SKOOG - SOLUCIONÁRIO CAPÍTULO 9

Prévia do material em texto

Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
Chapter 9 
9-1 (a) A weak electrolyte only partially ionizes when dissolved in water. NaHCO3 is an 
example of a weak electrolyte. 
(b) A Brønsted-Lowry acid is a molecule that donates a proton when it encounters a base 
(proton acceptor). By this definition, NH4+ can be a Brønsted-Lowry acid. 
(c) The conjugate acid of a Brønsted-Lowry base is the species formed when a Brønsted-
Lowry base accepts a proton. For example, the NH4+ is the conjugate acid of NH3. 
(d) Neutralization, according to the Brønsted-Lowry concept, occurs when a reaction 
involving an acid and its conjugate base is combined with a second reaction involving a 
base and its conjugate acid. Thus, 
−+
←
→ ++ OHNHOHNH 423 
In the example above, NH3 acts as a base with NH4+ as its conjugate acid. H2O acts as an 
acid with OH- as its conjugate base. 
(e) An amphiprotic solvent can act either as an acid or a base depending on the solute. 
Water is an example of an amphiprotic chemical species. 
(f) A zwitterion is a chemical species that bears both positive and negative charges. Free 
amino acids, such as glycine, can exist as zwitterions in solution. 
zwitterion
COOCHNH
glycine
COOHCHNH 2322
−+
←
→
 
(g) Autoprotolysis is the act of self-ionization of a solvent to produce both a conjugate 
acid and a conjugate base. 
(h) A strong acid dissociates completely such that no undissociated molecules are left in 
aqueous solution. Hydrochloric acid, HCl, is an example of a strong acid. 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
(i) The Le Châtelier principle states that the position of an equilibrium always shifts in 
such a direction to relieve a stress applied to the system. 
(j) The common-ion effect is responsible for the reduced solubility of an ionic precipitate 
when one of the soluble components reacting to form the precipitate is added to the 
solution in equilibrium with the precipitate. 
9-2 (a) An amphiprotic solute is a chemical species that possesses both acidic and basic 
properties. The dihydrogen phosphate ion, H2PO4-, is an example of an amphiprotic 
solute. 
(b) A differentiating solvent reveals different strengths of acids. By this definition, 
anhydrous acetic acid is a differentiating solvent because perchloric acid dissociates 5000 
times more than hydrochloric acid. 
(c) A leveling solvent is one in which a series of acids (or bases) all dissociate completely. 
Water is an example, since strong acids like HCl and HClO4 ionize completely. 
(d) A mass-action effect arises when a shift in the chemical equilibrium occurs due to the 
introduction of one of the participating chemical species (i.e., addition of one of the 
reactants or products. 
9-3 For an aqueous equilibrium in which water is a participant, the concentration of water is 
normally so much larger than the concentrations of the other reactants or products that it 
can be assumed to be constant and independent of the equilibrium position. Thus, its 
concentration is included within the equilibrium constant. For a pure solid, the 
concentration of the chemical species in the solid phase is constant. As long as some 
solid exists as a second phase, its effect on the equilibrium is constant and is included 
within the equilibrium constant. 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
9-4 
 Acid Conjugate Base 
(a) HOCl OCl- 
(b) H2O OH- 
(c) NH4+ NH3 
(d) HCO3- CO32- 
(e) H2PO4- HPO42- 
9-5 
 Base Conjugate Acid 
(a) H2O H3O+ 
(b) HONH2 HONH3+ 
(c) H2O H3O+ 
(d) HCO3- H2CO3 
(e) PO43- HPO42- 
9-6 (a) −+←
→ + OHOHOH2 32 
(b) −+←
→ + COOCHCOOHCHCOOHCH2 3233 
(c) −+←
→ + NHCHNHCHNHCH2 33323 
(d) −+←
→ + OCHOHCHOHCH2 3233 
9-7 (a) −+←
→ ++ OHNHHCOHNHHC 2522252 
4
252
252
11
14
a
w
b 1033.4]NHH[C
]][OHNHH[C
1031.2
1000.1 −−
+
−
−
×==×
×==
K
KK 
(b) +−←
→ ++ OHCNOHHCN 32 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
103
b
w
a 102.6[HCN]
]O][H[CN −+− ×===
K
KK 
(c) +←
→+ ++ OHNHCOHNHHC 355255 
6
55
355
b
w
a 1090.5]NHH[C
]ON][HH[C −
+
+
×===
K
KK 
(d) −←
→− ++ OHHCNOHCN 2 
5
14
10
a
w
b 106.1][CN
][HCN][OH
102.6
1000.1 −
−
−−
− ×==×
×==
K
KK 
(e) 
+−
←
→
+−
←
→−
+−
←
→−
+−
←
→
++
++
++
++
OH3AsOOH3AsOH
OHAsOOHHAsO
OHHAsOOHAsOH
OHAsOHOHAsOH
3
3
4243
3
3
42
2
4
3
2
4242
342243
 
213712
1a2a3a
43
3422a3a
43
2
3
2
43a
43
3
3
3
4
3
2
4
3
3
4
3a
42
3
2
4
2a
43
342
1a
100.2)108.5()101.1()102.3(
]AsO[H
]O][HAsO[H
]AsOH[
]O][H[HAsO
]AsO[H
]O][H[AsO
][HAsO
]O][H[AsO
]AsO[H
]O][H[HAsO
]AsO[H
]O][HAsO[H
−−−−
3
+−+−+−
−
+−
−
+−+−
×=×××××=β
====β
===
KKKKKK
KKK
 
(f) 
−
→
←−
−
←
→−
−−
←
→−
++
++
++
OH2COHOH2CO
OHCOHOHHCO
OHHCOOHCO
322
2
3
3223
32
2
3
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
 
14
114
214
2a1a
2
w
1b2b2
3
32b
2
3
2
32
1a
w
33
w32
3
32
2b
2a
w
2
33
w3
2
3
3
1b
104.1
)1069.4()105.1(
)100.1(
]CO[
]][OHHCO[
][CO
]][OHCO[H
]][HCOOH[
]COH[
][HCO
]][OHCO[H
]][COOH[
]HCO[
][CO
]][OH[HCO
−
−−
−
2
−
−−
−
−
2
−+−
−
−+
−
−
−−
×=×××
×=β
====β
===
===
KK
KKKK
K
KKK
K
KKK
 
9-8 (a) ]][I[CuICu)CuI( sp
−+−+
←
→ =+ Ks 
(b) ]][F][Cl[PbFClPb)PbClF( 2sp
2 −−+−−+
←
→ =++ Ks 
(c) 22sp
2
2 ]][I[Pb2IPb)(PbI
−+−+
←
→ =+ Ks 
(d) 33sp
3
3 ]][I[Bi3IBi)(BiI
−+−+
←
→ =+ Ks 
(e) ]][PO][NHMg[PONHMg)(POMgNH 344
2
sp
3
44
2
44
−++−++
←
→ =++ Ks 
9-9 (a) 2sp S]][I[Cu][I][CuS ==== −+−+ K 
(b) S = [Pb2+] = [Cl-] = [F-] Ksp = [Pb2+][Cl-][F-] = S3 
(c) S = [Pb2+] = 1/2[I-] Ksp = [Pb2+][I-]2 = (S)(2S)2 = 4S3 
(d) S = [Bi3+] = 1/3[I-] Ksp = [Bi2+][I-]3 = (S)(3S)3 = 27S4 
(e) S = [Mg2+] = [NH4+] = [PO43-] Ksp = [Mg2+][NH4+][PO43-] = S3 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
9-10 (a) 
8
442
3
2
sp
2
3
2
3
42
3
2
1002.2
)1042.1()1042.1(]][SeO[CuSeOCu)(CuSeO
L
mole101.42][SeO]Cu[
−
−−−+−+
←
→
−−+
×=
×××==+
×==
Ks 
(b) 
132552
3
2
sp3
2
23
552
3
52
102.3)106.8()103.4(]][IOPb[2IOPb)()Pb(IO
L
mole108.6)L
mole10(4.32]2[Pb][IOL
mole104.3][Pb
−−−−+−+
←
→
−−+−−+
×=×××==+
×=××==×=
Ks
(c) 
923422
sp
2
2
34242
105.2)107.1()106.8(]][F[Sr2FSr)(SrF
L
mole101.7)L
mole10(8.62]2[Sr][FL
mole108.6][Sr
−−−−+−+
←
→
−−+−−+
×=×××==+
×=××==×=
Ks
 
(d) 
1543444
sp
4
4
34444
100.1)103.1()103.3(]][OH[Th4OHTh)(Th(OH)
L
mole101.3)L
mole10(3.34]4[Th][OHL
mole103.3][Th
−−−−+−+
←
→
−−+−−+
×=×××==+
×=××==×=
Ks
 
9-11 (a) 
Ksp = 2.02×10-8 = [Cu2+][SeO32-] = (0.050 M)[SeO32-] 
S = [SeO32-] = M100.4050.0
100.2 78 −− ×=× 
(b) 
Ksp = 3.2×10-13 = [Pb2+][IO3-]2 = (0.050 M)[IO3-]2 
[IO3-] = 6
13
105.2
050.0
102.3 −− ×=× M 
S = ½[IO3-] = ½(2.5×10-6 M) = 1.3×10-6 M 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
 
(c) 
M101.1)M102.2(
2
1]F[
2
1S
M102.2
050.0
105.2]F[
]M)[F050.0(]][F[Sr105.2
44
4
9
2229
sp
−−−
−
−
−
−−+−
×=××==
×=×=
==×=K
 
(d) 
M104.9)M108.3(
4
1]OH[
4
1S
M108.3
050.0
100.1]OH[
]M)[OH050.0(]][OHTh[100.1
54
44
15
44415
sp
−−−
−
−
−
−−+−
×=××==
×=×=
==×=K
 
9-12 (a)M100.4
050.0
1002.2]Cu[S
)](0.050MCu[]][SeOCu[1002.2
7
8
2
22
3
28
sp
−
−
+
+−+−
×=×==
==×=K
 
(b) 
M103.1
)050.0(
102.3]Pb[S
)](0.050MPb[]][IOPb[102.3
10
2
13
2
222
3
213
sp
−
−
+
+−+−
×=×==
==×=K
 
(c) 
Ksp = 2.5×10-9 = [Sr2+][F-]2 = [Sr2+](0.050 M)2 
S = [Sr2+] = M100.1
)050.0(
105.2 6
2
9
−
−
×=× 
 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
(d) 
Ksp = 1.0×10-15 = [Th4+][OH-]4 = [Th4+](0.050 M)4 
S = [Th4+] = M106.1
)050.0(
100.1 10
4
15
−
−
×=× 
9-13 ][CrO]Ag[102.1CrO2Ag)(CrOAg 24
212
sp
2
442
−+−−+
←
→ =×=+ Ks 
(a) M100.1
)1041.3(
102.1]CrO[ 922
12
2
4
−
−
−− ×=×
×= 
(b) M30.0
)1000.2(
102.1]CrO[ 26
12
2
4 =×
×= −
−− 
9-14 3334sp
3
3 ]][OHAl[100.33OHAl)(Al(OH)
−+−−+
←
→ =×=+ Ks
3432
sp
23
3
100.3][OH][OH
3
1M105.0
][OH
3
1M105.0][Al
][OH
3
1Al(OH)ofsolubilityMolar
−−−−
−−+
−
×=⎟⎠
⎞⎜⎝
⎛ +×=
+×=
=
K
 
(a) Because the Ksp is so small, we can assume the solubility of Al(OH)3 is not large. 
Therefore, it is reasonable that the [Al3+] ≈ 5.0 X 10-2 M. The Ksp equation then 
simplifies to 
Ksp = (5.0×10-2 M)[OH-]3 = 3.0×10-34 
[OH-] = M100.2
M100.5
100.3 113
1
2
34
−
−
−
×=⎟⎟⎠
⎞
⎜⎜⎝
⎛
×
× 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
(b) As in part (a), 
M101.1
M1000.2
100.3]OH[ 9
3
1
7
34
−
−
−
− ×=⎟⎟⎠
⎞
⎜⎜⎝
⎛
×
×= 
9-15 10333
3
sp 102.3)S3(S]][IOCe[
−−+ ×===K 
(a) 
M0125.0
L
mL1000
)mL0.50mL0.50(
mole1025.1
mole1025.1mL0.50
mL1000
L
L
mole105.2CeM0250.0
3
323
=×+
×
×=×××≡
−
−−+
 
(b) mole100.2mL0.50
mL1000
L
L
mole100.4IOM040.0 323
−−− ×=×××≡ 
( ) mole10833.5mole100.2
3
1mole1025.1unreactedCemoles 4333 −−−+ ×=×−×= 
( ) SM10833.5SL
mL1000
mL0.50mL0.50
mole10833.5]Ce[ 3
4
3 +×=+⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×+
×= −
−
+ 
where S is derived iteratively using the equation 
33
sp )S3()S10833.5( +×= −K 
We start by solving for S assuming no contribution to [Ce3+] other than from the 
dissociation of Ce(IO3)3. In this case, S equals 1.855 × 10-3. Now, we substitute 1.855 × 
10-3 back into the Ksp equation above and solve for Kiterative. Kiterative equals 1.3259 × 10-9. 
S is too large; we choose a smaller S (i.e., 1 × 10-3) and recalculate Kiterative. Iteration 
continues until Kiterative ≈ Ksp. The results of this approach are shown in the Table below. 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
Iteration S Kiterative 
1 1.855×10-3 1.3259×10-9 
2 1×10-3 1.8449×10-10 
3 1.2×10-3 3.2813×10-10 
4 1.19×10-3 3.1954×10-10 
 
We now substitute S = 1.19 × 10-3 into the equation below, 
M100.71019.1M10833.5]Ce[ 3333 −−−+ ×=×+×= 
(c) 
mole0125.0mL0.50
mL1000
L
L
mole105.2IOM250.0 13 =×××≡ −− 
The Ce3+ is completely consumed; thus, 
103
sp
3
3
33
3
102.3)S3M0875.0(S
S3M0875.0S3
L
mL1000
mL0.100
mole1075.8]IO[
mole1075.8)mole1025.1(3mole0125.0unreactedIOmoles
−
−−
−−−
×=+=
+=+⎟⎟⎠
⎞
⎜⎜⎝
⎛ ××=
×=×−=
K
 
Make the assumption that 3S << 0.0375 M. 
M108.4
)M0875.0(
102.3S]Ce[
102.3)M0875.0(S
7
3
10
3
103
sp
−
−
+
−
×=⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×==
×==K
 
 
(d) 
mole105.7mL0.50
mL1000
L
L
mole105.1IOM150.0 313
−−− ×=×××≡ 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
The Ce3+ is completely consumed; thus, 
103
sp
3
3
333
3
102.3)S3M0375.0(S
S3M0375.0S3
L
mL1000
mL0.100
mole1075.3]IO[
mole1075.3)mole1025.1(3mole105.7unreactedIOmoles
−
−−
−−−−
×=+=
+=+⎟⎟⎠
⎞
⎜⎜⎝
⎛ ××=
×=×−×=
K
 
Make the assumption that 3S << 0.0375 M. 
M101.6
)M0375.0(
102.3S]Ce[
102.3)M0375.0(S
6
3
10
3
103
sp
−
−
+
−
×=⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×==
×==K
 
9-16 6226
2
sp 100.6)S()S2(][PdCl]K[
−−+ ×===K 
M100.0
L
mL1000
)mL0.50mL0.50(
mole1000.1
mole0100.0mL0.50
mL1000
L
L
mole100.2KM200.0
2
1
=×+
×
=×××≡
−
−+
 
(a) mole105.2mL0.50
mL1000
L
L
mole100.5PdClM0500.0 3226
−−− ×=×××≡ 
( )
M050.0
L
mL1000
)mL0.50mL0.50(
mole100.5
mole100.5mole105.22mole1000.1unreactedKmoles
3
332
=×+
×
×=×−×=
−
−−−+
 
The [K+] is 0.05 M plus the contribution from the dissociation of the precipitate, x, or 
[K+] = 0.05 M + 2x 
Substituting the equation above into the equilibrium expression, we find 
6323
22
6
2
sp
100.6)105.2(20.04
)205.0(]PdCl[]K[
−−
−+
×=×++
+==
xxx
xxK
 
We use an iterative approach to solve for x. We begin by ignoring the contribution to [K+] 
from the dissociation of the precipitate. 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
011447.0
4
100.6
3
6
=×=
−
x 
Substituting this value back into the Ksp equation above, we obtain Kiterative equal to 
6.0824 × 10-5, which is too large. We now proceed by choosing a smaller value of x equal 
1 × 10-3 and obtain Kiterative equals to 2.7040× 10-6. We continue the iterative approach 
until Kiterative ≈ Ksp as shown in the table below. 
Iteration x Kiterative 
1 0.011447 6.0814 × 10-5 
2 1 × 10-3 2.7040 ×10-6 
3 2 × 10-3 5.8320 ×10-6 
 
Substituting x equals 2 × 10-3 into the [K+] equation, we find 
[K+] = 0.05 M + 2(2 × 10-3M) = 0.054 M 
(b) 
mole1000.5mL0.50
mL1000
L
L
mole100.1PdClM100.0 3126
−−− ×=×××≡ 
The [K+] is determined directly from Ksp; thus, 
M022.0)M011.0(2S2]K[
011.0
4
100.6S
100.6S4S)S2(
S]PdCl[S2]K[
3
1
6
632
sp
2
6
===
=⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×=
×===
==
+
−
−
−+
K
 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
(c) 
mole010.0mL0.50
mL1000
L
L
mole100.2PdClM200.0 126 =×××≡ −− 
The K+ is completely consumed; thus, 
M011.0)M10477.5(2S2]K[
M10477.5
)05.0(4
100.6S
100.6)M05.0(S4
M05.0]PdCl[
M05.0SthatAssume
SM05.0S
L
mL1000
mL100.0
mole105.0][PdCl
mole105.0mole)10(1.00
2
1mole101.0unreactedPdClmoles
3
3
2
1
6
62
sp
2
6
3
2
6
3-222
6
=×==
×=⎟⎟⎠
⎞
⎜⎜⎝
⎛ ×=
×==
=
<<
+=+⎟⎟⎠
⎞
⎜⎜⎝
⎛ ××=
×=×−×=
−+
−
−
−
−
−−
−−−
K
 
9-17 CuI(s) ' Cu+ + I- Ksp = [Cu+][I-] = 1×10-12 
AgI(s) ' Ag+ + I- Ksp = [Ag+][I-] = 8.3×10-17 
PbI2(s) ' Pb2+ + 2I- Ksp = [Pb2+][I-]2 = 7.1×10-9 = S(2S)2 = 4S3 
BiI3(s) ' Bi3+ + 3I- Ksp = [Bi3+][I-]3 = 8.1×10-19 = S(3S)3 = 27S4 
(a) 
waterinAgICuIBiIPbI
103.1
27
)101.8(S]I[
3
1]Bi[S
102.1
4
)101.7(S]I[
2
1]Pb[S
101.9103.8S][I]Ag[S
101101S][I]Cu[S
32
54
19
3
BiI
33
9
2
PbI
917
AgI
612
CuI
3
2
>>>
×=×===
×=×===
×=×===
×=×===
−
−
−+
−
−
−+
−−−+
−−−+
 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
(b) 
NaIM10.0inBiIAgICuIPbI
101.8
)M10.0(
101.8S
101.7
)M10.0(
101.7S
103.8
M10.0
103.8S
101
M10.0
101S
32
16
3
19
BiI
7
2
9
PbI
16
17
AgI
11
12
CuI
3
2
>>>
×=×=
×=×=
×=×=
×=×=
−
−
−
−
−
−
−
−
 
(c) 
cationsolutetheofsolutionM0.010ainAgICuIBiIPbI
104.1
M010.0
101.8
3
1S
102.4
M010.0
101.7
2
1S
103.8
M010.0
103.8S
101
M010.0
101S
32
63
19
BiI
4
9
PbI
15
17
AgI
10
12
CuI
3
2
>>>
×=×=
×=×=
×=×=
×=×=
−
−
−
−
−
−
−
−
 
9-18 BiOOH(s) ' BiO+ + OH- Ksp = [BiO+][OH-] = 4.0×10-10 
Be(OH)2(s) ' Be2+ + 2OH- Ksp = [Be2+][OH-]2 = 7.0×10-22 = S(2S)2 = 4S3 
Tm(OH)3(s) ' Tm3+ + 3OH- Ksp = [Tm3+][OH-]3= 3.0×10-24 = S(3S)3 = 27S4 
Hf(OH)4(s) ' Hf4+ + 4OH- Ksp = [Hf4+][OH-]4 = 4.0×10-26 = S(4S)4 = 256S5 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
(a) 
water.insolubilitylowestthehasBe(OH)
107.2
256
100.4S][OH
4
1]Hf[S
108.5
27
100.3S][OH
3
1]Tm[S
106.5
4
100.7S][OH
2
1]Be[S
100.2100.4S][OH]BiO[S
2
65
26
4
)OH(Hf
74
24
3
)OH(Tm
83
22
2
)OH(Be
510
BiOOH
4
3
2
−
−
−+
−
−
−+
−
−
−+
−−−+
×=×===
×=×===
×=×===
×=×===
 
(b) 
.NaOHM0.10insolubilitylowestthehasHf(OH)
100.4
)M10.0(
100.4S
100.3
)M10.0(
100.3S
100.7
)M10.0(
100.7S
100.4
M10.0
100.4S
4
22
4
26
)OH(Hf
21
3
24
)OH(Tm
20
2
22
)OH(Be
9
10
BiOOH
4
3
2
−
−
−
−
−
−
−
−
×=×=
×=×=
×=×=
×=×=
 
9-19 At 0ºC, 
7.472M)10log(3.38]Olog[HpH
M103.38100.144][OH]O[H100.114]][OHOH[
8
3
814
3
14
3sp
=×−=−=
×=×==×==
−+
−−−+−−+K
 
At 100ºC, 
155.6M)10log(7.00]Olog[HpH
M107.001049][OH]O[H1049]][OHOH[
7
3
714
3
14
3sp
=×−=−=
×=×==×==
−+
−−−+−−+K
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
9-20 (a) 
M103.3
M103.0
101.0][OH
M103.0
2
)104(9.0)10(3.0103.0
]O[H
0109.0]O[H103.0]O[H
])O[HM(0.0300103.0]O[H103.0
])O[HM(0.0300
]O[H
]O[HM0.0300[HOCl]][OCl]OH[
100.3
[HOCl]
]O][H[OClOHOClOHHOCl
10
5
14
5
10288
3
10
3
82
3
3
82
3
8
3
2
3
33
83
a32
−
−
−
−
−
−−−
+
−+−+
+−+−
+
+
+−+
−
+−
+−
←
→
×=×
×=
×=×+×+×−=
=×−×+
−×=×=−
−==
×==++ K
 
(b) 
M101.06
M109.47
101.0][OH
M109.47
2
)104(9.12)10(1.52101.52
]O[H
0109.12]O[H101.52]O[H
])O[HM(0.0600101.52]O[H101.52
])O[HM(0.0600
]O[H
]O[HM0.0600COOH]CHCH[CH]COOCHCH[CH]O[H
1052.1
COOH]CHCH[CH
]O][HCOOCHCH[CH
OHCOOCHCHCHOHCOOHCHCHCH
11
4
14
4
7255
3
7
3
52
3
3
52
3
5
3
2
3
32232233
5
223
3223
a
32232223
−
−
−
−
−
−−−
+
−+−+
+−+−
+
+
+−+
−
+−
+−
←
→
×=×
×=
×=×+×+×−=
=×−×+
−×=×=−
−==
×==
++
K
 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
(c) 
M101.6
M106.37
101.0]O[H
M106.4
2
)104(4.33)10(4.33104.33
][OH
0104.33][OH104.33][OH
])[OHM(0.100104.33][OH104.33
])[OHM(0.100
][OH
][OHM0.100]NHH[C]NHH[C][OH
1033.4
1031.2
100.1
]NHH[C
]][OHNHH[C
OHNHHCOHNHHC
12
3
14
3
3
5244
542
424
2
252352
4
11
14
a
w
252
352
b
3522252
−
−
−
+
−
−−−
−
−−−−
−−−−
−
−
−+−
−
−
−−+
−+
←
→
×=×
×=
×=×+×+×−=
=×−×+
−×=×=−
−==
×=×
×===
++
K
KK
 
(d) 
M102.83
M103.53
101.0]O[H
M103.53
2
)104(1.27)10(6.33106.33
][OH
0101.27][OH106.33][OH
])[OHM(0.200106.33][OH106.33
])[OHM(0.200
][OH
][OHM0.200N])[(CH]NH)[(CH][OH
1033.6
1058.1
100.1
N])[(CH
]][OHNH)[(CH
OHNH)(CHOHN)(CH
12
3
14
3
3
5255
552
525
2
3333
5
10
14
a
w
33
33
b
33233
−
−
−
+
−
−−−
−
−−−−
−−−−
−
−
−+−
−
−
−−+
−+
←
→
×=×
×=
×=×+×+×−=
=×−×+
−×=×=−
−==
×=×
×===
++
K
KK
 
 
 
 
 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
(e) 
M103.9
M102.6
101.0]O[H
M102.6
2
)104(6.7)10(3.3103.3
][OH
0106.7][OH103.3][OH
])[OHM(0.200103.3][OH103.3
])[OHM(0.200
][OH
][OHM0.200][OCl[HOCl]][OH
103.3
100.3
100.1
][OCl
][HOCl][OHOHHOClOHOCl
11
4
14
3
4
8277
872
727
2
7
8
14
a
w
b2
−
−
−
+
−
−−−
−
−−−−
−−−−
−
−
−−−
−
−
−
−
−
−
←
→−
×=×
×=
×=×+×+×−=
=×−×+
−×=×=−
−==
×=×
×===++
K
KK
 
(f) 
M101.25
M108.01
101.0]O[H
M108.01
2
)104(6.42)10(7.42107.42
][OH
0106.42][OH107.42][OH
])[OHM(0.0860107.42][OH107.42
])[OHM(0.0860
][OH
][OHM0.0860]COOH[CCOOH]H[C][OH
1042.7
1034.1
100.1
]COOH[C
]COOH][OHH[C
OHCOOHHCOHCOOHC
9
6
14
3
6
1121010
11102
10210
2
5252
10
5
14
a
w
52
52
b
52252
−
−
−
+
−
−−−
−
−−−−
−−−−
−
−
−−−
−
−
−
−
−
−
←
→−
×=×
×=
×=×+×+×−=
=×−×+
−×=×=−
−==
×=×
×===
++
K
KK
 
 
 
 
 
 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
(g) 
M101.91
M105.24
101.0][OH
M105.24
2
)104(2.75)10(1.10101.10
]O[H
0102.75]O[H101.10]O[H
])O[HM(0.250101.10]O[H101.10
])O[HM(0.250
]O[H
]O[HM0.250][HONH][HONH]O[H
1010.1
][HONH
]O][H[HONHOHHONHOHHONH
11
4
14
4
7266
3
7
3
62
3
3
62
3
6
3
2
3
3323
6
3
32
a3223
−
−
−
−
−
−−−
+
−+−+
+−+−
+
+
+++
−
+
+
+
←
→+
×=×
×=
×=×+×+×−=
=×−×+
−×=×=−
−==
×==++ K
 
(h) 
M103.55
M102.82
101.0][OH
M102.82
2
)104(7.95)10(3.18103.18
]O[H
0107.95]O[H103.18]O[H
])O[HM(0.0250103.18]O[H103.18
])O[HM(0.0250
]O[H
]O[HM0.0250]NHH[HOC]NHH[HOC]O[H
1018.3
]NHH[HOC
]O][HNHH[OHCOHNHHHOCOHNHHHOC
9
6
14
6
1221010
3
12
3
102
3
3
102
3
10
3
2
3
33422423
10
342
3242
a32422342
−
−
−
−
−
−−−
+
−+−+
+−+−
+
+
+++
−
+
+
+
←
→+
×=×
×=
×=×+×+×−=
=×−×+
−×=×=−
−==
×==++ K
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
9-21 (a) 
M1100.0
2
)104(1.36)10(1.36101.36
]O[H
0101.36]O[H101.36]O[H
])O[HM(0.100101.36]O[H101.36
])O[HM(0.100
]O[H
]O[HM0.100COOH][ClCH]COO[ClCH]O[H
1036.1
COOH][ClCH
]O][HCOO[ClCHOHCOOClCHOHCOOHClCH
4233
3
4
3
32
3
3
32
3
3
3
2
3
3223
3
2
32
a3222
=×+×+×−=
=×−×+
−×=×=−
−==
×==++
−−−
+
−+−+
+−+−
+
+
+−+
−
+−
+−
←
→ K
 
(b) 
M101.17
M108.57
101.0]O[H
M108.57
2
)104(7.35)10(7.35107.35
][OH
0107.35][OH107.35][OH
])[OHM(0.100107.35][OH107.35
])[OHM(0.100
][OH
][OHM0.100]COO[ClCHCOOH][ClCH][OH
1035.7
1036.1
100.1
]COO[ClCH
]COOH][OH[ClCH
OHCOOHClCHOHCOOClCH
8
7
14
3
7
1321212
13122
12212
2
22
12
3
14
a
w
2
2
b
222
−
−
−
+
−
−−−
−
−−−−
−−−−
−
−
−−−
−
−
−
−
−
−
←
→−
×=×
×=
×=×+×+×−=
=×−×+
−×=×=−
−==
×=×
×===
++
K
KK
 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
(c) 
M105.32
M101.88
101.0]O[H
M101.88
2
)104(4.35)10(4.35104.35
][OH
0104.35][OH104.35][OH
])[OHM(0.0100104.35][OH104.35
])[OHM(0.0100
][OH
][OHM0.0100]NH[CH]NH[CH][OH
1035.4
103.2
100.1
]NH[CH
]][OHNH[CH
OHNHCHOHNHCH
12
3
14
3
3
6244
642
424
2
2333
4
11
14
a
w
23
33
b
33223
−
−
−
+
−
−−−
−
−−−−
−−−−
−
−
−+−
−
−
−−+
−+
←
→
×=×
×=
×=×+×+×−=
=×−×+
−×=×=−
−==
×=×
×===
++
K
KK
 
(d) 
M104.8
2
)104(2.3)10(2.3102.3
]O[H
0102.3]O[H102.3]O[H
])O[HM(0.0100102.3]O[H102.3
])O[HM(0.0100
]O[H
]O[HM0.0100]NH[CH]NH[CH]O[H
103.2
]NH[CH
]O][HNH[CHOHNHCHOHNHCH
7
1321111
3
13
3
112
3
3
112
3
11
3
2
3
333233
11
33
323
a323233
−
−−−
+
−+−+
+−+−
+
+
+++
−
+
+
+
←
→+
×=×+×+×−=
=×−×+
−×=×=−
−==
×==++ K
 
(e) 
M101.46
2
)104(2.51)10(2.51102.51
]O[H
0102.51]O[H102.51]O[H
])O[HM(0.0010102.51]O[H102.51
])O[HM(0.0010
]O[H
]O[HM0.0010]NHH[C]NHH[C]O[H
1051.2
]NHH[C
]O][HNHH[COHNHHCOHNHHC
4
8-2553
8
3
52
3
3
52
3
5
3
2
3
33562563
5
356
3256
a32562356
−
−−
+
−+−+
+−+−
+
+
+++
−
+
+
+
←
→+
×=×+×+×−=
=×−×+
−×=×=−
−==
×==++ K
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
(f) 
0.12M
2
4(0.034))10(1.7101.7
]O[H
00.034]O[H101.7]O[H
])O[HM(0.200101.7]O[H101.7
])O[HM(0.200
]O[H
]O[HM0.200][HIO][IO]OH[
107.1
][HIO
]O][H[IOOHIOOHHIO
211
3
3
12
3
3
12
3
1
3
2
3
3333
1
3
33
a3323
=+×+×−=
=−×+
−×=×=−
−==
×==++
−−
+
+−+
+−+−
+
+
+−+
−
+−
+−
←
→ K
 
9-22 A buffer solution resists changes in pH with dilution or with addition of acids or bases. A 
buffer is composed of a mixture of a weak acid and its conjugate base. 
9-23 Buffer capacity of a solution is defined as the number of moles of a strong acid (or a 
strong base) that causes 1.00 L of a buffer to undergo a 1.00-unit change in pH. 
9-24 (a) 943.8
(0.200M)
(0.100M)log)107.5log(
][NH
][NHlogppH 10
4
3
a =+×−=+= −+K 
(b) 943.8
(0.100M)
(0.050M)log)107.5log(
][NH
][NHlogppH 10
4
3
a =+×−=+= −+K 
The solutions have identical pH values, but the solution in part (a) has the greater buffer 
capacity because it has the higher concentration of weak acid and conjugate base. 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
9-25 (a) 
359.4
020.0
008.0log)1075.1log(pH
mole020.0mL200
mL1000
L
L
mole100.0HOAcM100.0
OHHOAcOHOAc
OAcOHOHHOAc
5
2
32
=+×−=
=××≡
++
++
−
−
←
→−
−+
←
→
 
(b) 
4.359
106.25
102.50log)10log(1.75pH
M102.50
L
mL1000
mL200
mole0.0050][OAc
M106.25
L
mL1000
mL200
e0.0050)mol(0.0175[HOAc]
mole0.005mL100
mL1000
L
L
mole0.0500NaOHM0.0500
mole0.0175mL100
mL1000
L
L
mole0.175HOAcM0.175
2
2
5
2
2
=×
×+×−=
×=×=
×=×−=
=××≡
=××≡
−
−
−
−−
− 
(c) 
4.359
102.4
109.6log)10log(1.75pH
M102.4
L
mL1000
mL200
mole0.0048[HOAc]
M109.6
L
mL1000
mL200
e0.0048)mol(0.00672][OAc
mole0.0048mL40.0
mL1000
L
L
mole0.1200HClM0.1200
mole0.00672mL160
mL1000
L
L
mole0.042OAcM0.0420
2
3
5
2
3
=×
×+×−=
×=×=
×=×−=
=××≡
=××≡
−
−
−
−
−−
−
 
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
 
Since the ratios of the amounts of weak acid to conjugate base are identical the three 
solutions will have the same pH. They will differ in buffer capacity, however, with (a) 
having the greatest and (c) the least. 
9-26 (a) Malic acid / sodium hydrogen malate 
(b) HOCl / OCl- 
(c) NH4Cl / NH3 
(d) pyridine / pyridinium 
9-27 
( )
g15.5
mole1
g67.997
HCOOmole1
HCOONamole1HCOOmole228.0HCOONaweight
mole228.0HCOOH)mole(0.400(0.569)HCOOmole
HCOOHmole400.0mL400.0
mL1000
L
L
mole1.00HCOOHM00.1
569.0
HCOOHmole
HCOOmole569.010
[HCOOH]
][HCOO245.0
[HCOOH]
][HCOOlog
[HCOOH]
][HCOOlog74.3
[HCOOH]
][HCOOlog108.1log
[HCOOH]
][HCOOlogp50.3pH
245.0
4
a
=××=
=×=
=××≡
===−=
+=+×−=+==
−
−
−
−
−
−−
−−
−
−
K
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
 
9-28 
( )
g43.2
mole1
98.01g
COOHOCHmole1
COONaHOCHmole1HCOOmole441.0COONaHOCHweight
mole441.0COOH)HOCHmole(0.300(1.47)COOHOCHmole
COOHHOCHmole300.0mL300.0
mL1000
L
L
mole1.00COOHHOCHM1.00
1.47
COOHHOCHmole
COOHOCHmole
1.4710
COOH][HOCH
]COO[HOCH167.0
COOH][HOCH
]COO[HOCHlog
COOH][HOCH
]COO[HOCHlog83.3
COOH][HOCH
]COO[HOCHlog1047.1log
COOH][HOCH
]COO[HOCHlogp00.4pH
2
2
2
22
22
2
2
0.167
2
2
2
2
2
2
2
24
2
2
a
=××=
=×=
=××≡
=
===
+=+×−=
+==
−
−
−
−
−−
−−
−
−
K
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
9-29 
( )
HClmL194
L
mL1000
mole0.200
Lmole3880.0HClvolume
3880.0
1.933
mole0.075
1.933
CHOHCOOHCmoleHClmole
HCl)mole(933.0HClmoleCHOHCOOHCmole
933.0
HClmole
HCl)moleCHOHCOOHC(mole
933.0
CHOHCOOHHCmole
CHOHCOOHCmole
933.010
CHOHCOOH]H[C
]CHOHCOOH[C03.0
CHOHCOOH]H[C
]CHOHCOOH[Clog
CHOHCOOH]H[C
]CHOHCOOH[Clog40.3
CHOHCOOH]H[C
]CHOHCOOH[Clog100.4log
CHOHCOOH]H[C
]CHOHCOOH[ClogKp37.3pH
CHOHCOONaHCmole0750.0
mL250.0
mL1000
L
L
mole0.300CHOHCOONaHCM300.0
56
56
56
56
56
03.0
56
56
56
56
56
56
56
564
56
56
a
56
56
=××=
===
=−
=−
=
==−=
+=+×−=
+==
=
××≡
−
−
−
−
−
−−
−−
−
−
x
xx
x
x
 
Fundamentals of Analytical Chemistry: 8th ed. Chapter 9 
9-30 
( )
NaOHmL3.89
L
mL1000
mole00.2
Lmole179.0NaOHvolume
mole179.0
47.2
)mole300.0(47.1
47.2
)COOHOCHmole(47.1NaOHmole
)NaOHmoleCOOHHOCHmole(47.1NaOHmole
47.1
)NaOHmoleCOOHHOCHmole(
NaOHmole
47.1
COOHHOCHmole
COOHOCHmole
47.110
]COOHHOCH[
]COOHOCH[167.0
]COOHHOCH[
]COOHOCH[log
]COOHHOCH[
]COOHOCH[log83.3
]COOHHOCH[
]COOHOCH[log1047.1log
]COOHHOCH[
]COOHOCH[logpK00.4pH
COOHHOCHmole300.0mL0.300
mL1000
L
L
mole00.1COOHHOCHM00.1
2
2
2
2
2
1067.1
2
2
2
2
2
2
2
24
2
2
a
22
1
=××=
=×=×=
−×=
=−
=
===
+=+×−=
+==
=××≡
−
−
×
−−
−−
−
−
−
x
xx
x
x
 
9-31 The statement “A buffer maintains the pH of a solution constant” is false. The change in 
pH of a buffered solution is relatively small with the addition of a small volume of acid 
or base as shown in the example below. 
]HA[
]NaA[logppH a += K 
mL of 0.050M NaOH ]HA[
]NaA[ ∆pH 
 1.48 0.170 
100 1.59 0.200 
200 1.70 0.230 
300 1.83 0.262