Capítulo 10

Prévia do material em texto

CHAPTER 10 
ACIDS AND BASES 
 
 
 
 
10.1 (a) (b) 3 3CH NH
+
2 3NH NH
+ (c) (d) 2H CO3
2
3CO
−
(e) (f) 6 5C H O
−
3 2CH CO
− 
 
10.3 For all parts (a) – (e), H2O and H3O+ form a conjugate acid-base pair in 
which H2O is the base and H3O is the acid 
(a) +2 4 2 3 4H SO (aq) H O(l) H O (aq) HSO (aq)
−+ +U
2 4 4H SO and HSO (aq)
− form a conjugate acid-base pair in which 
 2 4 4H SO is the acid and HSO (aq) is the base.
−
(b) 6 5 3 2 3 6 5 2C H NH (aq) H O(l) H O (aq) C H NH (aq)
+ ++ +U
6 5 3 6 5 2C H NH and C H NH (aq)
+ form a conjugate acid-base pair in which 
 6 5 3 6 5 2C H NH is the acid and C H NH (aq) is the base.
+
(c) 22 4 2 3 4H PO (aq) H O(l) H O (aq) HPO (aq)
− ++ +U −
−2
2 4 4H PO (aq) and HPO (aq)
− form a conjugate acid-base pair in which 
 22 4 4H PO (aq) is the acid and HPO (aq) is the base.
− −
(d) 2 3 2HCOOH(aq) H O(l) H O (aq) HCO (aq)
+ −+ +U
2HCOOH(aq) and HCO (aq)
− form a conjugate acid-base pair in which 
 2HCOOH(aq) is the acid and HCO (aq) is the base.
−
 (e) 2 3 2 3 2 2NH NH (aq) H O(l) H O (aq) NH NH (aq)
+ ++ +U
2 3 2 2NH NH (aq) and NH NH (aq)
+ form a conjugate acid-base pair in 
which 2 3 2 2NH NH (aq) is the acid and NH NH (aq) is the base.
+
 
SM-274 
10.5 (a) Brønsted acid: 3HNO
Brønsted base: 24HPO
− 
(b) conjugate base to 3 3HNO :NO
− 
conjugate acid to 24 2 4HPO :H PO
− − 
 
10.7 (a) as an acid: 
HCO
3HCO ,
− 2
3 2 3 3HCO (aq) H O(l) H O (aq) CO (aq)
− ++ +U − . 
−
3−
−
4
−
3
- and CO32- form a conjugate acid-base pair in which HCO3- is the 
acid and CO32- is the base. 
3HCO ,
− as a base: . 
CHO
2 3 2 3H O(l) HCO (aq) H CO (aq) OH (aq)
− −+ +U
3
- and H2CO3 form a conjugate acid-base pair in which HCO3- is the 
base and H2CO3 is the acid. H2O and OH- form a conjugate acid-base pair 
in which H2O is the acid and OH- is the base. 
(b) , as an acid: 
. 
form a conjugate acid-base pair in which 
H
2
4HPO
−
2 3
4 2 3 4HPO (aq) H O(l) H O (aq) PO (aq)
− ++ +U
2
4 4HPO (aq) and PO (aq)
−
2 3
4 4HPO (aq) is the acid and PO (aq) is the base. 
− −
2O and H3O+ form a 
conjugate acid-base pair in which H2O is the base and H3O+ is the acid. 
2
4HPO ,
− as a base: . 
 form a conjugate acid-base pair in which HPO
2
4 2 2 4HPO (aq) H O(l) H PO (aq) OH (aq)
− −+ +U
2
4 2HPO and H PO
−
4
2- is 
the base and H2PO4- is the acid. H2O and OH- form a conjugate acid-base 
pair in which H2O is the acid and OH- is the base. 
 
10.9 (a) basic; (b) acidic; (c) amphoteric; (d) basic 
 
10.11 (a) ; 3 2 4 2 2 7SO (g) H SO H S O (l)+ →
H
S
+
O
O
O
O
S
+
O
O
O
H
(b) 
 
 
SM-275 
 
(c) sulfuric acid acts as the Lewis base while SO3 acts as a Lewis acid. 
 
10.13 In each case, use then 14w 3[H O ][OH ] 1.0 10 ,K
+ − −= = ×
14
w
3 3
1.0 10[OH ]
[H O ] [H O ]
K −−
+ +
×= = 
(a) 
14
13 11.0 10[OH ] 5.0 10 mol L
0.02
−
− −×= = × −⋅ 
(b) 
14
9 1
5
1.0 10[OH ] 1.0 10 mol L
1.0 10
−
− −
−
×= = ××
−⋅ 
(c) 
14
12 1
3
1.0 10[OH ] 3.2 10 mol L
3.1 10
−
− −
−
×= = ××
−⋅
−
 
 
10.15 (a) 14 2w 3 32.1 10 [H O ][OH ] , where [H O ] [OH ]K x x
− + − += × = = = =
14 7 1
3
2.1 10 1.4 10 mol L
pH log[H O ] 6.80
x − −
+
= × = × ⋅
= − =
−
−
=
 
(b) 
7 1
3[OH ] [H O ] 1.4 10 mol L
pOH log[OH ] 6.80
− + −
−
= = × ⋅
= − =
 
10.17 Because is a strong base, 
 
2Ba(OH)
2
2Ba(OH) (aq) Ba (aq) 2 OH (aq), 100%.
+ −⎯⎯→ +
Then 
 22 0 2 0 2 0[Ba(OH) ] [Ba ], [OH ] 2 [Ba(OH) ] , where [Ba(OH) ]
+ −= = ×
nominal concentration of 2Ba(OH) .
moles of 32 1
0.25 gBa(OH) 1.5 10 mol
171.36 g mol
−
−= = ×⋅ 
3
2 1
2 0
1.5 10 mol[Ba(OH) ] 1.5 10 mol L [Ba ]
0.100 L
−
2− − +×= = × ⋅ = 
2 1 2 1
2 0[OH ] 2 [Ba(OH) ] 2 1.5 10 mol L 2.9 10 mol L
− − −= × = × × ⋅ = × ⋅− − 
SM-276 
14
13 1w
3 2
1.0 10[H O ] 3.4 10 mol L
[OH ] 2.9 10
K −+ −
− −
× −= = = ×× ⋅ 
 
10.19 Because if σε δ Taking the antilogs 
of both sides gives 
3 3pH log[H O ], log[H O ] pH.
+ += − = −
pH 1
3[H O ] 10 mol L
+ − −= ⋅ 
(a) 3.3 4 13[H O ] 10 5 10 mol L
+ − −= = × ⋅ −
−⋅
−⋅
−⋅
(b) 6.7 1 7 13[H O ] 10 mol L 2 10 mol L
+ − − −= ⋅ = ×
(c) 4.4 1 5 13[H O ] 10 mol L 4 10 mol L
+ − − −= ⋅ = ×
(d) 5.3 1 6 13[H O ] 10 mol L 5 10 mol L
+ − − −= ⋅ = ×
 
10.21 (a) 13 3[HNO ] [H O ] 0.0146 mol L
+ −= = ⋅
pH log(0.0146) 1.84, pOH 14.00 ( 1.84) 12.16= − = = − − = 
(b) 13[HCl] [H O ] 0.11 mol L
+ −= = ⋅
pH log(0.11) 0.96, pOH 14.00 0.96 13.04= − = = − = 
(c) M 2[OH ] 2 [Ba(OH) ] 2 0.0092
− = × = × 10.018 mol L−= ⋅ 
pOH log(0.018) 1.74, pH 14.00 1.74 12.26= − = = − = 
(d) 0[KOH] [OH ]
−=
1 42.00 mL[OH ] (0.175 mol L ) 7.0 10 mol L
500 mL
− −⎛ ⎞= × ⋅ = ×⎜ ⎟⎝ ⎠
1− −⋅ 
4pOH log(7.0 10 ) 3.15, pH 14.00 3.15 10.85−= − × = = − = 
(e) 0[NaOH] [OH ]
−=
4
1
0.0136 gnumber of moles of NaOH 3.40 10 mol
40.00 g mol
−
−= = ×⋅ 
4
4 1
0
3.40 10 mol[NaOH] 9.71 10 mol L [OH ]
0.350 L
−
− − −×= = × ⋅ = 
4pOH log(9.71 10 ) 3.01, pH 14.00 3.01 10.99−= − × = = − = 
(f) 0 3[HBr] [H O ]
+=
SM-277 
4 1 5
3
75.0 mL[H O ] (3.5 10 mol L ) 5.3 10 mol L
500 mL
+ − −⎛ ⎞ 1− −= × × ⋅ = × ⋅⎜ ⎟⎝ ⎠
 
5pH log(5.3 10 ) 4.28, pOH 14.00 4.28 9.72−= − × = = − = 
 
10.23 therefore, after taking antilogs, a1 a1p log ;K = − K a1pa1 10 KK −= 
 Acid a1pK a1K
 (a) 2.12 3H PO4
37.6 10−× 
 (b) 2.00 3H PO3
21.0 10−× 
 (c) 2.46 2H SeO3
33.5 10−× 
 (d) 1.92 4HSeO
21.2 10−× 
(e) The larger , the stronger the acid; therefore a1K
2 3 3 4 3 3 4H SeO H PO H PO HSeO
−< < < 
 
10.25 (a) 
 
2 2 3 2HClO (aq) H O(l) H O (aq) ClO (aq)
+ −+ +U
3 2
a
2
[H O ][ClO ]
[HClO ]
K
+ −
= 
 
 
2 2 2ClO (aq) H O(l) HClO (aq) OH (aq)
− −+ +U
2
b
2
[HClO ][OH ]
[ClO ]
K
−
−= 
 (b) 
 
2 3HCN(aq) H O(l) H O (aq) CN (aq)
+ −+ +U
3
a
[H O ][CN ]
[HCN]
K
+ −
= 
 
 
2CN (aq) H O(l) HCN(aq) OH (aq)
− −+ +U
b
[HCN][OH ]
[CN ]
K
−
−= 
SM-278 
 (c) 
 
6 5 2 3 6 5C H OH(aq) H O(l) H O (aq) C H O (aq)
+ −+ +U
3 6 5
a
6 5
[H O ][C H O ]
[C H OH]
K
+ −
= 
 
 
6 5 2 6 5C H O (aq) H O(l) C H OH(aq) OH (aq)
− −+ +U
6 5
b
6 5
[C H OH][OH ]
[C H O ]
K
−
−= 
 
10.27 Decreasing will correspond to increasing acid strength because 
. The values (given in parentheses) determine the 
following ordering: 
apK
a ap logK K= − apK
 3 2 2 3(CH ) NH (14.00 3.27 10.73) NH OH (14.00 7.97 6.03)
+ +− = < − =
 2 2HNO (3.37) HClO (2.00).< <
Remember that the for the conjugate acid of a weak base will be 
given by 
apK
a bp p 1K K+ = 4.
 
10.29 Decreasing bpK will correspond to increasing base strength because 
b bp log .K K= − The bpK values (given in parentheses) determine the 
following ordering: 
3
5 5 3
(14.00 3.45 10.55) CH COO (14.00 4.75 9.25)
C H N (8.75) NH (4.75).
F − −− = < − =
< << 
Remember that the bpK for the conjugate base of a weak acid will be 
given by a bp p 1K K+ = 4.
 
10.31 Any acid whose conjugate base lies above water in Table 10.3 will be a 
strong acid; that is, the conjugate base of the acid will be a weaker base 
than water, and so water will accept the H+ preferentially. Based upon 
this information, we obtain the following analysis: (a) , strong; 
(b) weak; (c) 
3HClO
2H S, 4HSO
− , weak (Note: even though is a 2H SO4
SM-279 
strong acid, is a weak acid. Its conjugate base is ); (d) 
 weak acid;(e) 
4HSO
− 2
4SO
−
3 3CH NH ,
+
3HCO ,
− weak; (f) strong; (g) 
 weak. 
3HNO ,
4CH ,
 
10.33 For oxoacids, the greater the number of highly electronegative O atoms 
attached to the central atom, the stronger the acid. This effect is related to 
the increased oxidation number of the central atom as the number of O 
atoms increases. Therefore, is the stronger acid, with the lower 3HIO ap .K
 
10.35 (a) HCl is the stronger acid, because its bond strength is much weaker 
than the bond in HF, and bond strength is the dominant factor in 
determining the strength of binary acids. 
(b) is stronger; there is one more O atom attached to the Cl atom 
in than in HClO. The additional O in helps to pull the 
electron of the H atom out of the H—O bond. The oxidation state of Cl is 
higher in than in HClO. 
2HClO
2HClO 2HClO
2HClO
(c) is stronger; Cl has a greater electronegativity than Br, making 
the H—O bond more polar than in 
2HClO
2HClO 2HBrO .
(d) is stronger; Cl has a greater electronegativity than P. 4HClO
(e) is stronger. The explanation is the same as that for part (b). 
 has one more O atom. 
3HNO
3HNO
(f) is stronger; C has greater electronegativity than Ge. See part 
(c). 
2H CO3
 
10.37 (a) The — group that is bonded to the carboxyl group, —COOH, in 
trichloroacetic acid, is more electron withdrawing than the — group 
in acetic acid. Thus, trichloroacetic acid is the stronger acid. 
3CCl
3CH
SM-280 
(b) The — group in acetic acid has electron-donating properties, 
which means that it is less electron withdrawing than the —H attached to 
the carboxyl group in formic acid, HCOOH. Thus, formic acid is a slightly 
stronger acid than acetic acid. However, it is not nearly as strong as 
trichloroacetic acid. The order is 
3CH
3 3CCl COOH HCOOH CH COOH.>> >
 
10.39 (a) Nitrous acid is a stronger acid than acetic acid and, therefore, the 
acetate ion is a stronger base than the nitrite ion. Since the definition of a 
strong acid is one that favors products in the deprotonation reaction, the 
presence of acetate ions will shift the nitrous acid deprotonation reaction 
toward products by consuming protons making nitrous acid behave like a 
strong acid. Carbonic acid is a weaker acid and, therefore, will not behave 
like a strong acid in the presence of acetic acid. 
(b) Ammonia will act like a strong base because it’s conjugate acid, the 
ammonium ion, is a weaker acid than acetic acid. The presence of acetic 
acid will shift the reaction: 3 2 4NH H O NH OH
+ −+ U + toward products. 
 
10.41 The larger the the stronger the corresponding acid. 2,4,6-
Trichlorophenol is the stronger acid because the chlorine atoms have a 
greater electron-withdrawing power than the hydrogen atoms present in 
the unsubstituted phenol. 
a ,K
 
10.43 The larger the of an acid, the stronger the corresponding conjugate 
base; hence, the order is aniline < ammonia < methylamine < ethylamine. 
Although we should not draw conclusions from such a small data set, we 
might suggest the possibility that 
apK
(1) arylamines < ammonia < alkylamines 
(2) methyl < ethyl < etc. 
(Arylamines are amines in which the nitrogen of the amine is attached to a 
benzene ring.) 
SM-281 
 
10.45 (a) Concentration 
 1 3 2 3(mol L ) CH COOH H O H O CH CO
− −⋅ + +U 3 2
 initial — 0 0 0.29
 change x− — x+ x+ 
 equilibrium 0.29 x− — x x
2 2
5 3 3 2
a
3
3 1
3
3
[H O ][CH CO ]
1.8 10
[CH COOH] 0.29 0.29
[H O ] 2.3 10 mol L
pH log(1.6 10 ) 2.64, pOH 14.00 2.64 11.36
x xK
x
x
+ −
−
+ − −
−
= × = = ≈−
= = × ⋅
= − × = = − =
 
(b) The equilibrium table for (b) is similar to that for (a). 
2
1 3 3 2
a
3
2 1
1 1 2
[H O ][CCl CO ]
3.0 10
[CCl COOH] 0.29
or 3.0 10 0.087 0
3.0 10 (3.0 10 ) (4) ( 0.087)
0.18, 0.48
2
xK
x
x x
x
+ −
−
−
− −
= × = = −
+ × − =
− × ± × − −= = −
2
 
The negative root is not possible and can be eliminated. 
1
3[H O ] 0.18 mol L
pH log(0.18) 0.74, pOH 14.00 0.74 13.26
x + −= = ⋅
= − = = − = 
 (c) Concentration 1 2 3(mol L ) HCOOH H O H O HCO
− + −⋅ + +U 
 initial — 0 0.29 0
 change x− — x+ x+
 equilibrium 0.29 x− — x x
2
43 2
a
4 3 1
3
3
[H O ][HCO ]
1.8 10
[HCOOH] 0.29 0.29
[H O ] 0.29 1.8 10 7.2 10 mol L
pH log(7.2 10 ) 2.14, pOH 14.00 2.14 11.86
x x xK
x
x
+ −
−
+ − −
−
⋅= = ≈ = ×−
= = × × = × ⋅
= − × = = − =
− 
 
 
 
SM-282 
10.47 (a) Concentration 
 1 2 3 4(mol L ) H O NH NH OH
− +⋅ + +U −
 initial — 0.057 0 0
 change — x− x+ x+
 equilibrium — 0.057 x− x x
2
54
b
3
5 3 1
3
3
[NH ][OH ]
1.8 10
[NH ] 0.057 0.057
[OH ] 0.057 1.8 10 1.0 10 mol L
pOH log(1.0 10 ) 3.00, pH 14.00 3.00 11.00
1.0 10percentage protonation 100% 1.8%
0.057
x x xK
x
x
+ −
−
− − −
−
−
⋅= = ≈ = ×−
= = × × = × ⋅
= − × = = − =
×= × =
−
−+
 
(b) Concentration 
 1 2 2 3(mol L ) NH OH H O NH OH OH
− +⋅ + U
 initial 0.162 — 0 0
 change x− — x+ x+
 equilibrium 0.162 x− — x x
2 2
8
b
5 1
5
5
1.1 10
0.162 0.162
[OH ] 4.2 10 mol L
pOH log(4.2 10 ) 4.38, pH 14.00 4.38 9.62
4.2 10percentage protonation 100% 0.026%
0.162
x xK
x
x
−
− − −
−
−
= × = ≈−
= = × ⋅
= − × = = − =
×= × =
 
(c) Concentration 
 1 3 3 2 3 3(mol L ) (CH ) N H O (CH ) NH OH
− + −⋅ + U +
 initial — 0.35 0 0
 change x− — x+ x+
 equilibrium 0.35 x− — x+ x+
SM-283 
2
5
3 1
3 1
3
3
6.5 10
0.35
Assume 0.35
Then 4.8 10 mol L
[OH ] 4.8 10 mol L
pOH log(4.8 10 ) 2.32, pH 14.00 2.32 11.68
4.8 10percentage protonation 100% 1.4%
0.35
x
x
x
x
−
− −
− − −
−
−
× = −
<<
= × ⋅
= × ⋅
= − × = = − =
×= × =
 
(d) 6b a bp 14.00 p 14.00 8.21 5.79, 1.6 10K K K
−= − = − = = ×
2
2 2
6
b
4 1
4
4
codeine H O codeineH OH
1.6 10
0.073 0.073
[OH ] 1.1 10 mol L
pOH log(1.1 10 ) 3.96, pH 14.00 3.96 10.04
1.1 10percentage protonation 100% 2.5%
0.0073
x xK
x
x
+ −
−
− − −
−
−
+ +
= × = ≈−
= = × ⋅
= − × = = − =
×= × =
L
 
 
10.49 (a) 2 2 3 2HClO H O H O ClO
+ −+ → + 
pH 1.2
3 2
2
3 2
a
2
a
[H O ] [ClO ] 10 10 0.06 mol L
[H O ][ClO ] (0.06) 0.09 (1 sf )
[HClO ] 0.10 0.06
p log(0.09) 1.0
K
K
+ − − −
+ −
1−= = = = ⋅
= = =−
= − =
 
(b) +3 7 2 2 3 7 3C H NH H O C H NH +OH
−+ L 
2.14 3 1
3 7 3
3 2
43 7 3
b 3
3 7 2
4
b
pOH 14.00 11.86 2.14
[C H NH ] [OH ] 10 7.2 10 mol L
[C H NH ][OH ] (7.2 10 ) 5.6 10
[C H NH ] 0.10 7.2 10
p log(5.6 10 ) 3.25
K
K
+ − − − −
+ − −
−
−
−
= − =
= = = × ⋅
×= = =− ×
= − × =
× 
 
10.51 (a) pH 4.60 53pH 4.60, [H O ] 10 10 2.5 10 mol L
1+ − − −= = = = × −⋅ 
Let nominal concentration of HClO, thenx = 
SM-284 
 Concentration 
 + 1(mol L )−⋅ HClO 2 3H O H O ClO+ −+U 
 nominal — x 0 0
 equilibrium 52.5 10x −− × — 52.5 10−× 52.5 10−×
 
5 2
8
a 5
(2.5 10 )3.0 10
2.5 10
K
x
−
−
−
×= × = − × 
5 2 5 8
8
2 1 1
(2.5 10 ) (2.5 10 ) (3.0 10 )Solve for ;
3.0 10
2.1 10 mol L 0.021 mol L
x x
− −
−
− − −
× + × ×= ×
= × ⋅ = ⋅
−
 
(b) pOH 14.00 pH 14.00 10.20 3.80= − = − =
pOH 3.80 4
2 2
[OH ] 10 10 1.6 10
Let nominal concentration of NH NH , thenx
− − − −= = = ×
= 
 Concentration
 1(mol L )−⋅ 2 2 2 2 3NH NH H O NH NH OH+ −+ +U 
 nominal — 0 0 x
 equilibrium41.6 10x −− × — 41.6 10−× 41.6 10−×
4 2
6
b 4
2 1
(1.6 10 )1.7 10
1.6 10
Solve for ; 1.5 10 mol L
K
x
x x
−
−
−
− −
×= × = − ×
= × ⋅
 
 
10.53 Concentration 
1
6 5 2 3 6 5 2(mol L ) C H COOH H O H O C H CO
− +⋅ + +U − 
 initial 0. — 0 110 0
 change x− — x+ x+
 equilibrium 0.110 x− — 
 
x x
2H O octylamine octylamineH OH
+ −+ +U 
SM-285 
1
3 6 5 2
2
53 6 5
a
6 5
3
0.024 0.110 mol L [H O ] [C H CO ]
[H O ][C H COO ] (0.024 0.110) 6.3 10
[C H COOH] (1 0.024) 0.110
pH log(2.6 10 ) 2.58
x
K
− + −
+ −
−
−
= × ⋅ = =
×= = =− ×
= − × =
× 
 
10.55 The change in the concentration of octylamine is 
 10.067 0.10 0.0067 mol L .x −= × = ⋅ Thus the equilibrium table is 
Concentration 
1(mol L )−⋅ 2H O octylamine octylamineH OH+ −+ +U 
 initial — 0.100 0 0
 change — 0.0067− 0.0067+ 0.0067+
 equilibrium — 0.100 0.0067− 0.0067 0.0067
The equilibrium concentrations are 
[octylamine] 10.100 0.067 0.10 0.093 mol L−= − × = ⋅ 
1
3 2
4
b
[OH ] [octylamineH ] 0.0067 mol L
pOH log(0.0067) 2.17, pH 14.00 2.17 11.83
[octylamineH ][OH ] (6.7 10 ) 4.8 10
[octylamine] 0.093
K
− + −
+ − −
−
= = ⋅
= − = = − =
×= = = ×
 
 
10.57 - +3 2 2 3 2 3CH CH COOH(aq) H O(1) CH CH COO (aq) H O (aq)+ +U
Concentration 1(mol L )−⋅
 initial 0.0147 — 0 0 
 change x− — x+ x+ 
 equilibrium0.0147 x− — x x
+ 2 2
53 2 3
a
3 2
[CH CH COO ][H O ]
1.3 10
[CH CH COOH] 0.0147 0.0147
x xK
x
−
−= = × = − ≈ 
4
4 4.4 104.4 10 , Percent deprotonated 100% 3.0%
0.0147
x
−
− ×= × = × = 
 
SM-286 
10.59 (a) 4 2 3 3less than 7, NH (aq) H O(l) H O (aq) NH (aq)
+ ++ +U
(b) 22 3 3greater than 7, H O(l) CO (aq) HCO (aq) OH (aq)
− −+ +U −
+
+
(c) 2greater than 7, H O(l) F (aq) HF(aq) OH (aq)
− −+ +U
(d) neutral 
(e) less than 7,
3 2
2 6 2 3 2 5Al(H O) (aq) H O(l) H O (aq) Al(H O) OH (aq)
+ ++ +U 
(f) less than 7,
2
2 6 2 3 2 5Cu(H O) (aq) H O(l) H O (aq) Cu(H O) OH (aq)
+ ++ +U 
 
10.61 (a) 
14
10w
b 5
a
1.00 10 5.6 10
1.8 10
K
K
K
−
−
−
×= = = ×× 
Concentration
 1 3 2 2 3 2(mol L ) CH CO (aq) H O(1) HCH CO (aq) OH (aq)
− − −⋅ + +U
initial 0.63 — 0 0 
 change x− — x+ x+
 equilibrium 0.63 x− — x x
 
2 2
103 2
b
3 2
[HCH CO ][OH ]
5.6 10
0.63 0.63[CH CO ]
x xK
x
−
−
−= = × = − ≈ 
5 51.9 10 [OH ], pOH log(1.9 10 ) 4.72x − − −= × = = − × = 
pH 14.00 pOH 14.00 4.72 9.28= − = − = 
(b) 
14
10w
a 15
b
1.00 10 5.6 10
1.8 10
K
K
K
−
−
−
×= = = ×× 
1
4 2 3Concentration (mol L ) NH (aq) H O(l) H O (aq) NH (aq)
− + +⋅ + +U 3
 initial 0.19 — 0 0 
 change x− — x+ x+
 equilibrium 0.19 x− — x x
2 2
103 3
a
4
[H O ][NH ]
5.6 10
[NH Cl] 0.19 0.19
x xK
x
+
−= = × = − ≈ 
SM-287 
5 1
31.0 10 mol L [H O ]x
− −= × ⋅ = +
+
 
5pH log(1.0 10 ) 5.00−= − × = 
 (c) Concentration 
1 3 2
2 6 2 3 2 5(mol L ) Al(H O) (aq) H O(l) H O (aq) Al(H O) OH (aq)
− + +⋅ + +U 
 initial 0.055 — 0 0 
 change x− — x+ x+ 
 equilibrium 0.055 x− — x x
2 2 2
53 2 5
a 3
2 6
[H O ][Al(H O) OH ]
1.4 10
0.055 0.055[Al(H O) ]
x xK
x
+ +
−
+= = × = − ≈
+
−
 
4 1
38.8 10 mol L [H O ]x
− −= × ⋅ = 
4pH log(8.8 10 ) 3.06−= − × = 
(d) Concentration 
 1 2(mol L ) H O(l) CN (aq) HCN(aq) OH (aq)
− −⋅ + +U
 initial — 0.065 0 0 
 change — x− x+ x+
 equilibrium — 0.065 x− x x
14 2 2
5w
b 10
a
1.00 10 [HCN][OH ]2.0 10
0.065 0.0654.9 10 [CN ]
K x xK
K x
− −
−
− −
×= = = × = = ≈−× 
3 1[OH ] 1.1 10 mol Lx − −= = × ⋅ − 
3pOH log(1.1 10 ) 2.96, pH 11.04−= − × = = 
 
10.63 Concentration 
1
3 3 2 3 3 2(mol L ) CH NH (aq) H O(l) H O (aq) CH NH (aq)
− + +⋅ + +U 
 initial 0.510 — 0 0 
 change x− — x+ x+
 equilibrium 0.510 x− — x x
SM-288 
2 2
113 3 2
a
3 3
[H O ][CH NH ]
2.8 10
0.510 0.510[CH NH ]
x xK
x
+
−
+= = × = − ≈
+
 
6 1
33.8 10 mol L [H O ]x
− −= × ⋅ = 
6pH log(3.8 10 ) 5.4−= − × = 
 
10.65 (a) 250 mL of solution contains , molar 
mass= 
2 3 25.34 g KC H O
198.14 g mol−⋅
2 3 2
2 3 2 2 3 2
2 3 2
1 mol KC H O 1(5.34 g KC H O ) 0218 KC H O
98.14 g KC H O 0.250 L
M⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
 
Concentration 
1
2 2 3 2 2 3 2(mol L ) H O(l) C H O (aq) HC H O (aq) OH (aq)
− −⋅ + +U − 
 initial — 0.218 0 0 
 change — x− x+ x+
 equilibrium — 0.218 x− x x
14 2 2
5
1.0 10
0.218 0.2181.8 10
x x
x
−
−
× = ≈−× 
5 1[OH ] 1.1 10 mol L− −= × ⋅ −
−
 
10 1
3[H O ] 9.1 10 mol L
+ −= × ⋅ 
10pH log(9.1 10 ) 9.04−= − × = 
(b) 100 mL of solution contains molar mass = 
 
45.75 g NH Br,
197.95 g mol−⋅
4
4 4
4
M
1 mol NH Br 1(5.75 g NH Br) 0.587 NH Br
97.95 g NH Br 0.100 L
⎛ ⎞⎛ ⎞ =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠
 
Concentration 
1
4 2 3 3(mol L ) NH (aq) H O(l) NH (aq) H O (aq)
− + +⋅ + +U 
 initial 0.587 — 0 0 
 change x− — x+ x+ 
SM-289 
 equilibrium 0.587 x− — x x
14 2 2
5
1.0 10
0.587 0.5871.8 10
x x
x
−
−
× = ≈−× 
5 1
3[H O ] 1.8 10 mol L
+ − −= × ⋅ 
5pH log(1.8 10 ) 4.74−= − × = 
 
10.67 (a) 
1
13 20.020 mol L NaCH CO 0.150 L 0.0400 mol L
0.500 L
−
−⋅ × = ⋅ 
Concentration 
1(mol L )−⋅ 2 3 2 3H O(l) CH CO (aq) CH COOH(aq) OH (aq)− −+ +U
 initial — 0.0400 0 0 
 change — x− x+ x+
 equilibrium — 0.0400 x− x x
14
10w 3
b 5
a 3 2
[CH COOH][OH ]1.00 10 5.6 10
1.8 10 [CH CO ]
K
K
K
−−
−
− −
×= = = × =× 
2 2
105.6 10
0.0400 0.0400
x x
x
−× = ≈− 
6 1
34.7 10 mol L [CH COOH]x
− −= × ⋅ = 
(b) 4 43
4
2.16 g NH Br 1 mol NH Br1 mL
400 mL 97.95 g NH Br10 L−
⎛ ⎞⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠
 
1
40.0551 (mol NH Br) L
−= ⋅ 
 Concentration 
 1(mol L )−⋅ 4 2 3 3NH (aq) H O(l) H O (aq) NH (aq)+ ++ +U
 initial 0.0551 — 0 0 
 change x− — x+ x+ 
 equilibrium 0.0551 x− — x x
14
10w 3
a 5
b 4
[NH ][H O ]1.00 10 5.6 10
1.8 10 [NH ]
K
K
K
3
+−
−
− +
×= = = × =× 
SM-290 
2 2
105.6 10
0.0551 0.0551
x x
x
−× = ≈− 
6 1
35.5 10 mol L [H O ]x
− −= × ⋅ = + and 6pH log(5.5 10 ) 5.26−= − × =
 
10.69 (a) 
C C C
H
H
H
N
H
H
O
O
Na+ 
 
 
 (b) 
1 2
1 1
2 2pH = + (p p ) (2.34 9.89) 6.1a aK K = + =
−
−
−
−
 
10.71 (a) 2 4 2 3 4H SO (aq) H O(l) H O (aq) HSO (aq)
+ −+ +U
2
4 2 3 4HSO (aq) H O(l) H O (aq) SO (aq)
− ++ +U 
(b) 3 4 2 3 2 4H AsO (aq) H O(l) H O (aq) H AsO (aq)
+ −+ +U
2
2 4 2 3 4
2 3
4 2 3 4
H AsO (aq) H O(l) H O (aq) HAsO (aq)
HAsO (aq) H O(l) H O (aq) AsO (aq)
− +
− +
+ +
+ +
U
U 
(c) 
 6 4 2 2 3 6 4 2C H (COOH) (aq) H O(l) H O (aq) C H (COOH)CO (aq)
+ −+ +U
2
6 4 2 2 3 6 4 2 2C H (COOH)CO (aq) H O(l) H O (aq) C H (CO ) (aq)
− ++ +U 
 
10.73 The initial concentrations of 14 3HSO and H O are both 0.15 mol L
− + −⋅ as a 
result of the complete ionization of in the first step. The second 
ionization is incomplete. 
2H SO4
4
−
Concentration 
1 2
4 2 3(mol L ) HSO H O H O SO
− − +⋅ + +U 
 initial — 0.15 0.15 0
 change x− — x+ x+
 equilibrium 0.15 x− — 0.15 x+ x
SM-291 
2
2 3 4
a2
4
2 3
2 3
1
1 1
3
[H O ][SO ] (0.15 ) ( )1.2 10
0.15[HSO ]
0.162 1.8 10 0
0.162 (0.162) (4) (1.8 10 )
0.0104 mol L
2
[H O ] 0.15 (0.15 0.0104) mol L 0.16 molL
pH log(0.16) 0.80
x xK
x
x x
x
x
+ −
−
−
−
−
−
+ −
+= × = = −
+ − × =
− + + ×= =
−
⋅
= + = + ⋅ = ⋅
= − =
 
 
10.75 (a) Because the second ionization can be ignored. a2 a1 ,K K<<
 Concentration 
1
2 3 2 3 3(mol L ) H CO H O H O HCO
− +⋅ + +U − 
 initial — 0 0.010 0
 change x− — x+ x+
 equilibrium 0.010 x− — x x
2 2
73 3
a1
2 3
5 1
3
5
[H O ][HCO ]
4.3 10
[H CO ] 0.010 0.010
[H O ] 6.6 10 mol L
pH log(6.6 10 ) 4.18
x xK
x
x
+ −
−
+ − −
−
= = ≈ =−
= = × ⋅
= − × =
×
2
−
 
(b) Because the second ionization can be ignored. a2 a1 ,K K<<
Concentration 
 1 2 2 3(mol L ) (COOH) H O H O (COOH)CO
− +⋅ + +U
 initial 0.10 — 0 0 
 change x− — x+ x+
 equilibrium 0.10 x− — x x
2
2 3 2
a1
2
2 2 3
2 2 2 3
1
[H O ][(COOH)CO ]
5.9 10
[(COOH) ] 0.10
5.9 10 5.9 10 0
5.9 10 (5.9 10 ) (4)(5.9 10 )
0.053 mol L
2
pH log(0.053) 1.28
xK
x
x x
x
+ −
−
− −
− − −
−
= × = = −
+ × − × =
− × + × + ×= =
= − =
⋅
 
SM-292 
(c) Because the second ionization can be ignored. a2 a1 ,K K<<
Concentration 1 2 2 3(mol L ) H S H O H O HS
− + −⋅ + +U 
 equilibrium 0.20 x− — x x
2 2
7 3
a1
2
4 1
3
4
[H O ][HS ]
1.3 10
[H S] 0.20 0.20
[H O ] 1.6 10 mol L
pH log(1.6 10 ) 3.80
x xK
x
x
+ −
−
+ − −
−
= × = = ≈−
= = × ⋅
= − × =
 
 
10.77 (a) The pH is given by 1 a1 a22pH (p p ).K K= + From Table 10.9, we find 
2
a1 a1
7
a2 a2
1
2
1.5 10 p 1.82
1.2 10 p 6.92
pH (1.82 6.92) 4.37
K K
K K
−
−
= × =
= × =
= + =
 
(b) The pH of a salt solution of a polyprotic acid is independent of the 
concentration of the salt, therefore pH 4.37.= 
 
10.79 (a) The pH is given by 1 a1 a22pH (p p ).K K= + For the monosodium salt, 
the pertinent values are a1 a2p and pK K :
1
2pH (3.14 5.95) 4.55= + = 
(b) For the disodium salt, the pertinent values are a2 a3p and pK K :
1
2pH (5.95 6.39) 6.17= + = 
 
10.81 The equilibrium reactions of interest are 
7
2 3 2 3 3 a1
2 1
3 2 3 3 a2
H CO (aq) H O(l) H O (aq) HCO (aq) 4.3 10
HCO (aq) H O(l) H O (aq) CO (aq) 5.6 10
K
K 1
+ − −
− + −
+ + =
+ + =
U
U −
×
× 
Because the second ionization constant is much smaller than the first, we 
can assume that the first step dominates: 
 
 
SM-293 
Concentration 
 1 2 3 2 3 3(mol L ) H CO (aq) H O(l) H O (aq) HCO (aq)
− +⋅ + +U −
initial 0.0456 — 0 0 
 change x− — x+ x+
 final 0.0456 x− — x+ x+
3 3
a1
2 3
2
7
2 7
4
4 1
3 3
[H O ][HCO ]
[H CO ]
( ) ( )4.3 10
0.0456 0.0456
Assume that 0.0456
Then (4.3 10 ) (0.0456)
1.4 10
Because 1% of 0.0456, the assumption was valid.
[H O ] [HCO ] 1.4 10 mol L
K
x x x
x x
x
x
x
x
x
+ −
−
−
−
+ − − −
=
× = =− −
<<
= ×
= ×
<
= = = × ⋅
 
This means that the concentration of 12 3H CO is 0.0456 mol L
−⋅ −
10.00014 mol L 0.0455 mol L .−⋅ = ⋅ 1− We can then use the other equilibria 
to determine the remaining concentrations: 
2
3 3
a2
3
4 2
11 3
4
2 11
3
[H O ][CO ]
[HCO ]
(1.4 10 )[CO ]
5.6 10
(1.4 10 )
[CO ] 5.6 10 mol L
K
+ −
−
1
− −
−
−
− −
=
×× = ×
= × ⋅ −
−
 
Because the initial assumption that the first 
ionization would dominate is valid. 
11 45.6 10 1.4 10 ,−× << ×
To calculate [ we use the relationship: OH ],− wK
w 3
14
11 1w
4
3
1
2 3 3 3
4 1
2 11 1 11
3
[H O ][OH ]
1.00 10[OH ] 7.1 10 mol L
[H O ] 1.4 10
In summary, [H CO ] 0.0455 mol L , [H O ] [HCO ]
 1.4 10 mol L ,
[CO ] 5.6 10 mol L , [OH ] 7.1 10 mol L .
K
K
+ −
−
− −
+ −
1
−
− + −
− −
− − − − −
=
×= = = × ⋅×
= ⋅ =
= × ⋅
= × ⋅ = × ⋅ −
 
SM-294 
10.83 The equilibrium reactions of interest are now the base forms of the 
carbonic acid equilibria, so bK values should be calculated for the 
following changes: 
2
3 2 3
14
4w
b1 11
a2
CO (aq) H O(l) HCO (aq) OH (aq)
1.00 10 1.8 10
5.6 10
K
K
K
− − −
−
−
−
+ +
×= = = ××
U
 
3 2 2 3
14
8w
b2 7
a1
HCO (aq) H O(l) H CO (aq) OH (aq)
1.00 10 2.3 10
4.3 10
K
K
K
− −
−
−
−
+ +
×= = = ××
L
 
Because the second hydrolysis constant is much smaller than the first, we 
can assume that the first step dominates: 
Concentration 
1 2
3 2 3(mol L ) CO (aq) H O(l) HCO (aq) OH (aq)
− − − −⋅ + +U 
 initial 0.0456 — 0 0 
 change — x− x+ x+
 final 0.0456 x− — x+ x+
3
b1 2
3
[HCO ][OH ]
[CO ]
K
− −
−= 
2
4 ( ) ( )1.8 10
0.0456 [0.0456 ]
x x x
x x
−× = =− − 
Assume that 0.0456x <<
Then 2 4(1.8 10 ) (0.0456)x −= ×
32.9 10x −= × 
Because of 0.0456, the assumption was not valid and the full 
expression should be solved using the quadratic equation: 
5%x >
2 4 41.8 10 (1.8 10 ) (0.0456) 0x x− −+ × − × = 
Solving using the quadratic equation gives 10.0028 mol L .x −= ⋅ 
1
3[HCO ] [OH ] 0.0028 mol Lx
− −= = = −⋅
1−
 
Therefore, 2 1 13[CO ] 0.0456 mol L 0.0028 mol L 0.0428 mol L
− − −= ⋅ − ⋅ = ⋅
SM-295 
We can then use the other equilibria to determine the remaining 
concentrations: 
2 3
b2
3
[H CO ][OH ]
[HCO ]
K
−
−= 
8 2 3[H CO ](0.0028)2.3 10
(0.0028)
−× = 
8 1
2 3[H CO ] 2.3 10 mol L
− −= × ⋅ 
Because , the initial assumption that the first 
hydrolysis would dominate is valid. To calculate 
82.3 10 0.0028−× <<
3[H O ]
+ , we use the 
relationship: 
wK
w 3[H O ][OH ]K
+ −= 
14
12 1w
3
1.00 10[H O ] 3.6 10 mol L
0.0028[OH ]
K −+ −
−
× −= = = × ⋅ 
In summary, 8 12 3 3[H CO ] 2.3 10 mol L , [OH ] [HCO ]
− − − −= × ⋅ = =
1−⋅
,
 
1 2 1 12
3 30.0028 mol L , [CO ] 0.0428 mol L , [H O ] 3.6 10 mol L
− − − + −⋅ = ⋅ = × 
 
10.85 (a) phosphorous acid: The two values are 2.00 and 6.59. Because 
 lies between the dominant form will be the 
singly deprotonated HA
apK
pH 6.30= a1 a2p and pK K
− ion. 
(b) oxalic acid: The two values are 1.23 and 4.19. Because 
 lies above the species present in largest concentration 
will be the doubly deprotonated 
apK
pH 6.30= a 2pK ,
2A − ion. 
(c) hydrosulfuric acid: The two values are 6.89 and 14.15. Because 
 lies below both values, the species present in highest 
concentrations will be the fully protonated form. 
apK
pH 6.30= apK
2H A
 
10.87 The equilibria present in solution are 
SM-296 
2
2 3 2 3 3 a1
2 7
3 2 3 3 a2
H SO (aq) H O(l) H O (aq) HSO (aq) 1.5 10
HSO (aq) H O(l) H O (aq) SO (aq) 1.2 10
K
K
+ − −
− + −
+ + =
+ + =
U
U −
×
×
3
2−
−
 
The calculation of the desired concentrations follows exactly after the 
method derived in Eq. 25, substituting for 
 First, calculate the 
quantity 
2H SO
2
2 3 3 3 3 3H CO , HSO for HCO , and SO for CO .
− − −
5.5 6 1
3(at pH 5.50 [H O ] 10 3.2 10 mol L ) :f
+ − − −= = = × ⋅
2
3 3 a1 a1 a 2
6 2 6 2 2 7
8
[H O ] [H O ]
(3.2 10 ) (3.2 10 ) (1.5 10 ) (1.5 10 ) (1.2 10 )
5.0 10
f K K K+ +
− − − −
−
= + +
= × + × × + × ×
= ×
 
The fractions of the species present are then given by 
6 2
43
2 3 8
6 2
3 a1
3 8
2 7
2 a1 a2
3 8
[H O ] (3.2 10 )(H SO ) 2.1 10
5.0 10
[H O ] (3.2 10 ) (1.5 10 )(HSO ) 0.96
5.0 10
(1.5 10 ) (1.2 10 )(SO ) 0.036
5.0 10
f
K
f
K K
f
α
α
α
+ −
−
−
+ − −
−
−
− −
−
−
×= = = ××
× ×= = =×
× ×= = =×
 
Thus, in a 10.150 mol L−⋅ solution at the dominant species will 
be with a concentration of 
pH 5.50,3HSO
− 1 1(0.150 mol L ) (0.96) 0.14 mol L .− −⋅ = ⋅ 
The concentration of 
4 1 5
2 3H SO will be (2.1 10 ) (0.150 mol L ) 3.2 10 mol L
1− − −× ⋅ = × −⋅ and the 
concentration of 23SO
− will be 
1 1(0.036) (0.150 mol L ) 0.0054 mol L .− −⋅ = ⋅ 
 
10.89 (a) Concentration 
 1 3 2 3(mol L ) B(OH) 2 H O H O B(OH)
− +⋅ + +U 4−
 initial 41.0 10−× — 0 0 
 change x− — x+ x+
 equilibrium — 41.0 10 x−× − x x
SM-297 
2 2
10 3 4
a 4 4
3
7 1
3
7
[H O ][B(OH) ]
7.2 10
[B(OH) ] 1.0 10 1.0 10
[H O ] 2.7 10 mol L
pH log(2.7 10 ) 6.57
x xK
x
x
+ −
−
− −
+ − −
−
= × = = ≈× − ×
= = × ⋅
= − × =
 
Note: this value of 3[H O ]
+ is not much different from the value for pure 
water, therefore, it is at the lower limit of safely 
ignoring the contribution to 
71.0 10 mol L ;−× ⋅ 1−
3[H O ]
+ from the autoprotolysis of water. The 
exercise should be solved by simultaneously considering both equilibria. 
Concentration 
1
3 2 3(mol L ) B(OH) 2 H O H O B(OH)
− +⋅ + +U 4 − 
 equilibrium 41.0 10 x−× − — x y 
Concentration 1 2 3(mol L ) 2 H O H O OH
− + −⋅ +U 
 equilibrium — x z
Because there are now two contributions to 3 3[H O ], [H O ]
+ + is no longer 
equal to nor is it equal to [ as in pure water. To avoid a 
cubic equation, will again be ignored relative to 
4[B(OH) ],
− OH ],−
x 4 11.0 10 mol L .− −× ⋅ 
This approximation is justified by the approximate calculation above, and 
because is very small relative to 
concentration of then 
aK
41.0 10 . Let initiala−× =
3B(OH) ,
10 a
a
14
w
7.2 10 or
1.0 10
aKxy xyK y
a x a x
K xz
−
−
= × = ≈ =−
= × =
 
Electroneutrality requires 
wor ; hence, ( ).x y z z x y K xz x x y= + = − = = − 
Substituting for from above: y
SM-298 
a
w
2
a w
2
w a
14 4 10
w a
7 1
3
7
1 0 10 1 0 10 7 2 10
2 9 10 mol L [H O ]
pH log(2.9 10 ) 6.54
aK
x x K
x
x aK K
x K aK
x K aK . . .
x .
− −
− − +
−
⎛ ⎞× − =⎜ ⎟⎝ ⎠
− =
= +
= + = × + × × ×
= × ⋅ =
= − × =
−
 
This value is slightly, but measurably, different from the value 6.57 
obtained by ignoring the contribution to 3[H O ]
+ from water. 
(b) In this case, the second ionization can safely be ignored; a 2 a1.K K<<
Concentration 
1(mol L )−⋅ 3 4 2 3 2 4H PO H O H O H PO+ −+ +U 
 initial 0.015 — 0 0 
 change — x− x+ x+ 
 equilibrium 0.015 – x — x x 
2
3
a1
2 3 4
3 3 2
3
3 1
3
7.6 10
0.015
7.6 10 1.14 10 0
7.6 10 (7.6 10 ) 4.56 10
[H O ]
2
7.5 10 mol L
pH log(7.5 10 ) 2.12
xK
x
x x
x
−
− −
4− − −
+
− −
−
= × = −
+ × − × =
− × + × + ×= =
= × ⋅
= − × =
 
(c) In this case, the second ionization can safely be ignored; a2 a1.K K<<
Concentration 
 1 2 3 2 3 3(mol L ) H SO H O H O HSO
− +⋅ + +U −
 initial 0.1 — 0 0 
 change x− — x+ x+ 
 equilibrium 0.1 x− — x x
SM-299 
2
2
a1
2 2 3
2 2 2 3
1
3
1.5 10
0.10
1.5 10 1.5 10 0
1.5 10 (1.5 10 ) 6.0 10
[H O ] 0.032 mol L
2
pH log(0.032) 1.49
xK
x
x x
x
−
− −
− − −
+ −
= × = −
+ × − × =
− × + × + ×= = = ⋅
= − =
 
 
10.91 The three equilibria involved are: 
1
2
3
+ -
- + 3 3 2
3 4 2 4 3
3 4
+ 2
- 2- + 8 3 4
2 4 4 3 -
2 4
+
2- 3- + 13 3
4 4 3
[H O ][H PO ]
H PO (aq) H PO (aq) H O (aq), 7.6 10
[H PO ]
[H O ][HPO ]
H PO (aq) HPO (aq) H O (aq), 6.2 10
[H PO ]
[H O ][P
HPO (aq) PO (aq) H O (aq), 2.1 10
a
a
a
K
K
K
−
−
−
+ = × =
+ = × =
+ = × =
U
U
U
4
-
3-
4
2-
4
O ]
[HPO ]
We also know that the combined concentration of all the phosphate 
species is: 
[ ] - 2- 3-3 4 2 4 4 4
-pH 2.25 3 -1
 H PO +[H PO ] [HPO ] [PO ] 1.5 10
and the hydronium ion concentration is:
 [H] 10 10 5.62 10 mol L
2−
− −
+ + = ×
= = = × ⋅
 
At this point it is a matter of solving this set of simultaneous equations to 
obtain the concentrations of the phosphate containing species. We start by 
dividing both sides of the equilibrium constant expressions above by the 
given hydronium ion concentration to obtain three ratios: 
- 2-
5 12 4 4 4
- 2
3 4 2 4 4
[H PO ] [HPO ] [PO ]
1.35 , 1.10 10 , and 3.74 10
[H PO ] [H PO ] [HPO ]
− −= × = × =
3-
1
-
4
3 4
 
Through rearrangement and substitution of these three ratios, we can 
obtain the following expressions: 
-
2 4 3 4
2- -5 - -5
4 2 4 3
-5
3 4
3- 11 2- 11 -5
4 4
-16
3 4
[H PO ] 1.35 [H PO ],
[HPO ] 1.10×10 [H PO ] 1.10×10 1.35 [H PO ]
 1.48×10 [H PO ], and
[PO ] 3.74 10 [HPO ] 3.74 10 1.48×10 [H PO ]
 5.54×10 [H PO ].
− −
= ⋅
= ⋅ = ⋅ ⋅
= ⋅
= × ⋅ = × ⋅ ⋅
= ⋅
 
SM-300 
Substituting these expressions back into the sum: 
[ ]
[ ] ( ) ( ) (
[ ]
- 2- 3-
3 4 2 4 4 4
-5 -16
3 4 3 4 3 4 3 4
2
-3
3 4
- -3
2 4 3 4
2- -5 -8
4 3 4
3-
4
H PO +[H PO ] [HPO ] [PO ]
H PO + 1.35 [H PO ] 1.48×10 [H PO ] 5.54×10 [H PO ]
1.5 10
we find:
H PO 6.4×10 ,
[H PO ] 1.35 [H PO ] 8.6×10 ,
[HPO ] 1.48×10 [H PO ] 9.4×10 , and
[PO ]
−
+ + =
= ⋅ + ⋅ + ⋅
= ×
=
= ⋅ =
= ⋅ =
= -16 -183 45.54×10 [H PO ] 3.5×10 .⋅ =
)
=
4
=
4
 
10.93 We can use the relationship derived in the text: 
, in which HA is any strong acid. 23 initial 3 w[H O ] [HA] [H O ] 0K
+ +− −
2 7 1
3 3[H O ] (6.55 10 )[H O ] (1.00 10 ) 0
+ − + −− × − × = 
Solving using the quadratic equation gives 
 73[H O ] 6.70 10 , pH 6.174.
+ −= × =
This value is slightly lower than the value calculated, based on the acid 
concentration alone 7(pH log(6.55 10 ) 6.184).−= − × =
 
10.95 We can use the relationship derived in the text: 
, in which B is any strong base. 23 initial 3 w[H O ] [B] [H O ] 0K
+ ++ −
2 8 1
3 3[H O ] (9.78 10 )[H O ] (1.00 10 ) 0
+ − + −+ × − × = 
Solving using the quadratic equation gives 
 83[H O ] 6.24 10 , pH 7.205.
+ −= × =
This value is higher than the value calculated, based on the base 
concentration alone 8(pOH log(9.78 10 ) 7.009).−= − × =
 
10.97 (a) In the absence of a significant effect due to the autoprotolysis of 
water, the pH values of the 
4
1.00 10
−× M and 61.00 10−× M HBrO 
solutions can be calculated as described earlier. 
SM-301 
For 4 11.00 10 mol L :− −× ⋅
Concentration 
 1(mol L )−⋅ 2 3HBrO(aq) H O(l) H O (aq) BrO (aq)+ −+ +U
 initial — 0 0 41.00 10−×
 change x− — x+ x+
 final 41.00 10 x−× − — x+ x+
3
a
[H O ][BrO ]
[HBrO]
K
+ −
= 
2
9
4 4
( )( )2.0 10
1.00 10 [1.00 10 ]
x x x
x x
−
− −× = =× − × − 
Assume 41.00 10x −<< × 
2 9(2.0 10 ) (1.00 10 )x − −= × × 4 
74.5 10x −= × 
Because 41% of 1.00 10x −< × , the assumption was valid. Given this 
value, the pH is then calculated to be 7log(4.5 10 ) 6.35.−− × =
For 6 11.00 10 mol L :− −× ⋅
Concentration 
1(mol L )−⋅ HBrO(aq) + 2 3H O(l) H O (aq) BrO (aq)+ −+U
 initial 61.00 10−× — 0 0 
 change x− — x+ x+
 final 61.00 10 x−× − — x+ x+
3
a
[H O ][BrO ]
[HBrO]
K
+ −
= 
2
9
6 6
( )( )2.0 10
1.00 10 [1.00 10 ]
x x x
x x
−
− −× = =× − × − 
Assume 61.00 10x −<< × 
2 9
8
(2.0 10 ) (1.00 10 )
4.5 10
x
x
− −
−
= × ×
= ×
6
 
SM-302 
6is 4.5% of 1.00 10 ,x −× so the assumption is less acceptable. The pH is 
calculated to be Because this predicts a basic 
solution, it is not reasonable. 
8log(4.5 10 ) 7.35.−− × =
(b) To calculate the value taking into account the autoprotolysis of water, 
we can use equation(22): 
3 2
a w a initial w a 3( [HA] ) 0, where [H O ].x K x K K x K K x
++ − + ⋅ − ⋅ = = 
To solve the expression, you substitute the values of 
the initial concentration of acid, and 
14
w 1.00 10 ,K
−= ×
9
a 2.0 10K
−= × into this equation and 
then solve the expression either by trial and error or, preferably, using a 
graphing calculator such as the one found on the CD accompanying this 
text. 
Alternatively, you can use a computer program designed to solve 
simultaneous equations. Because the unknowns include 
 you will need four equations. As 
seen in the text, pertinent equations are 
3[H O ], [OH ], [HBrO], and [BrO ],
+ − −
3
a
w 3
3
initial
[H O ][BrO ]
[HBrO]
[H O ][OH ]
[H O ] [OH ] [BrO ]
[HBrO] [HBrO] [BrO ]
K
K
+ −
+ −
+ − −
−
=
=
= +
= +
 
Both methods should produce the same result. 
The values obtained are 
7 1
3
7 1
5 1
8 1
[H O ] 4.6 10 mol L , pH 6.34 (compare to 6.35 obtained in (a))
[BrO ] 4.4 10 mol L
[HBrO] 1.0 10 mol L
[OH ] 2.2 10 mol L
+ − −
− − −
− −
− − −
= × ⋅ =
= × ⋅
≅ × ⋅
= × ⋅
 
Similarly, for 6initial[HBrO] 1.00 10 :
−= ×
SM-303 
7 1
3
8 1
7 1
8 1
[H O ] 1.1 10 mol L , pH 6.96 (compare to 7.32 obtained in (a))
[BrO ] 1.8 10 mol L
[HBrO] 9.8 10 mol L
[OH ] 9.1 10 mol L
+ − −
− − −
− −
− − −
= × ⋅ =
= × ⋅
≅ × ⋅
= × ⋅
 
Note that for the more concentrated solution, the effect of the 
autoprotolysis of water is very small. Notice also that the less concentrated 
solution is more acidic, due to the autoprotolysis of water, than would be 
predicted if this effect were not operating. 
 
10.99 (a) In the absence of a significant effect due to the autoprotolysis of 
water, the pH values of the 
solutions can be calculated as described earlier. 
5 6M M8.50 10 and 7.37 10 HCN− −× ×
For 5 18.50 10 mol L :− −× ⋅
 Concentration 
1
2 3(mol L ) HCN(aq) H O(l) H O (aq) CN (aq)
− +⋅ + +U − 
 initial 58.50 10−× — 0 0 
 change x− — x+ x+
 final 58.50 10 x−× − — x+ x+
[ ]3a
2
10
5 5
5
2 10 5
7
[H O ][CN ]
HCN
( ) ( )4.9 10
8.5 10 [8.5 10 ]
Assume 8.5 10
(4.9 10 ) (8.5 10 )
2.0 10
K
x x x
x x
x
x
x
+ −
−
− −
−
− −
−
=
× = =× − × −
<< ×
= × ×
= ×
 
5Because 1% of 8.50 10 ,x −< × the assumption was valid. Given this 
value, the pH is then calculated to be 7log(2.0 10 ) 6.69.−− × =
6 1For 7.37 10 mol L :− −× ⋅ 
 
 
SM-304 
Concentration 
2 3(mol L ) HCN(aq) H O(l) H O (aq) CN (aq)
− +⋅ + +U − 
 initial 67.37 10−× — 0 0
 change x− — x+ x+
 final 67.37 10 x−× − — x+ x+
3
a
2
10
6 6
6
2 10 6
8
6
[H O ][CN ]
[HCN]
( ) ( )4.9 10
7.37 10 [7.37 10 ]
Assume 7.37 10
(4.9 10 ) (7.37 10 )
6.0 10
is 1% of 7.37 10 , so the assumption is still reasonable. The pH
is then calculated to be log(6.0
K
x x x
x x
x
x
x
x
+ −
−
− −
−
− −
−
−
=
× = =× − × −
<< ×
= × ×
= ×
< ×
− × 810 ) 7.22. This answer is
not reasonable because we know HCN is an acid.
− =
 
(b) To calculate the value, taking into account the autoprotolysis of water, 
we can use equation (21): 
3 2
a w a initial w a 3( [HA] ) 0, where [H O ].x K x K K x K K x
++ − + ⋅ − ⋅ = = 
To solve the expression, you substitute the values of 
the initial concentration of acid, and 
141.00 10 ,wK
−= ×
10
a 4.9 10K
−= × into this equation 
and then solve the expression either by trial and error or, preferably, using 
a graphing calculator such as the one found on the CD accompanying this 
text. 
Alternatively, you can use a computer program designed to solve 
simultaneous equations. Because the unknowns include 
 you will need four equations. As 
seen in the text, the pertinent equations are 
3[H O ], [OH ], [HBrO], and [BrO ],
+ − −
SM-305 
[ ]3a
w 3
3
initial
[H O ][CN ]
HCN
[H O ][OH ]
[H O ] [OH ] [CN ]
[HCN] [HCN] [CN ]
K
K
+ −
+ −
+ − −
−
=
=
= +
= +
 
Both methods should produce the same result. 
For 5[HCN] 8.5 10 mol L ,1− −= × ⋅ the values obtained are 
7 1
3[H O ] 2.3 10 mol L , pH 6.64 (compare to 6.69 obtained in (a))
+ − −= × ⋅ = 
7 1
5 1
8 1
6
initial
7 1
3
8 1
6 1
[CN ] 1.8 10 mol L
[HCN] 8.5 10 mol L
[OH ] 4.4 10 mol L
Similarly, for [HCN] 7.37 10 :
[H O ] 1.2 10 mol L , pH 6.92 (compare to 7.22 obtained in (a))
[CN ] 3.1 10 mol L
[HCN] 7.3 10 mol L
[
− − −
− −
− − −
−
+ − −
− − −
− −
= × ⋅
≅ × ⋅
= × ⋅
= ×
= × ⋅ =
= × ⋅
≅ × ⋅
8 1OH ] 8.6 10 mol L− − −= × ⋅
 
Note that for the more concentrated solution, the effect of the 
autoprotolysis of water is smaller. Notice also that the less concentrated 
solution is more acidic, due to the autoprotolysis of water, than would be 
predicted if this effect were not operating. 
 
10.101 (a) Assuming all sulfur is converted to SO2, the amount of SO2 produced 
is: 
( )( )( )
( )( )
3 -1
-1
-1
2
1.00 10 kg 0.025 1000 g kg
780 mol of S
32.07 g mol
Therefore, 780 mol 64.07 g mol 50,000 g or 50 kg of SO is produced
× ⋅ =⋅
⋅ =
(b) To determine the pH we first must calculate the volume of water in 
which this 50 kg of SO2 is dissolved: 
 ( )( ) 2 2 731000 m 100 cm 1 L2 cm 2.6 Km 5.2 10 L1 Km 1 m 1000 cmV
⎛ ⎞ ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ × 
SM-306 
The concentration of SO2(aq) is then: 
( ) ( )7780 mol 5.2 10 L 1.5 10 M−× = × 5
−
. We can assume that upon 
solution the SO2(aq) is converted to sulfurous acid and the pH is 
calculated as described earlier: 
Concentration 
 1 2 3 2 3 3(mol L ) H SO (aq) H O(l) H O (aq) HSO (aq)
− +⋅ + +U
 initial 51.50 10−× — 0 0
 change x− — x+ x+
 final 51.50 10 x−× − — x+ x+
[ ]3 3a 2 3
2
5
5
3
5
[H O ][HSO ]
H SO
( ) ( )1.55 10
1.5 10
Employing the quadratic formula we find
[H O ] 1.5 10 , giving a pH of
pH log(1.5 10 ) 4.8
K
x x
x
x
+ −
−
−
+ −
−
=
× = × −
= = ×
= × =
 
(c) When dissolved in water, SO3 will form sulfuric acid: 
3 2 3 4SO (aq) 2 H O(l) H O (aq) HSO (aq)
+ −+ → + 
The fist deprotonation of sulfuric acid is complete, the concentration of 
hydronium ion due to the second deprotonation of sulfuric acid may be 
found as described earlier: 
Concentration 
 
4 4
1 - 2-
2 3(mol L ) HSO (aq) H O(l) H O (aq) SO (aq)
− +⋅ + +U
 initial 51.50 10−× — 51.50 10−× 0 
 change x− — x+ x+
 final 51.50 10 x−× − — 51.50 10 x−× − x+
SM-307 
2
3 4
a
4
5
2
5
5
3
5 5 5
[H O ][SO ]
HSO
(1.5 10 ) ( )1.2 10
1.5 10
Employing the quadratic formula we find
[H O ] 1.495 10 ,
and the total hydronium ion concentration to be
1.495 10 1.5 10 3.0 10 .
Theref
K
x x
x
x
+ −
−
−
−
−
+ −
− − −
= ⎡ ⎤⎣ ⎦
× +× = × −
= = ×
× + × = ×
5ore, pH = log(3.0 10 ) 4.5−− × =
 
 
10.103 (a) 2 7w 3[H O ][OH ] 3.8 10K x
+ −= = = ×
7 1
3
3
[H O ] 1.9 10 mol L
pH log[H O ] 6.72
x+ −
+
= = × ⋅
= − = 
(b) There are three data points available: 
 
 
14
w w25 C ( 1.0 10 ), 40 C (K K
−° = × ° =
14 14
w3.8 10 ), and 37 C (K 2.1 10 )
− −× ° = ×
 ln T 11/ (K )−T wK
 0.003356 25 C° 32.2362− 
 0.003226 37 C° 31.4943− 
 0.003195 40 C° 31.1376− 
The slope is equal to /H R−∆ ° and the intercept equals / .S R∆ ° 
1 1
1 1 1
( 6502 K)(8.314 J K mol ) 54 kJ mol
( 10.43)(8.314 J K mol ) 87 J K mol
H
S
1
1
− − −
− − −
∆ ° = − − ⋅ ⋅ = ⋅
∆ ° = − − ⋅ ⋅ = ⋅ ⋅ − 
(c) The equation determined from the graph is for ln In order to write 
anequation for the pH dependence of pure water, we must rearrange the 
equation. 
w .K
First we note the relationship between and the pH of pure water. wK
SM-308 
1/2
3 w
3 w
w
w
w
[H O ]
1pH log[H O ] log
2
6501ln 10.43
65012.303 log 10.43
2823log 4.529
K
K
K
T
K
T
K
T
+
+
=
= − = −
= − −
= − −
= − −
 
w
1pH log
2
1411pH 2.264
K
T
= −
= +
 
 
10.105 (a) We begin by finding the empirical formula of the compound: 
 
2
2-1
-1
2
2-1
0.942 g CO
C : 0.0214 mol CO 0.214 mol C
44.011 g mol
 (0.214 mol C)(12.011 g mol ) 0.257 g C
0.0964 g H O
H: 0.00535 mol H O 0.0107 mol H
18.0158 g mol
 (0.0107 mol H)(1.008 
= ∴⋅
⋅ =
= ∴⋅
-1
2
-1
-1
g mol ) 0.0108 g H
0.246 g H O
Na: 0.0107 mol Na 
22.99 g mol
 (0.0107 mol Na)(22.99 g mol ) 0.0246 g Na
O: mass of O 1.200 g 0.257 g 0.0108 g 0.0246 g 0.686 g of O
0.686 g O 
16.00 
⋅ =
=⋅
⋅ =
= − − − =
-1 0.0429 mol Og mol
=⋅
 
Dividing through by 0.0107 moles, we find the empirical formula to be: 
C2HNaO4. 
A molar mass of 112.02 g⋅mol-1 indicates that this is also the molecular 
formula. 
(b) 
C C
O
O
O
O
H -
 
 
SM-309 
(c) The dissolved substance is sodium oxalate, it is capable of gaining or 
loosing a proton and, therefore, amphiprotic. ( )11 22pH p p 2.7a aK K= + = 
 
10.107 We wish to calculate for the reaction aK
2 3HF(aq) H O(l) H O (aq) F (aq)
+ −+ +U 
This equation is equivalent to 
HF(aq) H (aq) F (aq)+ −+U 
This latter writing of the expression is simpler for the purpose of the 
thermodynamic calculations. 
The value for this reaction is easily calculated from the free energies 
given in the appendix: 
G∆ °
1 1 1
1 1
/
(18030 J mol ) / [8.314 J K mol ) (298 K)] 4
( 278.79 kJ mol ) ( 296.82 kJ mol ) 18.03 kJ mol
ln
6.9 10
G RT
RT K
K e
K e
− − −
1− − −
−∆ °
− ⋅ ⋅ ⋅ −
= − ⋅ − − ⋅ = ⋅
−
=
= = ×
 
 
10.109 (a) 2 2 3D O D O D O OD
+ −+ +U 
(b) 
2 2 2
15
D 0 3 D 0 D 0[D O ][OD ] 1.35 10 , p log 14.870K K
+ − −= = × = − =K
(c) 15 8 13[D O ] [OD ] 1.35 10 3.67 10 mol L
+ − − −= = × = × ⋅ − 
(d) 8pD log(3.67 10 ) 7.435 pOD−= − × = =
(e) 
2D 0 2
pD pOD p (D O) 14.870K+ = =
 
10.111
Concentration 
+
3 2 3 3CH CH(OH)COOH(aq) + H O(l) H O (aq) + CH CH(OH)COO (aq) 
−U
1(mol L )−⋅
 initial 1.00 — 0 0 
 change x− — x+ x+
 equilibrium 1.00 x− — x x
SM-310 
+ 2
43 3
a
3
[H O ][CH CH(OH)COO ]
8.4 10
[CH CH(OH)COOH] 1.00-
0.029
0.029Percent deprotonation (100%) 2.9% deprotonated
1.00
xK
x
x
−
−= = =
=
= =
×
 
To calculate Tf for this solution we need the concentration in moles of 
solute per kilogram of solvent. Assuming a density of 1 g⋅cm-1, 1.00 L of 
solution will weigh 1000 g. One liter of solution will contain 1 mole 
(90.08 g) of lactic acid. Therefore, one liter of solution will contain 1000 g 
– 90.08 g = 910 g or 0.910 kg of solvent. 
The molarity of the solution is: 1.029 mol solute 1.13 
0.910 kg solvent
m= . 
( )( )-1 -11.86 K Kg mol 1.13 mol Kg 2.1 K
The temperature at which the solution will freeze is 271 K
fT∆ = ⋅ ⋅ ⋅ = 
 
10.113 30 C 303 K, 20 C 293 KT T= ° = = ° =�
a
a
1 1ln
K H H T T
K R T T R TT
′⎛ ⎞ ∆ ° ∆ ° −⎛ ⎞ ⎛= − =⎜ ⎟ ⎜ ⎟ ⎜′⎝ ⎠ ⎝⎝ ⎠
�
�
⎞⎟⎠ 
This is the van’t Hoff equation. 
4
4 1 1
1.768 10 303 K 293 Kln 0.0017
293 K 303 K1.765 10 8.314 J K mol
H−
− − −
⎛ ⎞ ⎛ ⎞× ∆ ° −= =⎜ ⎟ ⎜ ⎟×× ⋅ ⋅ ⎝ ⎠⎝ ⎠
 
 
10.115 (a) The equilibrium constant for the autoprotolysis of pure deuterium 
oxide is given by: 
 ( )( )
-1
-1 -1 15
84800 J mol
8.3145J K mol 298 K 1.37 10G R TK e e
−− ∆ ⋅
− ⋅
⋅ ⋅= = = × 
The concentration of D3O+(aq) at equilibrium is found in the familiar way: 
Concentration 
1(mol L )−⋅ 2 32 D O D O (aq) + OD (aq)+ −U 
 initial — 0 0 
 change — x+ x+ 
SM-311 
 equilibrium — x+ x+ 
15
3
8
3
8
[D O ][OD ] 1.37 10
[D O ] 3.7 10
pD log(3.7 10 ) 7.4
K + −
+ −
−
= =
= ×
= − × =
−×
O and H PO
 
(b) Given the expression for K in part (b) above, it is apparent that as T 
increases K will increase resulting in an increase in [D3O+(aq)] and a 
decrease in pD. 
 
10.117 (a) and (b) Buffer regions are marked A, B, and C. 
 
1.0 
 
 A B C0.5 
 
 
0 
14 7 0 
 
pH 
(c) Region A: H P3 4 2 4
− 
Region B: H P 22 4 4O and HPO
− −
and PO− −
4
Region C: HPO 2 34 4
(d) 
 
 
 
 
(e) The major species present are similar for both and 
 and 
3H PO
3 4 2 4H AsO : H EO
− 2
4HEO
− where E P= or As. For As, there is more 
 than 24HAsO
−
2 4H AsO ,
− with a ratio of approximately 0.63 to 0.37, or 
1.7 : 1. For P, the situation is reversed, with more 2 4H PO
− than HP 
in a ratio of about 0.61 to 0.39, or 2.2 : 1. 
2
4
1.0 
A CB0.5 
0 
14 7 0 
pH 
O −
 
SM-312 
10.119 Given that CO2 will react with water to form carbonic acid, H2CO3, it only 
remains to determine the concentration of H3O+ due to the deprotonation 
of H2CO3. 
( )( )2 -1 -1 4 62[CO ] 2.3 10 mol L atm 3.04 10 atm 7.0 10 mol L− −= × ⋅ ⋅ × = × ⋅ -1− 
+
2 3 2 3 3H CO (aq) + H O(l) H O (aq) + HCO (aq) 
−U 
 initial 7.0 × 10−6 — 0 0 
 change x− — x+ x+
 equilibrium -67.0 10 x× − — x x
 
2
a 6.37 7
6
6
3
6
a 10 10 4.27 10
7.0 10
[H O ] 1.73 10
pH log(1.73 10 ) 5.75
pK xK
x
− − −
−
+ −
−
⎛ ⎞= = = × = ⎜ ⎟×⎝ ⎠
= = ×
= − × =
 
SM-313 
	ACIDS AND BASES