H11NH+] = [OH-] and [C5H11N] = NHC 115c – [OH
-] 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
[OH-]2/( NHC 115c -[OH
-]) = 1.333×10-3 
rearranging gives the quadratic: 0 = [OH-]2 + 1.333×10-3[OH-] - NHC 115c ×1.333×10
-3 
 NHC 115c [OH
-] pOH pH 
(a) 0.100 1.090×10-2 1.96 12.04 
(b) 0.0100 3.045×10-3 2.52 11.48 
(c) 1.00×10-4 9.345×10-5 4.03 9.97 
14-26 HIO3 + H2O H3O+ + IO3- Ka = 1.7×10-1 
[H3O+] = [IO3-] and [HIO3] = 
3HIO
c – [H3O+] 
[H3O+]2/(
3HIO
c – [H3O+]) = 1.7×10-1 
rearranging gives the quadratic: 0 = [H3O+]2 + 1.7×10-1[H3O+] - 3HIOc ×1.7×10
-1 
 
3HIO
c [H3O+] pH 
(a) 0.100 7.064×10-2 1.15 
(b) 0.0100 9.472×10-3 2.02 
(c) 1.00×10-4 9.994×10-5 4.00 
14-27 (a) HAc = soln mL 500
1
HA g 090079.0
HA mmol 1HA g 0.43 ×× = 0.9547 M HA 
 HA + H2O H3O+ + A- Ka = 1.38×10-4 
 [H3O+] = [A-] and [HA] = 0.9547 – [H3O+] 
 [H3O+]2/(0.9547 – [H3O+]) = 1.38×10-4 
 rearranging and solving the quadratic gives: [H3O+] = 0.0114 and pH = 1.94 
(b) HAc = 0.9547×25.0/250.0 = 0.09547 M HA 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
 Proceeding as in part (a) we obtain: [H3O+] = 3.56×10-3 and pH = 2.45 
(c) HAc = 0.09547×10.0/1000.0 = 9.547×10
-4 M HA 
 Proceeding as in part (a) we obtain: [H3O+] = 3.00×10-4 and pH = 3.52 
14-28 (a) HAc = soln mL 100
1
HA g 22911.0
HA mmol 1HA g 05.1 ×× = 0.04583 M HA 
 HA + H2O H3O+ + A- Ka = 0.43 
 [H3O+] = [A-] and [HA] = 0.04583 – [H3O+] 
 [H3O+]2/(0.04583 – [H3O+]) = 0.43 
 rearranging and solving the quadratic gives: [H3O+] = 0.0418 and pH =1.38 
(b) HAc = 0.04583×10.0/100.0 = 0.004583 M HA 
 Proceeding as in part (a) we obtain: [H3O+] = 4.535×10-3 and pH = 2.34 
(c) HAc = 0.004583×10.0/1000.0 = 4.583×10
-5 M HA 
 Proceeding as in part (a) we obtain: [H3O+] = 4.583×10-5 and pH = 4.34 
14-29 Throughout 14-29: amount HA taken = 
mL
mmol 0.200mL 00.20 × = 4.00 mmol 
(a) HA + H2O H3O+ + A- Ka = 1.80×10-4 
 HAc = 4.00/45.0 = 8.89×10
-2 
 [H3O+] = [A-] and [HA] = 0.0889 – [H3O+] 
 [H3O+]2/(0.0889 – [H3O+]) = 1.80×10-4 
 rearranging and solving the quadratic gives: [H3O+] = 3.91×10-3 and pH = 2.41 
(b) amount NaOH added = 25.0 × 0.160 = 4.00 mmol 
 therefore, we have a solution of NaA 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
 A- + H2O OH- + HA Kb = 1.00×10-14/(1.80×10-4) = 5.56×10-11 
 -Ac = 4.00/45.0 = 8.89×10
-2 
 [OH-] = [HA] and [A-] = 0.0889 – [OH-] 
 [OH-]2/(0.0889 – [OH-]) = 5.56×10-11 
 rearranging and solving the quadratic gives: [OH-] = 2.22×10-6 and pH = 8.35 
(c) amount NaOH added = 25.0 × 0.200 = 5.00 mmol 
 therefore, we have an excess of NaOH and the pH is determined by its concentration 
 [OH-] = (5.00 - 4.00)/45.0 = 2.22×10-2 
 pH = 14 – pOH = 12.35 
(d) amount NaA added = 25.0 × 0.200 = 5.00 mmol 
 [HA] = 4.00/45.0 = 0.0889 
 [A-] = 5.00/45.00 = 0.1111 
 [H3O+]×0.1111/0.0889 = 1.80×10-4 
 [H3O+] = 1.440×10-4 and pH = 3.84 
14-30 Throughout 14-30 the amount of NH3 taken is 4.00 mmol 
(a) NH3 + H2O OH- + NH4+ Kb = 510
14
1075.1
107.5
1000.1 −
−
−
×=
×
× 
 
3NH
c = 4.00/60.0 = 6.67×10-2 
 [NH4+] = [OH-] and [NH3] = 0.0667 – [OH-] 
 [OH-]2/(0.0667 – [OH-]) = 1.75×10-5 
 rearranging and solving the quadratic gives: [OH-] = 1.07×10-3 and pH = 11.03 
(b) amount HCl added = 20.0 × 0.200 = 4.00 mmol 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
 therefore, we have a solution of NH4Cl 
 NH4+ + H2O H3O+ + NH3 Ka = 5.7×10-10 
 +
4NH
c = 4.00/60.0 = 6.67×10-2 
 [H3O+] = [NH3] and [NH4+] = 0.0667 – [H3O+] 
 [H3O+]2/(0.0667 – [H3O+]) = 5.7×10-10 
 rearranging and solving the quadratic gives: [H3O+] = 6.16×10-6 and pH = 5.21 
(c) amount HCl added = 20.0 × 0.250 = 5.00 mmol 
 therefore, we have an excess of HCl and the pH is determined by its concentration 
 [H3O+] = (5.00 - 4.00)/60.0 = 1.67×10-2 
 pH = 1.78 
(d) amount NH4Cl added = 20.0 × 0.200 = 4.00 mmol 
 [NH3] = 4.00/60.0 = 0.0667 [NH4+] = 4.00/60.0 = 0.0667 
 [H3O+]×0.0.0667/0.0667 = 5.70×10-10 
 [H3O+] = 5.70×10-10 and pH = 9.24 
(e) amount HCl added = 20.0 × 0.100 = 2.00 mmol 
 [NH3] = (4.00-2.00)/60.0 = 0.0333 [NH4+] = 2.00/60.0 = 0.0333 
 [H3O+]×0.0.0333/0.0333 = 5.70×10-10 
 [H3O+] = 5.70×10-10 and pH = 9.24 
14-31 (a) NH4+ + H2O H3O+ + NH3 5.70×10-5 = 
]NH[
]][NHOH[
4
33
+
+
 
 [NH3] = 0.0300 and [NH4+] = 0.0500 
 [H3O+] = 5.70×10-10 × 0.0500/0.0300 = 9.50×10-10 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
 [OH-] = 1.00×10-14/9.50×10-10 = 1.05×10-5 
 pH = -log (9.50×10-10) = 9.022 
(b) µ = ½ {(0.0500)(+1)2 + (0.0500)(-1)2} = 0.0500 
 From Table 10-2 +γ
4NH
 = 0.80 and 
3NH
γ = 1.0 
 
0300.000.1
0500.080.01070.5
]NH[
]NH[ 5
3NH
4NHa
OH
3
4
3 ×
×××
=
γ
γ
=
−+
+
+
K
a = 7.60×10-10 
 pH = -log (7.60×10-10) = 9.12 
14-32 In each part of this problem a buffer mixture of a weak acid, HA, and its conjugate base, 
NaA, is formed. In each case we will assume initially that [H3O+] and [OH-] are much 
smaller than the molar concentration of the acid and conjugate so that [A-] ≅ cNaA and 
[HA] ≅ cHA. These assumptions then lead to the following relationship: 
 [H3O+] = Ka cHA / cNaA 
(a) cHA = 
soln L 1.00
1
HA g 08.90
HA mol 1HA g 20.9 ×× = 0.1021 M 
 cNaA = 
soln L 1.00
1
NaA g 06.112
NaA mol 1HA g 15.11 ×× = 0.0995 M 
 [H3O+] = 1.38×10-4×0.1021/0.0995 = 1.416×10-4 
 Note that [H3O+] (and [OH-]) << cHA (and cNaA) as assumed. Therefore, 
 pH = -log (1.416×10-4) = 3.85 
(b) cHA = 0.0550 M and cNaA = 0.0110 M 
 [H3O+] = 1.75×10-5×0.0550/0.0110 = 8.75×10-5 
 pH = -log (8.75×10-5) = 4.06 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
(c) Original amount HA = 
g 0.13812
HA mmolg 00.3 × = 21.72 mmol HA 
Original amount NaOH = 
mL
HA mmol 0.1130mL 0.50 × = 5.65 mmol NaOH 
 cHA = (21.72 – 5.65)/500 = 3.214×10-2 M 
 cNaA = 5.65/500 = 1.130×10-2 M 
 [H3O+] = 1.06×10-3×3.214×10-2/(1.130×10-2) = 3.015×10-3 
Note, however, that [H3O+] is not << cHA (and cNaA) as assumed. Therefore, 
 [A-] = 1.130×10-2 + [H3O+] – [OH-] 
 [HA] = 3.214×10-2 – [H3O+] + [OH-] 
Certainly, [OH-] will be negligible since the solution is acidic. Substituting into the 
dissociation-constant expression gives 
 
( )
]OH[10214.3
]OH[10130.1]OH[
3
2
3
2
3
+−
+−+
−×
+×
 = 1.06×10-3 
Rearranging gives 
 [H3O+]2 + 1.236×10-2 [H3O+] – 3.407×10-5 = 0 
 [H3O+] = 2.321×10-3 M and pH = 2.63 
(d) Here we must again proceed as in part (c). This leads to 
 
( )
]OH[0100.0
]OH[100.0]OH[
3
33
+
++
−
+
 = 4.3×10-1 
 [H3O+]2 + 0.53 [H3O+] – 4.3×10-3 = 0 
 [H3O+] = 7.99×10-3 M and pH = 2.10 
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)(NH mmol
NH mmol 2
SO)(NH g 13214.0
SO)(NH mmol 1 +
× = 
 49.95 mmol 
Amount NaOH = 125.0 mL×0.1011 mmol/mL = 12.64 mmol 
mL0.500
1
NaOHmmol
NH mmol 1
NaOH mmol 64.12 3NH3 ××=c = 2.528×10
-2 M 
mL0.500
1NH mmol )64.1295.49( 4NH4 ×−=
+
+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×
B g 0.09313
B mmol 1 = 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 1075.1HOAc][
]][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 = Kb 
Here we can use an approximation solution because Kb 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 pH 
change resulting from dilution. Thus, we write 
 [HOAc] = cHOAc + [OH-] – [H3O+] ≈ cHOAc – [H3O+] 
 [OAc-] = cNaOAc – [OH-] + [H3O+] ≈ cNaOAc + [H3O+] 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
 Ka = 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, 
 cNaOH = (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 
 cNaOAc = (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 
 cNaOAc = 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 
 cNaOAc = 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) cNaOH = 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, cNaOH = (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. 
 cNaOAc = 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. 
 cNaOH = 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)] 
 cNaOAc = 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 
 cNaOAc = 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, Ka = 1.38×10-4 = [H3O+][A-]/[HA] 
Throughout this problem we will base calculations on Equations 9-25 and 9-26. 
 [A-] = cNaA + [H3O+] – [OH-] 
 [HA] = cHA – [H3O+] – [OH-] 
 
( )
]OH[
]OH[]OH[
3HA
3NaA3
+
++
−
+
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 
 cNaA = (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 
 cNaA = (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 
 cNaA = (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 Vi, NaOH 50.00 
2 ci, NaOH 0.1000M 
3 c, HCl 0.1000M 
4 Veq. pt. 50.00 
5 Kw 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
KOH mmol 1000.095.24
soln volumetotal
added KOHamount 
=
×
= = 0.03329 M 
 cHA ≅ [HA] = 
solnvolumetotal
added KOHamount -HAamount original 
 = 
solnmL74.95
HA mmol 0.1000)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+] = Ka 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
HAamount initial-added KOHamount 
 = 
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 = C2H5NH2 BH+ = C2H5NH3+ 
 
95.99
995.4
95.99
10000.095.49
soln volumetotal
HClO mmol no. 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] = 
( )
( )NaOHHA i
NaOHNaOH iHA iHA i
VV
VcVc
+
−
 - [H3O+] 
and [A-] = 
( )
( )NaOHHA i
NaOHNaOH i
VV
Vc
+
 + [H3O+] 
Substituting these expressions into the equilibrium expression for HA and rearranging gives 
 0 = [H3O+]2 + 
( )
( ) ⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛
+
+ aNaOHHA i
NaOHNaOH i K
VV
Vc
[H3O+] - 
( )
( )NaOHHA i
NaOHNaOH iHA iHA ia
VV
VcVcK
+
−
 
From which [H3O+] is directly determined. 
At and after the equivalence point: [A-] = 
( )
( )NaOHHA i
HAHA i
VV
Vc
+
 - [HA] 
and [OH-] = 
( )
( )NaOHHA i
HA iHA iNaOHNaOH i
VV
VcVc
+
− + [HA] 
Substituting these expressions into the equilibrium expression for A- and rearranging gives 
 0 = [HA]2 + 
( )
( ) ⎟
⎟
⎠
⎞
⎜⎜
⎝
⎛
+
+
−
a
w
NaOHHA i
HA iHA iNaOHNaOH i
K
K
VV
VcVc
[HA] - 
( )
( )NaOHHA ia
HAHA iw
VVK
VcK
+
 
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 Vi, HNO2 50.00 
3 ci, HNO2 0.1000 
4 Ka, HNO2 7.10E-04 
5 Kw, H2O 1.00E-14 
6 
7 c, NaOH 0.1000 
8 Veq. 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 Vi, Lactic Acid 50.00 
3 ci, Lactic Acid 0.1000 
4 Ka, Lactic Acid 1.38E-04 
5 Kw, H2O 1.00E-14 
6 
7 c, NaOH 0.1000 
8 Veq. 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 Vi, C5H5NH+ 50.00 
3 ci, C5H5NH+ 0.1000 
4 Ka, C5H5NH+ 5.90E-06 
5 Kw, H2O 1.00E-14 
6 
7 c, NaOH 0.1000 
8 Veq. 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 Vi, NH3 50.00 
3 ci, NH3 0.1000 
4 Ka, NH4+ 5.70E-10 
5 Kw, H2O 1.00E-14 
6 
7 c, HCl 0.1000 
8 Veq. 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 Vi, H2NNH2 50.00 
3 ci, H2NNH2 0.1000 
4 Ka, H2NNH3+ 1.05E-08 
5 Kw, H2O 1.00E-14 
6 
7 c, HCl 0.1000 
8 Veq. 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
1425.00 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 Vi, NaCN 50.00 
3 ci, NaCN 0.1000 
4 Ka, HCN 6.20E-10 
5 Kw, H2O 1.00E-14 
6 
7 c, HCl 0.1000 
8 Veq. 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 
14-43 
 A B C D E F 
1 Part (a) 
2 Vi, C6H5NH3+ 50.00 
3 ci, C6H5NH3+ 0.1000 
4 Ka, C6H5NH3+ 2.51E-05 
5 Kw, H2O 1.00E-14 
6 
7 c, NaOH 0.1000 
8 Veq. pt. 50.00 
9 
10 Vol. NaOH, mL b c [OH-] [H3O+] pH 
11 0.00 2.5100E-05 -2.5100E-06 1.5718E-03 2.8036
12 5.00 9.1160E-03 -2.0536E-06 2.1997E-04 3.6576
13 15.00 2.3102E-02 -1.3515E-06 5.8356E-05 4.2339
14 25.00 3.3358E-02 -8.3667E-07 2.5062E-05 4.6010
15 40.00 4.4470E-02 -2.7889E-07 6.2706E-06 5.2027
16 45.00 4.7394E-02 -1.3211E-07 2.7872E-06 5.5548
17 49.00 4.9520E-02 -2.5354E-08 5.1198E-07 6.2907
18 50.00 3.9841E-10 -1.9920E-11 4.4630E-06 2.2406E-09 8.6496
19 51.00 9.9010E-04 -1.9723E-11 9.9012E-04 1.0100E-11 10.9957
20 55.00 4.7619E-03 -1.8972E-11 4.7619E-03 2.1000E-12 11.6778
21 60.00 9.0909E-03 -1.8109E-11 9.0909E-03 1.1000E-12 11.9586
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Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
 
 A B C D E F 
1 Part (b) 
2 Vi, ClCH2COOH 50.00 
3 ci, ClCH2COOH 0.0100 
4 Ka, ClCH2COOH 1.36E-03 
5 Kw, H2O 1.00E-14 
6 
7 c, NaOH 0.0100 
8 Veq. pt. 50.00 
9 
10 Vol. NaOH, mL b c [OH-] [H3O+] pH 
11 0.00 1.3600E-03 -1.3600E-05 3.0700E-03 2.5129
12 5.00 2.2691E-03 -1.1127E-05 2.3889E-03 2.6218
13 15.00 3.6677E-03 -7.3231E-06 1.4351E-03 2.8431
14 25.00 4.6933E-03 -4.5333E-06 8.2196E-04 3.0852
15 40.00 5.8044E-03 -1.5111E-06 2.4960E-04 3.6027
16 45.00 6.0968E-03 -7.1579E-07 1.1523E-04 3.9385
17 49.00 6.3095E-03 -1.3737E-07 2.1698E-05 4.6636
18 50.00 7.3529E-12 -3.6765E-14 1.9174E-07 5.2155E-08 7.2827
19 51.00 9.9010E-05 -3.6401E-14 9.9010E-05 1.0100E-10 9.9957
20 55.00 4.7619E-04 -3.5014E-14 4.7619E-04 2.1000E-11 10.6778
21 60.00 9.0909E-04 -3.3422E-14 9.0909E-04 1.1000E-11 10.9586
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Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
 
 A B C D E F 
1 Part (c) 
2 Vi, HOCl 50.00 
3 ci, HOCl 0.1000 
4 Ka, HOCl 3.00E-08 
5 Kw, H2O 1.00E-14 
6 
7 c, NaOH 0.1000 
8 Veq. 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
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Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
 
 A B C D E F 
1 Part (d) 
2 Vi, HONH3+ 50.00 
3 ci, HONH3+ 0.1000 
4 Ka, HONH3+ 1.10E-06 
5 Kw, H2O 1.00E-14 
6 
7 c, HCl 0.1000 
8 Veq. 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 
 
 
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Fundamentals of Analytical Chemistry: 8th ed. Chapter 14 
14-44 
 A B C D E F G 
1 Species pH [H3O+] Ka α0 α1 
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, 
α0 = 44
4
1080.110310.6
10310.6
−−
−
×+×
× = 0.778 
[ ] [ ]
0850.0
HCOOHHCOOH
T
=
c
 = α0 
[HCOOH] = 0.778 × 0.0850 = 6.61×10-2 M 
14-46 [H3O+] = 3.38×10-12 M. For CH3NH3+, Equation 9-36 takes the form, 
α1 = 1112
11
a3
a
T
23
103.21038.3
103.2
]OH[
]NHCH[
−−
−
+ ×+×
×
=
+
=
K
K
c
 
 = 0.872 = 
120.0
]NHCH[ 23 
[CH3NH2] = 0.872 × 0.120 = 0.105 M 
14-47 For lactic acid, Ka = 1.38 × 10-4 
α0 = 
]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 
α1 = 1.000 – 0.640 = 0.360 
[A-] = α1 × 0.120 = (1.000 – 0.640) × 0.120 = 0.0432 M 
[H3O+] = Ka 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-] α0 α1 
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