Exercícios Resolvidos - Cap. 10 (Pares) - Ácidos e Bases - -Princípios de Química - Atkins
Transcript of Exercícios Resolvidos - Cap. 10 (Pares) - Ácidos e Bases - -Princípios de Química - Atkins
CHAPTER 10
ACIDS AND BASES 10.2 (a) (b) H3H O+
2O (c) 6 5 3C H NH + (d) HS− (e) 34PO −
(f) 4ClO −
10.4 (a) H2
ac
(b) H2
ac
(c) H2
ac
(d) H2
ac
(e) H2
ac
conjugate
O(l) + CN−(aq) HCN(l) + OH−(aq) id1 acid2 base1
O(l) + Nid1
id1
id1
id1
base2
conjugate
conjugateH2NH2(aq) NH2NH3+(aq) + OH−(aq)
base2 acid2 base1
conjugate
conjugate −(aq)
O(l) + CO32− (aq) HCO3− (aq) + OH
base2 acid2 base1
conjugateconjugate −(aq)
O(l) + HPO42− (aq) H2PO4− (aq) + OH
base2 acid2 base1
conjugateconjugate H−(aq)
O(l) + NH2CONH2 (aq) NH2CONH3+ (aq) + O
base2 acid2 base1conjugate
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10.6 (a) Brønsted acid: NH4+
Brønsted base: 3HSO −
(b) Conjugate base to NH4+ : NH3
Conjugate acid to 3HSO − : H2SO3
10.8 (a) as an acid: H2PO4−(aq) + H2O(l) H3O+(aq) + HPO4
2− (aq)
Conjugate acid/base pairs: H2PO4−(aq) / HPO4
2− (aq)
H3O+(aq) / H2O(l)
as a base: H2PO4−(aq) + H2O(l) OH−(aq) + H3PO4(aq)
Conjugate acid/base pairs: H3PO4(aq) / H2PO4−(aq)
H2O(l) / OH−(aq)
(c) as an acid: HC2O4−(aq) + H2O(l) H3O+(aq) + C2O4
2−(aq)
Conjugate acid/base pairs: HC2O4−(aq) / C2O4
2−(aq)
H3O+(aq) / H2O(l)
as a base: HC2O4−(aq) + H2O(l) OH−(aq) + H2C2O4(aq)
Conjugate acid/base pairs: H2C2O4(aq) / HC2O4−(aq)
H2O(l) / OH−(aq)
10.10 (a) acidic; (b) basic; (c) acidic; (d) amphoteric
10.12 (a) 2 OH −(aq) + SiO2(s) SiO32− (aq) + H2O(l)
(b) O
Si
O O
Si
O
O O
Si
O
O O
(c) Lewis Acid: SiO2(s), Lewis Base: OH −(aq)
10.14 In each case use then 14w 3[H O ][OH ] 1.0 10 ,+ − −= = ×K
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14w
31.0 10[H O ]
[OH ] [OH ]
−+
− −
×= =
K
(a) 14
133
1.0 10[H O ] 8.3 100.012
−+ −×
= = ×
(b) 14
103 5
1.0 10[H O ] 1.6 106.2 10
−+ −
−
×= = ×
×
(c) 14
123 3
1.0 10[H O ] 4.3 102.3 10
−+ −
−
×= = ×
×
10.16 (a) 8 13[H O ] [OH ] 3.9 10 mol L+ − −= = × ⋅ −
−×8 2 15w 3
15w w
[H O ][OH ] (3.9 10 ) 1.5 10
p log log(1.5 10 ) 14.82
+ − −
−
= = × =
= − = − × =
K
K K
(b) 8pH log(3.9 10 ) 7.41 pOH−= − × = =
10.18 [KNH2]0 = nominal concentration of KNH2
[KNH2]0 = 12 2
2
0.5 g KNH 1 mol KNH 0.036 mol L0.250 L 55.13 g KNH
−⎛ ⎞⎛ ⎞= ⋅⎜ ⎟⎜ ⎟
⎝ ⎠ ⎝ ⎠
Because KNH2 is a soluble salt, [KNH2]0 = [K+] = (where
is the nominal concentration of
2 0[NH ]−
2 0[NH ]−2NH − ); thus [K+] and =
0.036 .
2[NH ]−
1mol L−⋅
2NH − reacts with water:
2 2 3NH (aq) H O(l) NH (aq) OH (aq)− −+ ⎯⎯→ +
Because is a strong base, this reaction goes essentially to
completion; therefore
2NH −
12 0[NH ] [OH ] 0.036 mol L− −= = ⋅ −
1413 1w
31.0 10[H O ] 2.8 10 mol L
[OH ] 0.036K −
+ −−
×= = = × ⋅ −
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10.20 pH 13[H O ] 10 mol L (antilog pH, mol L )+ − − −= ⋅ ⋅ 1
−
−⋅
−⋅
−⋅
(a) 5 13[H O ] antilog( 5) 1 10 mol L+ −= − = × ⋅
(b) 2.3 1 3 13[H O ] 10 mol L ; antilog ( 2.3) 5 10 mol L+ − − −= ⋅ − = ×
(c) 7.4 1 8 13[H O ] 10 mol L ; antilog ( 7.4) 4 10 mol L+ − − −= ⋅ − = ×
(d) 10.5 1 11 13[H O ] 10 mol L ; antilog ( 10.5) 3 10 mol L+ − − −= ⋅ − = ×
Acidity increases as pH decreases. The order is thus:
milk of magnesia < blood < urine < lemon juice
10.22 (a) pH log(0.0356) 1.448pOH 14.00 1.448 12.55
= − == − =
(b) pH log(0.0725) 1.12pOH 14.00 1.12 12.86
= − == − =
(c) 22Ba(OH) Ba 2 OH+ −⎯⎯→ +
3 1 3 1
3
[OH ] 2 3.46 10 mol L 6.92 10 mol LpOH log(6.92 10 ) 2.160pH 14.00 2.160 11.84
− − − −
−
= × × ⋅ = × ⋅
= − × == − =
−
(d) 3
41
42 1
2
10.9 10 g 1.94 10 mol KOH56.11 g mol
1.94 10 mol[OH ] 1.94 10 mol L0.0100 L
pOH log(1.94 10 ) 1.712pH 14.00 1.712 12.29
−−
−
−− −
−
×= ×
⋅
×= = ×
= − × == − =
−⋅
(e) 1 110.0 mL[OH ] (5.00 mol L ) 0.0200 mol L2500 mL
pOH log(0.200) 1.70pH 14.00 0.699 12.30
− −⎛ ⎞= × ⋅ =⎜ ⎟
⎝ ⎠= − =
= − =
−⋅
(f) 4 1 53
5
5.0 mL[H O ] (3.5 10 mol L ) 7.0 10 mol L25.0 mL
pH log(7.0 10 ) 4.15pOH 14.00 4.15 9.85
+ − −
−
⎛ ⎞= × × ⋅ = ×⎜ ⎟
⎝ ⎠= − × =
= − =
1− −⋅
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10.24 Base Kb pKb
(a) NH3 51.8 10−× 4.74
(b) ND3 51.1 10−× 4.96
(c) NH2NH2 61.7 10−× 5.77
(d) NH2OH 81.1 10−× 7.96
(e) NH2OH < NH2NH2 < ND3 < NH3
10.26 (a) 3 2 2 3 2 2
3 2 2b
3 2
(CH ) NH(aq) H O(l) (CH ) NH (aq) OH (aq)
[(CH ) NH ][OH ] [(CH ) NH]
K
+ −
+ −
+ +
=
3 2 2 2 3 3 2
3 3 2a
3 2 2
(CH ) NH (aq) H O(l) H O (aq) (CH ) NH(aq)
[H O ][(CH ) NH] [(CH ) NH ]
K
+ +
+
+
+ +
=
(b) 14 10 2 2 14 10 2
14 10 2b
14 10 2
C H N (aq) H O(l) C H N H (aq) OH (aq)
[C H N H ][OH ] [C H N ]
K
+ −
+ −
+ +
=
14 10 2 2 3 14 10 2
3 14 10 2a
14 10 2
C H N H (aq) H O(l) H O (aq) C H N (aq)
[H O ][C H N ] [C H N H ]
K
+ +
+
+
+ +
=
(c) 6 5 2 2 6 5 3
6 5 3b
6 5 2
C H NH (aq) H O(l) C H NH (aq) OH (aq)
[C H NH ][OH ] [C H NH ]
K
+ −
+ −
+ +
=
6 5 3 2 3 6 5 2
3 6 5 2a
6 5 3
C H NH (aq) H O(l) H O (aq) C H NH (aq)
[H O ][C H NH ] [C H NH ]
K
+ +
+
+
+ +
=
10.28 Decreasing pKa will correspond to increasing acid strength because pKa
= The pKalog .− K a values (given in parentheses) determine the following
ordering:
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3 3 2 5(CH ) NH (14.00 4.19 9.71) N H (14.00 5.77 8.23)+ +− = < − =
< HCOOH (3.75) < HF (3.45)
Remember that the pKa for the conjugate acid of a weak base will be given
by
pKa + pKb = 14.
10.30 Decreasing pKb will correspond to increasing base strength because pKb
= blog− K . The pKb values (given in parentheses) determine the following
ordering:
2 4
2 5 3
N H (5.77) BrO (14.00 8.69 5.31) CN (14.00 9.31 4.69) (C H ) N(2.99)
− −< − = < − =<
Remember that the pKb for the conjugate base of a weak acid will be given
by
pKa + pKb = 14.
10.32 Any base whose conjugate acid lies below water in Table 10.3 will be a
strong base, that is, the conjugate acid of the base will be a weaker acid
than water, and so water will preferentially protonate the base. Based upon
this information, we obtain the following analysis: (a) , strong; 2O −
(b) , weak; (c) Br−4HSO − , weak; (d) 3HCO − , weak; (e) CH3NH2,
weak; (f) H , strong; (g) −3CH − , strong.
10.34 In oxoacids with the same number of oxygen atoms attached to the central
atom, the greater the electronegativity of the central atom, the more the
electrons of the O—H bond are withdrawn, making the bond more polar.
This allows the hydrogen of the OH group to be more readily donated as a
proton to H2O, due to the stronger hydrogen bonds that it forms with the
oxygen of water. Therefore, HClO is the stronger acid, with the lower pKa.
10.36 (a) H3PO4 is stronger; it has the more electronegative central atom.
296
(b) HBrO3 is stronger; there are more O atoms attached to the central
atom in HBrO3, making the H—O bond in HBrO3 more polar than in
HBrO.
(c) We would predict H3PO4 to be a stronger acid due to more oxygens on
the central atom.
(d) H2Te is the stronger acid, because the H—Te bond is weaker than the
H—Se bond.
(e) HCl is the stronger acid. Within a period, the acidities of the binary
acids are controlled by the bond polarity rather than the bond strength, and
HCl has the greater bond polarity, due to the greater electronegativity of
Cl relative to S.
(f) HClO is stronger because Cl has a greater electronegativity than I.
10.38 (a) Methylamine is CH3NH2, ammonia is NH3. Methylamine can be
thought of as being formed from NH3 by replacing one H atom with CH3.
Because CH3 is less electron withdrawing than H, CH3NH2 is a weaker
acid and therefore a stronger base.
(b) Hydroxylamine is HONH2, hydrazine is H2N—NH2. The former can
be thought of as being derived from NH3 by replacement of one H atom
with OH; the latter by replacement of one H atom with NH2. Because the
hydroxyl group is more electron withdrawing than the amino group, NH2,
hydroxylamine is the stronger acid and therefore a weaker base.
10.40 The solution of 0.10 M H2SO4 would have the higher pH (would be the
weaker acid) because the conjugate base, HSO4− ,is less electronegative
than the conjugate base of hydrobromic acid, namely Br−.
10.42 The smaller the value of pKb, the stronger the base; hence, aniline is the
stronger base. 4-Chloroaniline is the stronger acid due to the presence of
the electron-withdrawing Cl atom, making it the weaker base; and again
we see that aniline is the stronger base.
297
10.44 The higher the pKa of an acid, the stronger the corresponding conjugate
base; therefore, the order is
3-hydroxyaniline < aniline < 2-hydroxyaniline < 4-hydroxyaniline
No simple pattern exists, but the position of the —OH group does affect
the basicity.
10.46 (a) 2 2
4 3 3 2a
3
[H O ][CH CH(OH)CO ]8.4 10
[CH CH(OH)COOH] 0.12 0.12
+ −−= × = = ≈
−x xK
x
Here we have assumed that x is small enough to neglect it relative to 0.12
. This is a borderline case. We will also solve this exercise
without making this approximation and compare the results below.
1mol L−⋅
13[H O ] 0.010 mol L
pH log(0.010) 2.00pOH 14.00 2.00 12.00
+ −= = ⋅
= − == − =
x
Without the approximation, the quadratic equation that must be solved is 2
4
2 4 4
4 4 2
3
8.4 100.12
or 8.4 10 1.01 10 0
8.4 10 (8.4 10 ) 4( 1.01 10 )2
0.0096, 0.018.[H O ] 0.0096
−
− −
− −
+
× =−
+ × − × =
− × ± × − − ×=
= −
= =
xx
x x
x
xx
4−
pH log(0.0096) 2.02= − =
(b) 2
4a 38.4 10
1.2 10−
−= × =× −
xKx
or 2 4
3 3
8.4 10 1.0 10 01.5 10 , 1.1 10
− −
− −
+ × − × =
= − × ×
x xx
6
The negative root can be eliminated: 3
3
3
[H O ] 1.1 10
pH log(1.1 10 ) 2.96pOH 14.00 2.96 11.04
+ −
−
= = ×
= − × == − =
x
298
(c) 2
458.4 10
1.2 10−
−× =× −
xx
or 2 4
5 4
8.4 10 1.0 10 01.2 10 , 8.5 10
− −
− −
+ × − × =
= × − ×
x xx
8
The negative root can be eliminated. 5
3
5
[H O ] 1.2 10
pH log(1.2 10 )pH 4.92pOH 14.00 4.92 9.08
+ −
−
= ×
= − ×=
= − =
10.48 (a) 9b (pyridine) 1.8 10−= ×K
6 5 2 6 5C H N H O C H NH OH+ −+ +
Concentration C1(mol L )−⋅ 6H5N + H2O C6H5NH+ + OH−
initial 0.075 — 0 0
change −x — +x +x
final 0.075 − x — +x +x 2 2
9b
5 1
5
1.8 100.075 0.075
[OH ] 1.2 10 mol LpOH log(1.2 10 ) 4.92pH 14.00 4.92 9.00
−
− − −
−
= × = ≈−
= = × ⋅
= − × == − =
x xKx
x
Percentage ionized = 5
21.2 10 100% 1.6 10 %0.075
−−×
× = ×
(b) The setup is similar to that in (a). 2 2
6b 1.0 10 [OH ]
0.0112 0.0112− −= × = ≈ =
−x xK x
x
299
4 1
4
4
1.1 10 mol LpOH log(1.1 10 ) 3.97pH 14.00 3.97 10.03
3.2 10Percentage protonation 100% 2.6%0.0122
− −
−
−
= × ⋅
= − × == − =
×= × =
x
(c) bp 14.00 8.52 5.48= − =K
6b 3.3 10−= ×K
2quinine H O quinineH OH+ −+ +
2 26
b 3.3 100.021 0.021
−= × = ≈−
x xKx
4 1
4
4
[OH ] 2.6 10 mol LpOH log(2.6 10 ) 3.58pH 14.00 3.58 10.42
2.6 10Percentage protonation 100% 1.2%0.021
− − −
−
−
= = × ⋅
= − × == − =
×= × =
x
(d) 2strychnine H O strychnineH OH+ −+ +
146w
b 9a
2 26
b
4 1
4
4
[OH ]1.00 10 1.82 105.49 10
1.82 100.059 0.059
[OH ] 3.3 10 mol LpOH log(3.3 10 ) 3.48pH 14.00 3.48 10.52
3.3 10Percentage protonation 100% 0.56%0.059
−
−−
−
−
− − −
−
−
=
×= = = ×
×
= × = ≈−
= = × ⋅
= − × == − =
×= × =
xK
KK
x xKx
x
300
10.50 (a) HN 2 2 3 2O H O H O NO+ −+ +
2.63 3 13 2
3 24
a 3
4a
[H O ] [NO ] 10 antilog ( 2.63) 2.3 10 mol L
(2.3 10 ) 4.2 100.015 2.3 10
p log(4.3 10 ) 3.38
+ − − −
−−
−
−
−= = = − = × ⋅
×= = ×
− ×
= − × =
K
K
(b) 4 9 2 2 4 9 3C H NH H O C H NH OH+ −+ +
1.96 14 9 3
pOH 14.00 12.04 1.96[C H NH ] [OH ] 10 antilog ( 1.96) 0.011 mol L+ − −
= − =
= = = − = ⋅ −
234 9 3
b4 9 2
3b
[C H NH ][OH ] (0.011) 1.4 10[C H NH ] 0.10 0.011
p log(1.4 10 ) 2.85
+ −−
−
= = =−
= − × =
K
K
×
10.52 (a) 10a (HCN) 4.9 10−= ×K
pH 5.3 63[H O ] 10 10 antilog ( 5.3) 5 10 mol L+ − − − 1−= = = − = × ⋅
Let x = nominal concentration of HCN, then
Concentration HCN + H1(mol L )−⋅ 2O H3O+ + CN −
nominal x — 0 0
equilibrium 65 10−− ×x — 65 10−× 65 10−×
6 2−10
6
2 1 1
(5 10 )4.9 105 10
Solve for : 5 10 mol L 0.05 mol L
−−
− − −
×× =
− ×
= × ⋅ = ⋅
xx x
(b) 9b (pyridine) 1.8 10−= ×K
pOH 5.2 6 1
pOH 14.00 8.8 5.2[OH ] 10 10 antilog ( 5.2) 6 10 mol L− − − −
= − =
= = = − = × ⋅ −
Let x = nominal concentration of C5H5N, then
Concentration C1(mol L )−⋅ 5H5N + H2O C5H6N+ + OH−
nominal x
equilibrium 66 10−− ×x 66 10−× 66 10−×
301
6 29
6
2 1
(6 10 )1.8 106 10
Solve for : 2 10 mol L 0.02 mol L
−−
−
1− − −
×× =
− ×
= × ⋅ = ⋅
xx x
10.54 veronal + H2O H3O+ + veronalate ion
The equilibrium concentrations are 1
5 13
5
5 283
a
[veronal] 0.020 0.0014 0.020 0.020 mol L[H O ] [veronalate ion] 0.0014 0.020 2.8 10 mol L
pH log(2.8 10 ) 4.55[H O ][veronalate ion] (2.8 10 ) 3.9 10
[veronal] 0.020
−
+ − −
−
+ −−
= − × = ⋅
= = × = × ⋅
= − × =
×= = = ×K
10.56 cacodylic acid + H2O H3O+ + cacodylate ion
The equilibrium concentrations are 1
5 13
5
5 23
a
[cacodylic acid] 0.0110 (0.0077 0.0110) 0.0110 0.000 08 0.109 mol L[H O ] [cacodylate ion] 0.0077 0.0110 8.5 10 mol L
pH log(8.5 10 ) 4.07[H O ][cacodylate ion] (8.5 10 )
[cacodylic acid] 0.010
−
+ − −
−
+ −
= − × = − = ⋅
= = × = × ⋅
= − × =
×= =K 76.6 10
9−= ×
C N
H
H
H
H
H
C N+
H
H
H
H
H
H
methylamine conjugate acid
10.58 (a)
(b) First determine the concentration of methylamine following dilution.
Moles of methylamine are found by:
302
( ) ( )( ) ( )
( )( )
1
1
50 mL 0.85 g mL 42.5 g of solution
42.5 g 0.35 14.9 g of methylamine
14.9 g0.479 mol of methylamine
31.06 g mol
diluted to 1.000 L, the resulting solution is 0.479 M in methylamine
−
−
⋅ =
⋅ =
=⋅
Concentration CH1(mol L )−⋅ 3NH2 + H2O CH3NH3+ + OH−
initial 0.479 — 0 0
equilibrium 0.479 − x — x x
[ ]2
3 34b
3 2
2
2
CH NH OH3.6 10
CH NH 0.479solving for x we find:
1.3 10 M OH
pOH log(1.3 10 ) 1.9pH 14.0 1.9 12.1
+ −−
− −
−
⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= × = =−
⎡ ⎤= × = ⎣ ⎦= − × =
= − =
xKx
x
10.60 (a) 22 2 4 2 4pH 7, basic: H O(l) C O (aq) HC O (aq) OH (aq)− −> + + −
−
O (aq) Fe(H O) OH (aq)+ ++
(b) 23pH 7, neutral: Ca is not an acid and NO is not a base+ −=
(c) 3 3 2 3 3 2pH 7, acidic: CH NH (aq) H O(l) H O (aq) CH NH (aq)+ +< + +
(d) 3 22 4 4pH 7, basic: H O(l) PO (aq) HPO (aq) OH (aq)− −> + +
(e) 32 6 2pH 7, acidic: Fe(H O) (aq) H O(l)+< +
H 23 2 5
(f) 5 5 2 3 5 5pH 7, acidic: C H NH (aq) H O(l) H O (aq) C H N(aq)+ +< + +
10.62 (a) Concentration
CH1(mol L )−⋅ 3NH3+ + H2O(l) CH3NH2 + H3O+
initial 0.25 — 0 0
change −x — +x +x
equilibrium 0.25 − x — x x
303
(See Table 10.2 for Kb of CH3NH2, the conjugate base of CH3NH3+.)
1411w 3 2
a 4b 3 3
2 211
6 1 63
[CH NH ][H O ]1.00 10 2.8 103.6 10 [CH NH ]
2.8 100.25 0.25
2.6 10 mol L [H O ] and pH log(2.6 10 ) 5.58
+−−
− +
−
− − + −
×= = = × =
×
× = ≈−
= × ⋅ = = − × =
KK
K
x xx
x
3
(b) Concentration
SO1(mol L )−⋅ 23
− + H2O(l) 3HSO − + OH −
initial 0.13 — 0 0
change — +x +x −x
equilibrium 0.13 − x — x x
[See Table 10.9 for a 3 a2 2 3(HSO ) (H SO ).]− =K K
14 2 28w 3
b 7 2a2 3
28
[HSO ][OH ]1.00 10 8.3 101.2 10 [SO ] 0.13 0.13
8.3 100.13
K x xKK x
x
− −−−
− −
−
×= = = × = = ≈
× −
× =4 1
4
1.0 10 mol L [OH ]pOH log(1.0 10 ) 4.00pH 14.00 pOH 10.00
− −
−
= × ⋅ =
= − × == − =
x −
(c) Concentration
Fe(H1(mol L )−⋅ 2O)63+(aq) + H2O(l) H3O+(aq) + Fe(H2O)5OH+(aq)
initial 0.071 — 0 0
change — +x +x −x
equilibrium 0.071 − x — x x
233 2 5
a 32 6
[H O ][Fe(H O) OH (aq)]3.5 10
0.071[Fe(H O) ]
+ +−
+= = × =−
xKx
2 33.5 10−+ ×x x 42.5 10 0−× =
x = 0.014, 0.018−
The negative root can be discarded
3[H O ] 0.014, pH log 0.014 .85+ = = − =
304
10.64 (a) Initial concentration of C6H5NH3+ is:
( ) 1
17.8 g 94.133 g mol
0.24 M0.350 L
−
⎛ ⎞⎜ ⎟⎝ ⎠ =
Given this initial concentration and the Ka for this acid found in Table
10.7, the percent deprotonation is found as follows:
Concentration
C1(mol L )−⋅ 6H5NH3+ + H2O(l) C6H5NH2
2+ + H3O+
initial 0.24 — 0 0
change − — +x +x x
equilibrium 0.24 − x — x x 2+
5 6 5 2 3a +
6 5 3
2 25
3 13
3 1
1
[C H NH ][H O ]2.3 10
[C H NH ]
2.3 100.24 0.24
2.3 10 mol L [H O ]
2.3 10 mol Lpercent dissociation 100% 1.0%0.23 mol L
+−
−
− − +
− −
−
= × =
× = ≈−
= × ⋅ =
× ⋅= ⋅ =
⋅
K
x xx
x
10.66 (a) can act as both an acid and a base. Both actions need to be
considered simultaneously:
3HSO −
23 2 3 3
3 2 2 3
HSO H O H O SO
HSO H O H SO OH
− +
− −
+ +
+ +
−
3−
Summing, 2
3 3 2 3HSO HSO H SO SO− −+ +
The simplest approach is to recognize that there are two conjugate acid-
base pairs and to use the Henderson-Hasselbalch equation twice, once for
each pair.
305
3a1
2 3
23
a23
[HSO ]pH p log
[H SO ]
[SO ]pH p log
[HSO ]
−
−
−
⎛ ⎞= + ⎜ ⎟
⎝ ⎠⎛ ⎞
= + ⎜ ⎟⎝ ⎠
K
K
Summing, 2
3 3a1 a2
2 3 3
[HSO ] [SO ]2pH p p log
[H SO ] [HSO ]
− −
−
⎛ ⎞= + + ×⎜ ⎟
⎝ ⎠K K
Because [ 22 3 3 a1 a2H SO ] [SO ], 2pH p p−= = +K K
Therefore, 1 1a1 a 22 2pH (p p ) (1.81 6.91) 4.36= + = + =K K
Note that it is not necessary to know the concentration of because
it cancels out of the equation. At extremely low concentrations of
however, the approximations upon which the use of this equation are
based are no longer valid.
3HSO ,−
3HSO ,−
(b) Neither silver ion nor nitrate ion is acidic or basic in aqueous solution.
Therefore, the pH is that of neutral water, 7.00.
10.68 (a) 1
10
0.250 mol L KCN 0.0350 L 0.0875 mol L KCN [CN ]0.1000 L
−− −⋅ ×
= ⋅ =
Concentration
H1(mol L )−⋅ 2O(l) + HCN(aq) + CN (aq)− OH (aq)−
initial — 0.0875 0 0
change — −x +x +x
equilibrium — 0.0875 − x x x
145w
b 10a
2 25
3 1
1.00 10 [HCN][OH ]2.0 104.9 10 [CN ]
2.0 100.0875 0.0875
1.3 10 mol L [HCN]
− −−
− −
−
− −
×= = = × =
×
× = ≈−
= × ⋅ =
KK
K
x xx
x
306
(b) 3
3 3
2 13
1.59 g NaHCO 184.01 g NaHCO /mol NaHCO 0.200 L
9.46 10 mol L NaHCO− −
×
= × ⋅
Concentration
H1(mol L )−⋅ 2O(l) + H3HCO (aq)−2CO3(aq) + OH (aq)−
initial — 29.46 10−× 0 0
change — −x +x +x
equilibrium — 29.46 10−× − x x x
1482 3 w
b1 7a13
2 28 5 1
2 2
5
[H CO ][OH ] 1.00 10 2.3 10[HCO ] 4.3 10
2.3 10 , 4.7 10 mol L9.46 10 9.46 10
[OH ]pOH log(4.7 10 ) 4.33, pH 14.00 4.33 9.67
− −−
− −
− −− −
−
−
×= = = = ×
×
× = ≈ = × ⋅× − ×
=
= − × = = − =
KK
K
x x xx
−
10.70 The reaction of interest is:
C6H5CH2(CH3)NH3+(aq) + H2O(l) C6H5CH2(CH3)NH2(aq) +
H3O+(aq)
Ka for this reaction is: 14
11a 4
1 10 1.3 107.8 10
−−
−
×= = ×
×K
The initial concentration of C6H5CH2(CH3)NH3+(aq) is:
( ) ( )-1
-16.48 g 216.12 g mol
0.150 mol L0.200 L
⋅=
Concentration
H1(mol L )−⋅ 2O(l) + C6H5CH2(CH3)NH3+(aq) C6H5CH2(CH3)NH2(aq) + H3O+(aq)
initial — 9.46 210−× 0 0
change — −x +x +x
equilibrium — 9.46 210−× − x x x
307
2 211 6 1
+3
6
1.3 10 , 1.4 10 mol L0.145 0.145
[H O ]
pH log(1.4 10 ) 5.9
x x xx
− −
−
× = ≈ = × ⋅−
=
= − × =
−
O (aq) H O(l) H O (aq) H PO (aq)+ −+ +
−
−
−
−
−
10.72 (a) H P 3 4 2 3 2 4
22 4 2 3 4
2 34 2 3 4
H PO (aq) H O(l) H O (aq) HPO (aq)
HPO (aq) H O(l) H O (aq) PO (aq)
− +
− +
+ +
+ +
(b)
2 4 2 2 3 2 4 2
22 4 2 2 3 2 4 2 2
(CH ) (COOH) (aq) H O(l) H O (aq) (CH ) (COOH)CO (aq)
(CH ) (COOH)CO (aq) H O(l) H O (aq) (CH ) (CO ) (aq)
+ −
− +
+ +
+ +
(c)
2 2 2 2 3 2 2 2
22 2 2 2 3 2 2 2 2
(CH ) (COOH) (aq) H O(l) H O (aq) (CH ) (COOH)CO (aq)
(CH ) (COOH)CO (aq) H O(l) H O (aq) (CH ) (CO ) (aq)
+ −
− +
+ +
+ +
10.74 The reaction is (after the first, essentially complete ionization) 2
4 2 3 4HSeO H O H O SeO− ++ +
The initial concentrations of 14 3HSeO and H O are both 0.010 mol L− + −⋅
due to the complete ionization of H2SeO4 in the first step. The second
ionization is incomplete.
1(mol L )−⋅ 4HSeO − + H2O H3O+ + Se 24
Concentration
O −
initial 0.010 — 0.010 0
change −x — +x +x
equilibrium 0.010 − x — 0.010 + x x
308
22 3 4
a24
2 4
2 43
3 1 23
2
[H O ][SeO ] (0.010 )( )1.2 100.010[HSeO ]
0.022 1.2 10 0
0.022 (0.022) (4)( 1.2 10 )4.5 10
2[H O ] 0.010 (0.010 4.5 10 ) mol L 1.5 10 mol L
pH log(1.5 10 ) 1.82
+ −−
−
−
−−
+ − −
−
+= × = =
−
+ − × =
− + − − ×= = ×
1− −= + = + × ⋅ = × ⋅
= − × =
x xKx
x x
x
x
10.76 (a) Second ionization is ignored because a 2 a1<<K K
+2 2 3H S H O H O HS−+ +
2 27 3
a12
4 13
4
[H O ][HS ]1.3 10
[H S] 0.10 0.10
[H O ] 1.1 10 mol L
pH log(1.1 10 ) 3.96
+ −−
+ − −
−
= × = = ≈−
= = × ⋅
= − × =
x xKx
x
(b) This is a situation where it may not be justified to ignore the second
ionization. . This is a
marginal case; we can work it both ways, first without ignoring the second
ionization. Adopt the following notation:
4 5a1 a2 a2 a16.0 10 , 1.5 10 , and / 40− −= × = × =K K K K
H+ = H3O+
[H+] = [H3O+] = equilibrium concentration of H3O+
H2A = tartaric acid 1
0c solute molarity 0.15 mol L−= = ⋅
−
=
20 2c [H A] [HA ] [A ]−= + + (1)
2 0H total H present [H ] 2[H A] [HA ] 2c+ −= = + + (2)
The following equilibria occur:
2H A H HA+ −+ a12
[H ][HA ][H A]
+ −
=K (3)
2HA H A− + + − 2
a2[H ][A ]
[HA ]
+ −
−=K (4)
309
Examination of Eqs. 1 to 4 shows that there are four unknowns,
, to be determined. However, these four
simultaneous equations allow for their determination.
22[H ], [H A], [HA ], and [A ]+ − −
The unknowns 22[H A], [HA ], and [A ]− − may all be expressed in terms of
one unknown, [H+]. 2
0 2According to Eq.1, c [H A] [HA ] [A ]− −= + +
2a1
2
a 22 2
2a1 a 2
[H ][HA ]From Eq. 3, [H A]
[H ][A ]From Eq. 4, [HA ]
[H ] [A ]Substituting, [H A]
+ −
+ −−
+ −
=
=
=
K
K
K K
22
0a2 a1 a2
[H ] [H ]Then Eq.1 becomes c [A ] 1+ +
− ⎧ ⎫= + +⎨ ⎬
⎩ ⎭K K K (5)
Now, subtract 2 Eq.1 from Eq. 2 :×
0 22c [H ] 2[H A] [HA ]+ −= + + Eq. 2
20 2
2
2c 2[H A] 2[HA ] 2[A ][H ] [HA ] 2[A ] 0
− −
+ − −
= + +− − =
2 Eq.1×
2 2
a2
[H ][A ], [A ][H ] 2
+− −
+=+
K
Substitute this into Eq. 5 to give 2 3
0a2 a2 a1 a2
[H ] [H ] [H ]c 2 [H ]+ +
+⎛ ⎞+ = + +⎜ ⎟
⎝ ⎠K K K
+
K
Rearranging into standard cubic form gives 3 2
00
a1 a 2 a2 a2
c[H ] [H ] [H ] 1 2c 0+ +
+ ⎛ ⎞+ + − −⎜ ⎟
⎝ ⎠K K K K=
=
Now, put in the numerical values for c0, Ka1, and Ka2: 8 3 4 2 41.1 10 [H ] 6.7 10 [H ] 1.0 10 [H ] 0.30 0+ + +× + × − × −
Solution of this cubic by standard methods gives
310
3 13
3
[H ] [H O ] 9.2 10 mol L
pH log(9.2 10 ) 2.04
+ + −
−
= = × ⋅
= − × =
−
It is left as an exercise for the reader to show that, if the second ionization
is ignored, 3 1[H ] 9.5 10 mol L and pH 2.02+ − −= × ⋅ =
The difference is small but, perhaps, not within experimental error.
(c) The second ionization can be ignored because Ka2 << K a1.
2 4 2 3 4
2 28 3 4
a1 3 32 4
6 13
6
H TeO H O H O HTeO
[H O ][HTeO ]2.1 10
[H TeO ] 1.1 10 1.1 10[H O ] 4.8 10 mol L
pH log(4.8 10 ) 5.32
+ −
+ −−
− −
+ − −
−
+ +
= × = = ≈× − ×
= = × ⋅
= − × =
x xKx
x
10.78 (a) The pH is given by pH = 1a1 a22 (p p )+K K . From Table 10.9, we find
7a1 4.3 10−= ×K a1p 6.37=K
11a2 5.6 10−= ×K a2p 10.25=K
12pH (6.37 10.25) 8.31= + =
(b) The nature of the spectator counter ion does not affect the equilibrium
and the pH of a salt solution of a polyprotic acid is independent of the
concentration of the salt, therefore pH = 8.31.
10.80 The pH is given by pH = 1a1 a22 (p p )+K K .
12pH (2.46 7.31) 4.89= + =
10.82 The equilibrium reactions of interest are
2 3 2 3 3H SO (aq) H O(l) H O (aq) HSO (aq)+ −+ + 2a1 1.5 10−= ×K
23 2 3 3HSO (aq) H O(l) H O (aq) SO (aq)− + −+ + 7
a2 1.2 10−= ×K
Because the second ionization constant is much smaller than the first, we
can assume that the first step dominates:
311
Concentration 1(mol L )−⋅ H2SO3(aq) + H2O(l) H3O+(aq) + 3HSO (aq)−
initial 0.125 — 0 0
change — +x +x −x
final 0.125 − x — +x +x
3 3a1
2 32
2
[H O ][HSO ][H SO ]
( )( )1.5 100.125 0.125
+ −
−
=
× = =− −
K
x x xx x
Assume that x << 0.125, then 2 2(1.5 10 )(0.125)
0.043
−= ×=
xx
Because x > 5% of 0.0456, the assumption was not valid, and the full
expression must be evaluated using the quadratic: 2 2 21.5 10 (1.5 10 )(0.125) 0− −+ × − × =x x
Solving with the quadratic equation gives 10.036 mol L−= ⋅x . 1
3 3
1 1 12 3
[H O ] [HSO ] 0.036 mol L
[H SO ] 0.125 mol L 0.036 mol L 0.089 mol L
+ − −
− − −
= = = ⋅
= ⋅ − ⋅ = ⋅
x
We can then use the other equilibria to determine the remaining
concentrations: 2
3 3a2
32
7 3
2 73
[H O ][SO ][HSO ]
(0.036)[SO ]1.2 10
(0.036)[SO ] 1.2 10 mol L
+ −
−
−−
− −
=
× =
1−= × ⋅
K
Because 1.2 , the initial assumption that the first
dissociation would dominate is valid. To calculate [
710 0.036−× <<
OH ]− , we use the Kw
relationship:
312
w 314
13 1w
3
[H O ][OH ]
1.00 10[OH ] 2.8 10 mol L0.036[H O ]
+ −
−− −
+
=
×= = = × ⋅
KK −
2
In summary, 1 1
2 3 3 3 3[H SO ] 0.089 mol L , [H O ] [HSO ] 0.036 mol L , [SO ]− + − −= ⋅ = = ⋅ −
1−
7 1 131.2 10 mol L , [OH ] 2.8 10 mol L− − − −= × ⋅ = × ⋅
10.84 The equilibrium reactions of interest are now the base forms of the
carbonic acid equilibria, so Kb values should be calculated for the
following changes: 2
3 2 314
8wb1 7
a2
3 2 2 314
13wb2 2
a2
SO (aq) H O(l) HSO (aq) OH (aq)
1.00 10 8.3 101.2 10
HSO (aq) H O(l) H SO (aq) OH (aq)
1.00 10 6.7 101.5 10
− − −
−−
−
− −
−−
−
+ +
×= = = ×
×
+ +
×= = = ×
×
KK
K
KK
K
Because the second hydrolysis constant is much smaller than the first, we
can assume that the first step dominates:
Concentration 1(mol L )−⋅ + H2
3SO (aq)−2O(l) + 3HSO (aq)− OH (aq)−
initial 0.125 — 0 0
change — +x +x −x
final 0.125 − x — +x +x
3b1 2
3
28
[HSO ][OH ][SO ]
( )( )8.3 100.125 0.125
− −
−
−
=
× = =− −
K
x x xx x
Assume that x << 0.125, then 2 8
4
(8.3 10 )(0.125)1.0 10
−
−
= ×
= ×
xx
313
Because x < 1% of 0.125, the assumption was valid. 4 1
3
2 1 4 13
[HSO ] [OH ] 1.0 10 mol L
Therefore, [SO ] 0.125 mol L 1.0 10 mol L 0.125 mol L
− − − −
− − − −
= = = × ⋅
= ⋅ − × ⋅ ≅
x1−⋅
We can then use the other equilibria to determine the remaining
concentrations:
2 3b2
34
13 2 34
13 12 3
[H SO ][OH ][HCO ]
[H SO ](1.0 10 )6.7 10
(1.0 10 )[H SO ] 6.7 10 mol L
−
−
−−
−
− −
=
×× =
×
= × ⋅
K
Because , the initial assumption that the first
hydrolysis would dominate is valid.
13 46.7 10 1.0 10−× << × −
To calculate [H3O+], we use the Kw relationship:
w 314
10 1w3 14
[H O ][OH ]
1.00 10[H O ] 1.0 10 mol L[OH ] 1.0 10
+ −
−+ −
− −
=
× −= = = ××
KK
⋅
1
13 12 3
4 1 23 3
1 103
In summary, [H SO ] 6.7 10 mol L , [OH ]
[HSO ] 1.0 10 mol L , [SO ]
0.125 mol L , [H O ] 1.0 10 mol L
− − −
− − − −
− + − −
= × ⋅
= = × ⋅
= ⋅ = × ⋅
10.86 (a) tartaric acid: The two pKa values are 3.22 and 4.82. Because pH = 5.0
lies above both of these values, the major form present will be the doubly
deprotonated ion . (b) hydrosulfuric acid: The two pK2A −a values are
6.89 and 14.15. Because the pH of the solution lies below both of these
values, the dominant form will be the doubly protonated H2A. (c)
phosphoric acid: The three pKa values are 2.12, 7.21, and 12.68. The pH of
the solution lies between the first and second ionization, so the
predominant species should be the singly deprotonated ion 2 4H PO .−
314
10.88 The equilibria present in the solution are:
2 2 3
22 3
H (aq) H O(l) H O (aq) HS (aq)
HS (aq) H O(l) H O (aq) S (aq)
+ −
− + −
+ +
+ +
7a1
15a2
1.3 10
7.1 10
−
−
= ×
= ×
K
K
The calculation of the desired concentrations follows exactly after the
method derived in Eq. 25, substituting H2S for H2CO3, HS− for ,
and for . First, calculate the quantity
:
3HCO −
2S − 23CO −
9.35 10 13(at pH 9.35 [H O ] 10 4.5 10 mol L )+ − − −= = = ×f ⋅
−
23 3 a1 a1 a2
10 2 10 7 7 15
17
[H O ] [H O ]
(4.5 10 ) (4.5 10 )(1.3 10 ) (1.3 10 )(7.1 10 )5.9 10
+ +
− − − −
−
= + +
= × + × × + × ×
= ×
f K K K
The fractions of the species present are then given by 10 2
332 17
[H O ] (4.5 10 )(H S) 3.4 105.9 10
α+ −
−−
×= = = ×
×f
10 73 a1
17
[H O ] (4.5 10 )(1.3 10 )(HS ) 0.995.9 10
α+ − −
−−
× ×= = =
×K
f
17 152 15a1 a2
17
(1.3 10 )(7.1 10 )(S ) 1.6 105.9 10
α− −
− −−
× ×= = = ×
×K K
f
Thus, in a solution at pH 9.35, the dominant species will be with a
concentration of
HS−
1 1(0.250 mol L )(0.99) 0.25 mol L− −⋅ ≅ ⋅ . The
concentration of H2S will be
, and the concentration of
will be
3 1 4(3.4 10 )(0.250 mol L ) 8.5 10 mol L− − −× ⋅ = × 1−⋅
1−⋅2S − 5 1 6(1.6 10 )(0.250 mol L ) 4.0 10 mol L .− − −× ⋅ = ×
10.90 For the first ionization of (COOH)2—or H2C2O4—we write 2
2 2 4 2 3 2 4 a1H C O H O H O HC O , 5.9 10+ − −+ + =K ×
315
Concentration 1(mol L )−⋅ H2C2O4 + H2O H3O+ + 2 4HC O −
initial 0.10 — 0 0
change — +x +x −x
equilibrium 0.10 − x — x x 2 5
a1 a25.9 10 , 6.5 10− −= × = ×K K
Because Ka2 << Ka1, the second ionization can safely be ignored in the
calculation of [H3O+]. 2
2a1
2
21
3
1413 1
12 2 4
5.9 100.10
0.059 0.0059 0
0.059 (0.059) (4)(1)(0.0059)0.053 mol L [H O ]
21.0 10[OH ] 1.9 10 mol L
0.053[H C O ] 0.10 0.053 0.05 mol L
−
− +
−− − −
−
= × =−
+ − =
− + += =
×= = × ⋅
= − = ⋅
xKx
x x
x ⋅ =
1Concentration (mol L )−⋅ 2 4HC O − +H2O H3O+ + C O 2
2 4−
initial 0.053 — 0.053 0
change −x — +x +x
equilibrium 0.053 − x — 0.053 + x x
5a2
2 5 12 4
12 4
(0.053 )( )6.5 10 (because is small)0.053
or [C O ] 6.5 10 mol L , and
[HC O ] 0.053 0.053 0.000 065 0.053 mol L
−
− − −
− −
+= × = ≈
−
= = × ⋅
= − = − = ⋅
x xK xx
x
x
x
10.92 The equilibria involved are:
H2CO3(aq) + H2O(l) HCO3¯(aq) + H3O+ Ka1 = 4.3 × 10⎯7
HCO3⎯(aq) + H2O(l) CO3
2⎯(aq) + H3O+ Ka2 = 5.6 × 10⎯11
Given these two equilibria, two equations can be written:
316
[ ]
[ ]
+3 3
a12 3
2 +3 3
a23
2 32 3 3 3
+ 4.8 53
HCO H O
H CO
CO H O
HCO
also, enough information is provided to construct two other equations:
H CO HCO CO 4.5 10
H O 10 1.6 10
−
−
−
− − −
− −
⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦=
⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦=⎡ ⎤⎣ ⎦
⎡ ⎤ ⎡ ⎤+ + = ×⎣ ⎦ ⎣ ⎦⎡ ⎤ = = ×⎣ ⎦
K
K
Solving this set of simultaneous equations for the three unknown
concentrations, one finds:
[ ] 32 3
43
2 13
H CO 4.38 10
HCO 1.19 10
CO 4.23 10
−
− −
− −
= ×
⎡ ⎤ = ×⎣ ⎦⎡ ⎤ = ×⎣ ⎦
0
4
=
4
10.94 We can use the relationship derived in the text:
2
3 initial 3 w
2 8 13 3
[H O ] [HA] [H O ] 0
[H O ] (7.49 10 )[H O ] (1.00 10 ) 0
+ +
+ − + −
− − =
− × − × =
K
Solving using the quadratic equation gives
73[H O ] 1.44 10 ; pH 6.842+ −= × =
This value is lower than the value calculated, based on the acid
concentration alone 8(pH log(7.49 10 ) 7.126).−= − × =
10.96 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 ] 0,+ ++ − K
2 7 13 3[H O ] (8.23 10 )[H O ] (1.00 10 ) 0+ − + −+ × − × =
Solving using the quadratic equation gives
83[H O ] 1.20 10 , pH 7.922+ −= × =
This value is slightly higher than the value calculated, based on the base
concentration alone
7(pOH log(8.23 10 ) 6.084, pH 14.00 6.084 7.916).−= − × = = − =
317
10.98 (a) In the absence of a significant effect due to the autoprotolysis of
water, the pH values of the 42.50 10−× M and 62.50 10−× M C6H5OH
solutions can be calculated as described earlier.
For 4 12.50 10 mol L :− −× ⋅
Concentration
C1(mol L )−⋅ 6H5OH(aq) + H2O(l) + 3H O (aq)+6 5C H O (aq)−
initial 42.50 10−× — 0 0
change −x — +x +x
final 42.50 10−× − x — +x +x
3 6 5a
6 52
104 4
[H O ][C H O ][C H OH]
( )( )1.3 102.50 10 2.50 10
+ −
−− −
=
× = =× − × −
K
x x xx x
Assume x << 42.50 10−×
2 10
7
(1.3 10 )(2.50 10 )1.8 10
− −
−
= × ×
= ×
xx
4
10 , Because x < 1% of 2.50 4−×
10 mol L :− −
the assumption was valid. Given this
value, the pH is then calculated to be 8log(1.8 10 ) 6.74.−− × =
For 2.50 6 1× ⋅
O (aq)−
Concentration
C1(mol L )−⋅ 6H5OH(aq) + H2O(l) + C H
initial
3H O (aq)+6 5
62.50 10−× — 0 0
change −x — +x +x
final 62.50 10−× − x — +x +x
3 6 5a
6 52
106 6
[H O ][C H O ][C H OH]
( )( )1.3 102.50 10 2.50 10
+ −
−− −
=
× = =× − × −
K
x x xx x
318
Assume x << 62.50 10−×
2 10
8
(1.3 10 )(2.50 10 )1.8 10
− −
−
= × ×
= ×
xx
6
Because x < 1% of 62.50 10 ,−× the assumption is valid and the pH should
be 7.74. However, this number does not make sense because an acid was
added to the water.
(b) To calculate the value, taking into account the autoprotolysis of water,
we can use Eq. 22:
where x = [H3 2a w a initial w a( [HA] )+ − + ⋅ − ⋅ =x K x K K x K K 0, 3O+].
To solve the expression, you substitute the values of
the initial concentration of acid, and
14w 1.00 10 ,−= ×K
10a 1.3 10−= ×K into this equation
and then solve the expression either by trial and error or, preferably, using
a graphing calculator. Alternatively, you can use a computer program
designed to solve simultaneous equations. Because the unknowns include
[H3O+], [ [HClO], and [ you will need four equation. As
seen in the text, the pertinent equations are
OH ],− ClO ],−
3 6 5a
6 5
w 3
3 6 5
6 5 initial 6 5 6 5
[H O ][C H O ][C H OH]
[H O ][OH ]
[H O ] [OH ] [C H O ]
[C H OH] [C H OH] [C H O ]
+ −
+ −
+ − −
−
=
=
= +
= +
K
K
Using either method should produce the same result.
The values obtained for 4 12.50 10 mol L− −× ⋅ are
(compare to 6.74 obtained in (a)) 7 13[H O ] 2.1 10 mol L , pH 6.68+ − −= × ⋅ =
7 16 5
4 16 5
8 1
[C H O ] 1.6 10 mol L
[C H OH] 2.5 10 mol L
[OH ] 4.8 10 mol L
− − −
− −
− − −
= × ⋅
≅ × ⋅
= × ⋅
Similarly, for 66 5 initial[C H OH] 2.50 10 :−= ×
(compare to 7.74 obtained in (a)) 7 13[H O ] 1.0 10 mol L , pH 7.00+ − −= × ⋅ =
319
9 16 5
6 16 5
8 1
[C H O ] 3.2 10 mol L
[C H OH] 2.5 10 mol L
[OH ] 9.8 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.
10.100 (a) In the absence of a significant effect due to the autoprotolysis of
water, the pH values of the 51.89 10−× M and 79.64 10−× M HClO
solutions can be calculated as described earlier.
For : 5 11.89 10 mol L− −× ⋅
Concentration
HClO(aq) + H1(mol L )−⋅ 2O(l) H3O+(aq) + Cl O (aq)−
initial 51.89 10−× — 0 0
change — +x +x −x
final 51.89 10−× − x — +x +x
3a
28
5 5
[H O ][ClO ][HClO]
( )( )3.0 101.89 10 1.89 10
+ −
−− −
=
× = =× − × −
K
x x xx x
Assume 51.89 10−<< ×x
2 8
7
(3.0 10 )(1.89 10 )7.5 10
− −
−
= × ×
= ×
xx
5
10 Because x < 5% of 1.89 5−×
10 mol L− −
, the assumption was valid. Given this
value, the pH is then calculated to be 7log(7.5 10 ) 6.12.−− × =
For 9.64 7 1× ⋅ :
320
Concentration
HClO(aq) + H1(mol L )−⋅ 2O(l) H3O+(aq) + Cl O (aq)−
initial 79.64 10−× — 0 0
change — +x +x −x
final 79.64 10−× − x — +x +x
3a
28
7 7
[H O ][ClO ][HClO]
( )( )3.0 109.64 10 9.64 10
+ −
−− −
=
× = =× − × −
K
x x xx x
Assume 79.64 10−<< × −x x
2 8 7
7
(3.0 10 )(9.64 10 )1.7 10
− −
−
= × × −
= ×
x xx
x is approximately 10% of 79.64 10−× ; the assumption is not reasonable,
and so you must calculate the value explicitly for the following
expression, using the quadratic equation.
2 8 8 73.0 10 (3.0 10 )(9.64 10 ) 0− − −+ × − × × =x x
Upon solving the quadratic equation, a value of is obtained,
yielding pH = 6.80.
71.6 10−= ×x
(b) To calculate the value, taking into account the autoprotolysis of water,
we can use Eq. 21:
. 3 2a 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
14w 1.00 10−= ×K
8a 3.0 10−= ×K into this equation and
then solve the expression either by trial and error or, preferably, using a
graphing calculator. Alternatively, you can use a computer program
designed to solve simultaneous equations. Because the unknowns include
[H3O+], [ , [HClO], and [OH ]− ClO ]− , you will need four equations. As
seen in the text, the pertinent equations are
321
3a
[H O ][ClO ][HClO]
+ −
=K
w 3[H O ][OH ]+ −=K
3[H O ] [OH ] [ClO ]+ −= + −
initial[HClO] [HClO] [ClO ]−= +
Using either method should produce the same result.
The values obtained for 5 11.89 10 mol L− −× ⋅ are
7 13[H O ] 7.4 10 mol L , pH 6.13 (compare to 6.12 obtained in (a))+ − −= × ⋅ =
7 1[ClO ] 7.3 10 mol L− −= × ⋅ −
−
−
−
5 1[HClO] 1.8 10 mol L− −≅ × ⋅
8 1[OH ] 1.3 10 mol L− −= × ⋅
7initialSimilarly, for [HClO] 9.64 10 :−= ×
7 13[H O ] 1.9 10 mol L , pH 6.72 (compare to 6.80 obtained in (a))+ − −= × ⋅ =
7 1[ClO ] 1.3 10 mol L− −= × ⋅
7 1[HClO] 8.3 10 mol L− −≅ × ⋅
8 1[OH ] 5.3 10 mol L− −= × ⋅
Note than 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.
10.102 Because the process is only 90% efficient, to remove 50.0 kg of SO2 one
must supply enough CaCO3 to remove 55.6 kg of SO2 (50.0 kg is 90% of
55.6 kg). The moles of to be removed is given by: 2SO
21
55600 g 867 mol SO65.06 g mol− =
( )( )
2 3
13
1 mole of SO is consummed by 1 mole of CaCO . Therefore,
867 mol CaCO 100.09 g mol 86.8 kg− =
322
10.104 (1) 2 32 H O(l) H O (aq) OH (aq)+ −+ 14w 1.00 10−= ×K
(2) 3 2 4NH (aq) H O(l) NH (aq) OH (aq)+ −+ + 5b 3(NH ) 1.8 10−= ×K
or
(3)
4 2 3 3
10a 4 w b 3
NH (aq) H O(l) NH (aq) H O (l)
(NH ) / (NH ) 5.56 10
+ +
+ −
+ +
= =K K K ×
(4) 3 2 3 3
5a 3
CH COOH(aq) H O(l) H O (l) CH COO (aq)
(CH COOH) 1.8 10
+ −
−
+ +
= ×K
or
(5)
3 2 3
b 3 w a 3
10
CH COO (aq) H O(l) CH COOH(aq) OH (aq)
(CH COO) / (CH COOH)
5.56 10
− −
−
−
+ +
=
= ×
K K K
(6)
4 3 3 3
3 34 3
4 3
NH (aq) CH COO (aq) NH (aq) CH COOH(aq)[NH ][CH COOH]
(NH CH COO)[NH ][CH COO ]
+ −
+ −
+ +
=K
This equation is obtained by adding Equations (3) and (5) and subtracting
Equation (1):
4 2 3 3[NH (aq) H O(l) NH (aq) H O (l)]+ ++ + +
w b 3/ (NHK K )
)3 2 3[CH COO (aq) H O(l) CH COOH(aq) OH(aq)]−+ + +
w a 3/ (CH COOHK K
2 3[2 H O(l) H O (aq) OH (aq)]+ −− + wK
Because we are starting with pure ammonium acetate, we will use the last
relationship to calculate the concentrations of
4 3 3 3NH , CH COO , NH , and CH COOH in solution.+ −
323
Concentration 1(mol L )−⋅ 4NH (aq)+ + + CH3CH COO (aq) NH (aq)−
3 3COOH(aq)
initial 0.100 0.100 0 0
change −x −x +x +x
equilibrium 0.100 − x 0.100 − x +x +x
For the initial calculation, we will assume that the subsequent
deprotonation of acetic acid by water or the protonation of ammonia by
water will be small compared to the reaction of the ammonium ion with
the acetate ion.
53 3
4 3
[NH ][CH COOH]3.09 10
[NH ][CH COO ]−
−+= ×
25
2
5
5
5 5
4
3.09 10(0.100 )
3.09 100.100
(0.100 ) 3.09 10
(1 3.09 10 ) (0.100) 3.09 10
5.52 10
−
−
−
− −
−
= ×−
= ×−
= − ×
+ × = ×
= ×
xx
xx
x x
x
x
By using this as the value of the concentrations of NH3 and CH3COOH,
we can calculate the concentrations of NH4+, 3CH COO ,− H3O+, and OH−
in solution using the above equilibrium relationship. The values calculated
are
[NH4+] = [CH3COOH] = 0.099
73[H O ] [OH ] 1.00 10+ −= = × −
If we substitute these numbers back into each of the equilibrium constants
expressions, we get very good agreement. This justifies the assumption
that the subsequent hydrolysis reactions of the NH3 and CH3COOH were
small compared to the main reaction.
Alternatively (had this not been the case), we could have solved the
system of equations using a graphing calculator or suitable computer
324
software package by setting up a system of simultaneous equations. The
answer is the same in any case. This problem is simplified by the fact that
the tendency for subsequent hydrolysis of NH3 and CH3COOH have the
same magnitude.
10.106 According to Table 10.3, the amide ion that is formed from the
autoionization of ammonia, 2 3 4NH NH NH ,2+ −+ is a stronger base
than . Therefore, carbonic acid (and other weak acids) is expected to
be stronger in liquid ammonia than in water. Furthermore, carbonic acid is
a stronger acid than NH
OH−
4+; therefore, it will donate a proton to NH3 to
form NH4+. It can then be concluded that carbonic acid will be leveled in
liquid NH3 and will behave as a strong acid.
10.108 (a) The pH is given by pH = ( )1a1 a22 p p+K K . The first and second
dissociations of H3AsO4 are the pertinent values for NaH2AsO4:
( )12pH 2.25 6.77 4.51= + =
(b) For Na2HAsO4 the second and third acid dissociation constants must
be used:
( )12pH 6.77 11.60 9.19= + =
10.110 (a) 3 3 4NH NH NH NH2+ −+ +
(b) acid = 4 2NH , base NH+ −=
(c) 33am amp log log(1 10 ) 33.0−= − = − × =K K
(d) 2am 4 2 4 2
33 17 14
p [NH ][NH ] ( [NH ] [NH ])
[NH ] 1 10 3 10 mol L
+ − + −
+ − − −
= = = =
= × = × ⋅
K x x
(e) 174 2pNH pNH log(3.2 10 ) 17+ − −= = − × =
(f) 4 2 ampNH pNH p 33.0+ −= = ≈K
325
10.112 The osmotic pressure of 0.10 H2SO4 will be slightly higher than that of
0.10 M HCl. When H2SO4 is added to solution it completely deprotonates
into HSO4¯, a small fraction of which deprotonates further to SO4
2-.
Therefore, each mole of H2SO4 gives rise to slightly more than 2 moles of
dissolved ions. HCl, on the other hand, is a strong monoprotic acid giving
exactly two moles of dissolved ions per mole of HCl molecules.
To calculate the moles of dissolved ions in a 0.10 M solution of H2SO4
one assumes that H2SO4 is completely deprotonated, and solves the
equilibrium problem:
Concentration
HSO1(mol L )−⋅ 4¯ (aq) + H2O(l) H3O+(aq) + SO42¯ (aq)
initial 0.10 — 0.10 0
change — +x +x −x
final 0.10 − x — 0.10 + x +x
23 4
a4
2
[H O ][SO ][HSO ]
(0.10 )( )1.2 100.10
−
+ −
−
=
+× =
−
K
x xx
Using the quadratic equation one finds that x = 9.8 × 10−3 and, therefore:
[HSO4-] = 0.0902 M, [H3O+] = 0.1098 M, and [SO4
2¯] = 0.0098 M giving
a total dissolved ion concentration of 0.210 M (compared to a total
dissolved ion concentration of 0.20 for the 0.10 M HCl solution).
Assuming a temperature of 298 K, the osmotic pressure of the 0.10 M
H2SO4 solution is:
( )( )( )1 10.082057 L atm K mol 298 K 0.210 mol L
5.13 atm
− − −Π = = ⋅ ⋅ ⋅
=
RTc 1
While that of the HCl solution is:
( )( )( )1 10.082057 L atm K mol 298 K 0.200 mol L
4.89 atm
− − −Π = = ⋅ ⋅ ⋅ ⋅
=
RTc 1
326
10.114 (where 1 mol refers to the reaction as written) 157 kJ mol−∆ ° = + ⋅H
1
2 1 2
1
2 1
1
2 1
1 21 2
1 1ln
1 12.303 log
1 12.303 log
1 1p p2.303
⎛ ⎞∆ °= − −⎜ ⎟
⎝ ⎠⎛ ⎞∆ °
= − −⎜ ⎟⎝ ⎠⎛ ⎞∆ °
− = ⎜ ⎟⎝ ⎠
⎛ ⎞∆ °− = −⎜ ⎟
⎝ ⎠
K HK R T T
K HK R T T
K HK R T T
HK KR T T
2
2
−
4 :
Let condition 2 be where 25 C,° wp 1=K
1 1 11
1 1 11
57 000 J 1 1p 14298 K2.303 8.314 J K mol
57 000 J 1 1p 14298 K2.303 8.314 J K mol
− −
− −
⎛ ⎞− = −⎜ ⎟⋅ ⋅ ⋅ ⎝ ⎠
⎛ ⎞= + −⎜ ⎟⋅ ⋅ ⋅ ⎝ ⎠
KT
KT
We will not define K1 as the Kw value at some unknown temperature T: 1
w 1 1
3
w
57 000 J mol 1 1p 14298 K2.303 8.314 J K mol
3.0 10 Kp 4.00T
−
− −
⎛ ⎞⋅= + −⎜ ⎟⋅ ⋅ ⋅ ⎝ ⎠
×= +
KT
K
Substituting the value of T = 373 K gives pKw = 12.04. If the solution is
neutral, the pH = pOH = 6.02.
10.116 (a) The structures of alanine, glycine, phenylalanine, and cysteine are
C C
C
O O
N
H
H
HH
H
H
H
H C
C
O O
N
H
H
H
H
alanine glycine
327
C C
C
O O
N
H
H
HH
H
H
H
H C
C
O O
N
H
H
H
H
alanine glycine
All the amino acids have an amine —NH2 function as well as a carboxylic
acid —COOH group.
(b) The pKa value of the —COOH group of alanine is 2.34 and the pKb of
the —NH2 group is 4.31. Often, we find instead of the pKb value, the pKa
value of the conjugate acid of the —NH2 group. The relationship
is used to convert to the pKa bp p p⋅ =K K Kw b value.
(c) To find the equilibrium constant of the reaction of the acid function,
with the base function in the amino acid, we first write the known
equilibrium reactions and corresponding K values:
2 3R COOH(aq) H O(l) H O (aq) R COO (aq)++ + − pKa = 2.34
3a 4.6 10−= ×K
2 2 3R NH (aq) H O(l) RNH (aq) OH (aq)+ −+ + pKb = 4.31
5b 4.9 10−= ×K
If we sum these reactions, we obtain
2 2 3
73 a b
R COOH(aq) R NH (aq) 2 H O(l) H O (aq) OH (aq)
R COO (aq) RNH (aq) 2.3 10
+ −
− + −
+ + +
+ + ⋅ = ×K K
The presence of the H2O, H3O+, and OH− can be eliminated by
subtracting the autoprotolysis reaction of water to give the desired final
equation:
328
2 32 H O(l) H O (aq) OH (aq)+ −+ 14w 1.00 10−= ×K
2 3
1a b w
7
R COOH(aq) R NH (aq) R COO (aq) RNH (aq)
2.3 10
− +
−
+ = +
= ⋅ ⋅
= ×
K K K K
K
(d) A zwitterion is a compound that contains both a positive ion and a
negative ion in the same molecule. Overall, the molecule is neutral, but it
is ionic in that it possesses both positive and negative ions within itself.
This is a common occurrence for the amino acids in which the acid
function reacts to protonate the base site. Given the large value of the
equilibrium constant found in (c), it is more appropriate to write an amino
acid as then as R(NH3R(NH )(COO )+ −2)(COOH).
10.118 (a)
1.0
0.5
0 14 7 0
pH
(b) The major species at pH 7.5 is the doubly deprotonated form.
(c) pH = 5.97
(d) at pH values greater than about 11.4
(e) from approximately pH = 7.0 to 8.4
329