Chapter 14

Prévia do material em texto

Fundamentals of Analytical Chemistry: 8
th
 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: 8
th
 ed. Chapter 14 
14-6 The sharper end point will be observed with the solute having the larger Kb. 
(a) For NaOCl, 
7
8
14
b 103.3
100.3
1000.1 





K
 
 For hydroxylamine 
9
6
14
b 101.9
101.1
1000.1 





K
 Thus, NaOCl 
(b) For NH3, 
5
10
14
b 1075.1
107.5
1000.1 





K
 
 For sodium phenolate, 
4
10
14
b 1000.1
1000.1
1000.1 





K
 Thus, sodium phenolate 
(c) For hydroxyl amine Kb = 9.110
-9
 (part a) 
 For methyl amine, 
4
11
14
b 103.4
103.2
1000.1 





K
 Thus, methyl amine 
(d) For hydrazine 
7
8
14
b 105.9
1005.1
1000.1 





K
 
 For NaCN, 
3
10
14
b 106.1
102.6
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 pyruvic acid 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: 8
th
 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
 = -½ logKw 
At 0
o
C, pH = -½ log(1.1410-15) = 7.47 
At 50
o
C, pH = -½ log(5.4710-14) = 6.63 
At 100
o
C, pH = -½ log(4.910-13) = 6.16 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 14 
14-11 At 0oC, pKw = -log(1.1410
-15
) = 14.94 
At 50
o
C, pKw = -log(5.4710
-14
) = 13.26 
At 100
o
C, 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: 8
th
 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 
 [Mg
2+
] = 0.900 mmol HCl
soln mL 15.0
1
HCl mmol 2
Mg mmol 1 2

 = 0.0300 M 
 Ksp = 7.110
-12
 = [Mg
2+
][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+. 
 [Mg
2+
] = 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) [Mg
2+
] = 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: 8
th
 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)  = ½ {(0.0500)(+1)2 + (0.0500)(-1)2} = 0.0500 
 
 OH3
 = 0.85 (Table 10-2) 
 
OH3
a
 = 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)  = ½ {(0.0167)(+2)2 + (0.0334)(-1)2} = 0.050 
 
 OH
 = 0.81 (Table 10-2)OH
a
 = 0.810.0334 = 0.0271 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 14 
 
  OHOH 3
aa
 = 1.0010-14 
 
OH3
a
 = 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 = 
7
8
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: 8
th
 ed. Chapter 14 
14-23 NH3 + H2O  NH4
+
 + OH
-
 Kb = 
5
10
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 = 
3
12
14
10333.1
105.7
1000.1 





 
[C5H11NH
+
] = [OH
-
] and [C5H11N] = 
NHC 115
c
 – [OH-] 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 14 
[OH
-
]
2
/(
NHC 115
c
 -[OH
-
]) = 1.33310-3 
rearranging gives the quadratic: 0 = [OH
-
]
2
 + 1.33310-3[OH-] - 
NHC 115
c
1.33310-3 
 
NHC 115
c
 [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
+
] - 
3HIO
c
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 1
HA 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: 8
th
 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 1
HA 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 0.200
mL 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: 8
th
 ed. Chapter 14 
 A
-
 + H2O  OH
-
 + HA Kb = 1.0010
-14
/(1.8010-4) = 5.5610-11 
 
-A
c
 = 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 = 
5
10
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: 8
th
 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: 8
th
 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)  = ½ {(0.0500)(+1)2 + (0.0500)(-1)2} = 0.0500 
 From Table 10-2 

4NH
 = 0.80 and 
3NH

 = 1.0 
 
0300.000.1
0500.080.01070.5
]NH[
]NH[ 5
3NH
4NHa
OH
3
4
3 









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 1
HA g 20.9 
 = 0.1021 M 
 cNaA = 
soln L 1.00
1
NaA g 06.112
NaA mol 1
HA 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: 8
th
 ed. Chapter 14 
(c) Original amount HA = 
g 0.13812
HA mmol
g 00.3 
 = 21.72 mmol HA 
Original amount NaOH = 
mL
HA mmol 0.1130
mL 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: 8
th
 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 
mL 0.500
1
NaOH mmol
NH mmol 1
NaOH mmol 64.12 3NH3 c
 = 2.52810-2 M 
mL 0.500
1
NH 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: 8
th
 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 1075.1
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: 8
th
 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: 8
th
 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: 8
th
 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: 8
th
 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: 8
th
 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: 8
th
 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: 8
th
 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: 8
th
 ed. Chapter 14 
14-38 
 A B C D E F G H I 
1 Vi, NaOH 50.00 
 
2 ci, NaOH 0.1000 M 
3 c, HCl 0.1000 M 
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
soln mL 95.74
KOH mmol 1000.095.24
soln volumetotal
added KOHamount 



 = 0.03329 M 
 cHA  [HA] = 
soln volumetotal
added KOHamount -HAamount original
 
 = 
soln mL 74.95
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: 8
th
 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 = 
soln volumetotal
HAamount initial-added KOHamount 
 
 = 
soln mL 75.05
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: 8
th
 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
 +  
  








a
NaOHHA 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: 8
th
 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: 8
th
 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: 8
th
 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: 8
th
 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 
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 Vi, NaCN 50.003 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: 8
th
 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 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
 
Fundamentals of Analytical Chemistry: 8
th
 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 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
 
Fundamentals of Analytical Chemistry: 8
th
 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 
22 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
 
Fundamentals of Analytical Chemistry: 8
th
 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 
 
 
 
 
23 
24 
25 
26 
27 
28 
29 
30 
31 
32 
33 
34 
35 
36 
37 
38 
39 
 
Fundamentals of Analytical Chemistry: 8
th
 ed. Chapter 14 
14-44 
 A B C D E F G 
1 Species pH [H3O
+
] Ka 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 CH3NH3
+
, 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
 
[CH3NH2] = 0.872  0.120 = 0.105 M 
14-47 For lactic acid, Ka = 1.38  10
-4
 
0 = 
]OH[1038.1
]OH[
]OH[
]OH[
3
4
3
3a
3






K
 
Fundamentals of Analytical Chemistry: 8
th
 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
+
] = 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
-
] 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