resolução cap 10 atkins
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CHAPTER 10
ACIDS AND BASES
10.1 (a) (b)3 3CH NH + 2 3NH NH + (c) (d)2H CO3 23CO (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)+ +
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)
+ ++ +
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) ++ +
22 4 4H PO (aq) and HPO (aq)
form a conjugate acid-base pair in which
2
2 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)+ + +
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)+ ++ +
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.+
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10.5 (a) Brnsted acid: 3HNO
Brnsted 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) ++ + .
3
4
3-and CO3
2-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) + +
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) ++ +
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) + +
2
4 2HPO and H PO
42-
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)
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(c) sulfuric acid acts as the Lewis base while SO3 acts as a Lewis acid.
10.13 In each case, use then14w 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 L0.02
= =
(b)14
9 1
5
1.0 10[OH ] 1.0 10 mol L
1.0 10
= =
(c)
1412 1
3
1.0 10
[OH ] 3.2 10 mol L3.1 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
2
2 0 2 0 2 0[Ba(OH) ] [Ba ], [OH ] 2 [Ba(OH) ] , where [Ba(OH) ]+ = =
nominal concentration of 2Ba(OH) .
moles of 32 10.25 g
Ba(OH) 1.5 10 mol171.36 g mol
= =
32 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 = = =
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1413 1w
3 2
1.0 10[H O ] 3.4 10 mol L
[OH ] 2.9 10
K +
= = =
10.19 Because if Taking the antilogs
of both sides gives
3 3pH log[H O ], log[H O ] pH.+ += =
H 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) M2[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
= =
44 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 ]+=
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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 = Ka1p
a1 10K
K=
Acid a1pK a1K
(a) 2.123H PO437.6 10
(b) 2.003H PO321.0 10
(c) 2.462H SeO333.5 10
(d) 1.924HSeO
2
1.2 10
(e) The larger , the stronger the acid; thereforea1K
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)+ + +
3 2
a
2
[H O ][ClO ]
[HClO ]K
+
=
2 2 2ClO (aq) H O(l) HClO (aq) OH (aq) + +
2
b
2
[HClO ][OH ]
[ClO ]K
=
(b)2 3HCN(aq) H O(l) H O (aq) CN (aq)
+ + +
3
a
[H O ][CN ]
[HCN]K
+
=
2CN (aq) H O(l) HCN(aq) OH (aq) + +
b
[HCN][OH ]
[CN ]K
=
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(c) 6 5 2 3 6 5C H OH(aq) H O(l) H O (aq) C H O (aq)+ + +
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)
+ +
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).< > >
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 its 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+ + + 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.)
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10.45 (a) Concentration
1
3 2 3(mol L ) CH COOH H O H O CH CO + + 3 2
initial 0 00.29
change x+ x+
equilibrium 0.29 x x x
2 25 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).
21 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 + + +
initial 0 0.29 0
change x x+ x+
equilibrium 0.29 x x x
243 2
a
4 3 13
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
+
+
= = =
= = =
= = = =
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10.47 (a) Concentration1
2 3 4(mol L ) H O NH NH OH + + +
initial 0.057 0 0
change x
x+
x+
equilibrium 0.057 x x x
254
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 + +
initial 0.162 0 0
change x x+ x+
equilibrium 0.162 x x x
2 28
b
5 1
5
5
1.1 100.162 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 + + +
initial 0.35 0 0
change x x+ x+equilibrium 0.35 x+ x+
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25
3 1
3 1
3
3
6.5 100.35
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
=
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Concentration
+1(mol L ) HClO 2 3H O H O ClO+ +
nominal x 0 0
equilibrium
5
2.5 10x
5
2.5 10
5
2.5 10
5 2
8
a 5
(2.5 10 )3.0 10
2.5 10K
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 10Let nominal concentration of NH NH , thenx
= = = =
Concentration
1(mol L ) 2 2 2 2 3 NH NH H O NH NH OH+ + +
nominal 0 0x
equilibrium 41.6 10x 41.6 10 41.6 10
4 26
b 4
2 1
(1.6 10 )1.7 10
1.6 10Solve for ; 1.5 10 mol L
K
xx x
= =
=
10.53 Concentration
1
6 5 2 3 6 5 2(mol L ) C H COOH H O H O C H CO + + +
initial 0. 0 110 0
change x x+ x+
equilibrium 0.110 x x x
2H O octylamine octylamineH OH+ + +
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1
3 6 5 2
253 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+ + +
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 24
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] 0.093K
+
+
= =
= = = =
= = =
10.57- +
3 2 2 3 2 3CH CH COOH(aq) H O(1) CH CH COO (aq) H O (aq)+ +
Concentration 1(mol L )
initial 0.0147 0 0
change x x+ x+
equilibrium0.0147 x x
+ 2 2
53 2 3a
3 2
[CH CH COO ][H O ] 1.3 10[CH CH COOH] 0.0147 0.0147
x xKx
= = =
44 4.4 104.4 10 , Percent deprotonated 100% 3.0%
0.0147x
= = =
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10.59 (a) 4 2 3 3less than 7, NH (aq) H O(l) H O (aq) NH (aq)+ ++ +
(b) 22 3 3greater than 7, H O(l) CO (aq) HCO (aq) OH (aq)
+ +
+
+
(c) 2greater than 7, H O(l) F (aq) HF(aq) OH (aq) + +
(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)+ ++ +
(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)+ ++ +
10.61 (a)
1410w
b 5
a
1.00 105.6 101.8 10
KK K
= = =
Concentration
1 3 2 2 3 2(mol L ) CH CO (aq) H O(1) HCH CO (aq) OH (aq) + +
initial 0.63 0 0
change x+ x+
equilibrium 0.63 x x x
2 2
103 2b
3 2
[HCH CO ][OH ] 5.6 100.63 0.63[CH CO ]
x xKx
= = =
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
1.00 105.6 10
1.8 10
KK
K
= = =
1
4 2 3Concentration (mol L ) NH (aq) H O(l) H O (aq) NH (aq) + + + + 3
initial 0.19 0 0
change x x+ x+
equilibrium 0.19 x x
2 2103 3
a
4
[H O ][NH ]5.6 10
[NH Cl] 0.19 0.19
x xK
x
+= = =
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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) + +
+ +
initial 0.055 0 0
change x+ x+
equilibrium 0.055 x x x
2 2 253 2 5
a 3
2 6
[H O ][Al(H O) OH ]1.4 10
0.055 0.055[Al(H O) ]
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) + +
initial 0.065 0 0
change x x+ x+
equilibrium 0.065 x x x
14 2 2
5wb 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) + + + +
initial 0.510 0 0
change x x+ x+
equilibrium 0.510 x x
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2 2113 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 LM
=
Concentration1
2 2 3 2 2 3 2(mol L ) H O(l) C H O (aq) HC H O (aq) OH (aq) + +
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 =4
5.75 g NH Br,
197.95 g mol
4
4 4
4
M1mol 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) + + + +
initial 0.587 0 0
change x x+ x+
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equilibrium 0.587 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 L0.500 L
=
Concentration
1(mol L ) 2 3 2 3H O(l) CH CO (aq) CH COOH(aq) OH (aq) + +
initial 0.0400 0 0
change x+ x+
equilibrium 0.0400 x x x
1410w 3
b 5
a 3 2
[CH COOH][OH ]1.00 105.6 10
1.8 10 [CH CO ]
KK
K
= = = =
2 2105.6 10
0.0400 0.0400
x x
x
=
6 134.7 10 mol L [CH COOH]x = =
(b) 4 43
4
2.16 g NH Br 1 mol NH Br 1 mL
400 mL 97.95 g NH Br 10 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)+ ++ +
initial 0.0551 0 0
change x+ x+
equilibrium 0.0551 x x
1410w 3
a 5
b 4
[NH ][H O ]1.00 105.6 10
1.8 10 [NH ]
KK
K
3
+
+
= = = =
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2 2105.6 10
0.0551 0.0551
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 12 2
pH = + (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)
+ + +
2
4 2 3 4HSO (aq) H O(l) H O (aq) SO (aq) ++ +
(b)3 4 2 3 2 4H AsO (aq) H O(l) H O (aq) H AsO (aq)
+ + +
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)
+
+
+ +
+ +
(c)
6 4 2 2 3 6 4 2C H (COOH) (aq) H O(l) H O (aq) C H (COOH)CO (aq)+ + +
2
6 4 2 2 3 6 4 2 2C H (COOH)CO (aq) H O(l) H O (aq) C H (CO ) (aq) ++ +
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
+ + +
initial 0.15 0.15 0
change x x+ x+
equilibrium 0.15 x 0.15 x+ x
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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 L2
[H O ] 0.15 (0.15 0.0104) mol L 0.16 mol L
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
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(c) Because the second ionization can be ignored.a2 a1 ,K K
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Concentration
1
2 3 2 3 3(mol L ) H CO (aq) H O(l) H O (aq) HCO (aq) + + +
initial 0.0456 0 0
change x x+ x+
final 0.0456 x x+ x+
3 3
a1
2 3
27
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 10Because 1% of 0.0456, the assumption was valid.
[H O ] [HCO ] 1.4 10 mol L
K
x x x
x x
x
x
xx
x
+
+
=
= =
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10.83 The equilibrium reactions of interest are now the base forms of the
carbonic acid equilibria, so K values should be calculated for the
following changes:
2
3 2 3
144w
b1 11
a2
CO (aq) H O(l) HCO (aq) OH (aq)1.00 10
1.8 105.6 10
KK
K
+ +
= = =
3 2 2 3
148w
b2 7
a1
HCO (aq) H O(l) H CO (aq) OH (aq)
1.00 102.3 10
4.3 10
KK
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) + +
initial 0.0456 0 0
change x x+ x+
final 0.0456 x+ x+
3
b1 2
3
[HCO ][OH ]
[CO ]K
=
24 ( ) ( )1.8 10
0.0456 [0.0456 ]
x x x
x
= =
Assume that 0.0456x
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
= =
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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
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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
+
+
+ + =
+ + =
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 a2
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
6 243
2 3 8
6 2
3 a1
3 8
2 72 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
+
+
= = =
= = =
= = =
Thus, in a 10.150 mol L solution at the dominant species will
be with a concentration of
pH 5.50,
3
HSO 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 L1 = and the
concentration of 23SO
will be
1 1(0.036) (0.150 mol L ) 0.0054 mol L . =
10.89 (a) Concentration
13 2 3(mol L ) B(OH) 2 H O H O B(OH) + + + 4
initial 41.0 10 0 0
change x x+ x+
equilibrium 41.0 10 x x x
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2 210 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) + + + 4
equilibrium 41.0 10
x y
Concentration 12 3
(mol L ) 2 H O H O OH + +
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
aK41.0 10 . Let initiala =
3B(OH) ,
10 a
a
14
w
7.2 10 or
1.0 10
aK xy xyK y
a x a x
K xz
= = =
= =
Electroneutrality requires
wor ; hence, ( ).y z z x y K xz x x y= + = = = Substituting for from above:y
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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; a2 a1.K K
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22
a1
2 2 3
2 2 2 3
1
3
1.5 100.10
1.5 10 1.5 10 0
1.5 10 (1.5 10 ) 6.0 10
[H O ] 0.032 mol L2
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 ][PHPO (aq) PO (aq) H O (aq), 2.1 10
a
a
a
K
K
K
+ = =
+ = =
+ = =
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 toobtain 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 ] 1.1010 [H PO ] 1.1010 1.35 [H PO ]
1.4810 [H PO ], and
[PO ] 3.74 10 [HPO ] 3.74 10 1.4810 [H PO ]
5.5410 [H PO ].
=
= =
=
= =
=
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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 ] 1.4810 [H PO ] 5.5410 [H PO ]
1.5 10
we find:
H PO 6.410 ,
[H PO ] 1.35 [H PO ] 8.610 ,
[HPO ] 1.4810 [H PO ] 9.410 , and
[PO ]
+ + =
= + +
=
=
= =
= =
= -16 -183 45.5410 [H PO ] 3.510 . =
)
=4
=
4
10.93 We can use the relationship derived in the text:
, in which HA is any strong acid.2
3 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
7
3[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
8
3[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 the4
1.00 10
M and 61.00 10 M HBrO
solutions can be calculated as described earlier.
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For 4 11.00 10 mol L :
Concentration
1(mol L )2 3HBrO(aq) H O(l) H O (aq) BrO (aq)
+ + +
initial 0 04
1.00 10
change x+ x+
final 41.00 10 x x+ x+
3
a
[H O ][BrO ]
[HBrO]K
+
=
29
4 4
( )( )2.0 10
1.00 10 [1.00 10 ]
x x x
x x
= =
Assume 41.00 10x
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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 :=
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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) + + +
initial 58.50 10 0 0
change x x+ x+
final5
8.50 10 x
x+ x+
[ ]3
a
210
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
+
=
= =
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Concentration
2 3(mol L ) HCN(aq) H O(l) H O (aq) CN (aq) + + +
initial 67.37 10 0 0
change x x+ x+final 67.37 10 x x+ x+
3
a
210
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
+
=
= =
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[ ]3
a
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 kg780 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
7
3
1000 m 100 cm 1 L2 cm 2.6 Km 5.2 10 L
1 Km 1 m 1000 cmV
= =
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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) + + +
initial 51.50 10 0 0
change x+ x+
final 51.50 10 x x+ x+
[ ]
3 3
a2 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) + + +
initial 51.50 10 51.50 10 0
change x+ x+final 51.50 10 51.50 10 x+
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2
3 4
a
4
52
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 13
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 ) =
lnT 11/ (K )T wK
0.00335625 C 32.2362
0.00322637 C 31.4943
0.00319540 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
an equation 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
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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
KT
KT
+
+
=
= =
=
=
=
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
22-1
0.942 g COC : 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 OH: 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 Na22.99 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
=
=
=
= =
-10.0429 mol O
g mol=
Dividing through by 0.0107 moles, we find the empirical formula to be:
C2HNaO4.
A molar mass of 112.02 gmol-1
indicates that this is also the molecular
formula.
(b)
C C
O
O
O
O
H-
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(c) The dissolved substance is sodium oxalate, it is capable of gaining or
loosing a proton and, therefore, amphiprotic. ( )1
122
pH p p 2.7a aK K= + =
10.107 We wish to calculate for the reactionaK
2 3HF(aq) H O(l) H O (aq) F (aq)+ + +
This equation is equivalent to
HF(aq) H (aq) F (aq)+ +
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+ + +
(b)2 2 2
15D 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)
1(mol L )
initial 1.00 0 0
change x x+ x+
equilibrium 1.00 x x x
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+ 243 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 Tffor this solution we need the concentration in moles of
solute per kilogram of solvent. Assuming a density of 1 gcm-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
1.130.910 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 K T T= = = =
a
a
1 1ln
K H H T T
K R T T R TT
= =
This is the vant Hoff equation.4
4 1 1
1.768 10 303 K 293 K ln 0.0017
293 K 303 K 1.765 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 -115
84800 J mol8.3145J K mol 298 K
1.37 10G
R T K 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)+
initial 0 0
change x+ x+
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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 forKin part (b) above, it is apparent that as T
increasesKwill 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
01470
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
than24HAsO
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
01470
pH
O
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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 6
2[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)
initial 7.0 106 0 0
change x x+ x+
equilibrium -67.0 10 x x
2a 6.37 7
6
63
6
a 10 10 4.27 107.0 10
[H O ] 1.73 10
pH log(1.73 10 ) 5.75
pK xK
x
+
= = = =
= =
= =
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