(a) 16
16.2 (b) (c) 16.3
Acid-Base Equilibria
Visualizing Concepts
Plan. The stronger the acid, the greater the extent of ionization. The stronger the acid, the weaker its conjugate base. In an acid-base reaction, equilibrium will favor the side with the weaker acid and base. Solve. HY is stronger than HX. Starting with six HY molecules, four are dissociated; of six HX molecules, only two are dissociated. Because it is dissociated to a greater extent, HY is the stronger acid. If HY is the stronger acid, Y is the weaker base and X is the stronger base. HX and Y , the reactants, are the weaker acid and base. Equilibrium lies to the left, and K c < 1.
Plan. Strong acids are completely ionized. The acid that is least ionized is weakest, and has the smallest K a value. At equal concentrations, the weakest acid has the smallest [H + ] and highest pH. Solve. (a) (b) (c) HY is a strong acid. There are no neutral HY molecules in solution, only H + cations and Y anions. HX has the smallest K a value. It has most neutral acid molecules and fewest ions. HX has the fewest H + and highest pH.
[H ] [HA]Initial 100.
16.4
Plan. The definition of % ionization is (a) (b)
Solve.
Curve C shows the effect of initial concentration on % ionization of a weak acid. The % ionization is inversely related to initial acid concentration; only curve C shows a decrease in % ionization as acid concentration increases.
16.6
Plan. Write the molecular formula so we can count the correct number of valence electrons. Use the atom connectivity shown to draw the Lewis structure. Solve. (a) The molecular formula is (CH 3 ) 2 NH, or C 2 H 7 N. The number of valence electrons is 2(4) + 7 + 5 = 20 e , 10 e pr.
(b)
The compound is an amine. It is an ammonia molecule where two H atoms have been replaced by CH 3 groups.
241
16 Acid-Base Equilibria
16.7
Solutions to Black Exercises
Plan. Evaluate the interactions of Na + and X with H 2 O. Solve. Na + does not affect the [H + ] or [OH ] of an aqueous solution. It is a "negligible" acid in water (which can be thought of as the conjugate acid of the strong base NaOH). X is the conjugate base of HX. It is not a negligible base in water, because we see from the diagram that one X has gained an H + to form HX. In this solution, H 2 O acts as the Brnsted acid, according to the hydrolysis equilibrium:
X (aq) H 2 O(l)
HX(aq)
OH (aq).
The missing ion is OH (aq). According to the equilibrium reaction, the number of HX molecules and OH ions are equal. Since there is 1 HX molecule in the diagram, 1 OH should be shown. 16.9 The carboxyl group in the H-atom group at the "top" of the molecule.
The group on the right of the molecule is not a carboxyl group because it contains no ionizable H. 16.10 (a) Plan. Count valence electrons and draw the correct Lewis structures. Consider the definition of Lewis acids and bases. Solve.
PCl 4 + accepts an electron pair from Cl ; PCl 4 + is the Lewis acid and Cl is the Lewis base. (b) The hydrated cation is an oxyacid: the ionizable H is attached to O, which is bound to the central cation. As the charge on the cation increases, it attracts more electron density from the OH bond, which becomes weaker and more polar. The degree of ionization increases and the equilibrium constant (K a ) increases.
242
16 Acid-Base Equilibria
16.12
Solutions to Black Exercises
Arrhenius and Brnsted-Lowry Acids and Bases
When NaOH dissolves in water, it completely dissociates to form Na + (aq) and OH (aq). CaO is the oxide of a metal; it dissolves in water according to the following process: CaO(s) + H 2 O(l) Ca 2 +(aq) + 2OH (aq). Thus, the properties of both solutions are dominated by the presence of OH (aq). Both solutions taste bitter, turn litmus paper blue (are basic), neutralize solutions of acids, and conduct electricity. (a) According to the Arrhenius definition, a base when dissolved in water increases [OH ]. According to the Brnsted-Lowry theory, a base is an H + acceptor regardless of physical state. A Brnsted-Lowry base is not limited to aqueous solution and need not contain OH or produce it in aqueous solution.
NH 3 (g)
+
16.14
(b)
H 2 O(l)
NH
4
(aq )
OH (aq) When NH 3 dissolves in water, it
accepts H from H 2 O (B-L definition). In doing so, OH is produced (Arrhenius definition). Note that the OH produced was originally part of the H 2 O molecule, not part of the NH 3 molecule. 16.16 A conjugate acid has one more H + than its conjugate base. (a) 16.18 (a) (b) (c) 16.20 (a) (b) (c) HCN B-L acid HBrO(aq) HSO 4 (aq) H 3 O + (aq) (b) + OH B-L base H 2 O(l) HCO 3 (aq) HSO 3 (aq) (c) H 2 PO 4
(d) +
C 2 H 5 NH 3 + Conjugate base BrO (aq) SO 4 2 (aq) H 2 O(l)
Conjugate acid H 3 O + (aq) H 2 CO 3 (aq)
H 2 C 6 H 7 O 5 (aq) + H 2 O(l) H 2 C 6 H 7 O 5 (aq) + H 2 O(l)
H 2 SO 3 (aq) H 3 C 6 H 7 O 5 (aq) + OH (aq)
HC 6 H 7 O 5 2 (aq) + H 3 O + (aq)
H 3 C 6 H 7 O 5 is the conjugate acid of H 2 C 6 H 7 O 5 HC 6 H 7 O 5 2 is the conjugate base of H 2 C 6 H 7 O 5
16.22
(a) (b) (c) (d) (e)
C 2 H 3 O 2 , weak base; HC 2 H 3 O 2 , weak acid HCO 3 , weak base; H 2 CO 3 , weak acid O 2 , strong base; OH , negligible acid Cl , negligible base; HCl, strong acid NH 3 , weak base; NH 4 + , weak acid HNO 3 . It is one of the seven strong acids (Section 16.5). Also, in a series of oxyacids with the same central atom (N), the acid with more O atoms is stronger (Section 16.10). NH 3 . When NH 3 and H 2 O are combined, as in NH 3 (aq), NH 3 acts as the B-L base, accepting H + from H 2 O. NH 3 has the greater tendency to accept H + . For binary hydrides, base strength increases going to the left across a row of the periodic chart (Section 16.10).
16.24
(a)
(b)
243
16 Acid-Base Equilibria
16.26 (a) Base OH (aq) + + Acid NH 4 + (aq)
Solutions to Black Exercises
Conjugate acid H 2 O(l)
+ +
Conjugate base NH 3 (aq)
OH is a stronger base than NH 3 (Figure 16.4), so the equilibrium lies to the right. (b) C 2 H 3 O 2 (aq) + H 3 O + (aq) HC 2 H 3 O 2 (aq) + H 2 O(l)
H 3 O + is a stronger acid than HC 2 H 3 O 2 (Figure 16.4), so the equilibrium lies to the right. (c) F (aq) + HCO 3 (aq)
HC 2 H 3 O 2 (aq)
+
H 2 O(l)
CO 3 2 is a stronger base than F , so the equilibrium lies to the left.
Autoionization of Water
16.28 (a) (b) H 2 O(l)
H + (aq) + OH (aq)
K w = [H + ][OH ]. The [H 2 O(l)] is omitted because water is a pure liquid. The molarity (mol/L) of pure solids or liquids does not change as equilibrium is established, so it is usually omitted from equilibrium expressions. If a solution is basic, it contains more OH than H + ([OH ] > [H + ]). 10 7 M, the solution is
(c) 16.30
In pure water at 25 C, [H + ] = [OH ] = 1 10 7 M. If [OH ] > 1 basic; if [OH ] < 1 10 7 M, the solution is acidic. (a)
[OH ]
KW [H ]
Kw [H ]
1.0 7.5
1.0 6.5
10 10
3
14
M
14
1.3
10
12
M 1
10
7
M; acidic
(b) (c)
[OH ]
10 10
10
M
1.5
10
5
M 1
10
7
M; basic
[H + ] = 10[OH ]; K w = 10[OH ][OH ] = 10[OH ] 2 [OH ] = (K w /10) 1 /2 = 3.2 10 8 M < 1 10 7 M; acidic 10 16 = [D + ] 2 ;
16. 32
K w = [D + ][OD ]; for pure D 2 O, [D + ] = [OD ]; 8.9 [D + ] = [OD ] = 3.0 10 8 M
The pH Scale
16.34 [H + ] A = 500 [H + ] B From Solution 16.33,
pH log [H ]B 500 [H ]B log 1 500 2.70
pH
log
[H ]B [H ]A
The pH of solution A is 2.70 pH units lower than the pH of solution B, because [H + ] A is 500 times greater than [H + ] B . The greater [H + ], the lower the pH of the solution. 16.36 (a) K w = [H + ][OH ]. If HNO 3 is added to water, it ionizes to form H + (aq) and NO 3 (aq). This increases [H + ] and necessarily decreases [OH ]. When [H + ] increases, pH decreases.
244
16 Acid-Base Equilibria
(b) (c)
Solutions to Black Exercises
On Figure 16.5, 1.4 10 2 M OH is between pH = 12 (1 10 2 M OH ) and pH 13 (1 10 1 M OH ), slightly higher than pH = 12, so we estimate pH = 12.1. If pH > 7, the solution is basic. pH = 6.6 is midway between pH 6 and pH 7 on Figure 16.5. At pH = 7, [H + ] = 1 10 7; at pH = 6, [H + ] = 1 10 6 = 10 10 7.
A reasonable estimate is 5
10 7 M H + . By calculation: 10 7 10 7 = 10 10 8.
pH = 6.6, [H + ] = 10 pH = 10 6.6 = 3 At pH = 6, [OH ] = 1
10 9; at pH = 7, [OH ] = 1
A reasonable estimate is 5 16.38 pH 4.75 2.11 2.19 7.93 16.40 pOH 9.25 11.89 11.81 6.07 1.8 7.8 6.5 1.2
10 8 M OH . By calculation: 10 8 M OH . acidic or basic acidic acidic acidic basic
pOH = 14.0 6.6 = 7.4; [OH ] = 10 pOH = 10 7.4 = 4 [H + ] 10 5 M 10 3 M 10 3 M 10 8 M 5.6 1.3 1.5 8.6
[OH ] 10 10 M 10 12 M 10 12 M 10 7 M
The pH ranges from 5.25.6; pOH ranges from (14.05.2 =) 8.8 to (14.05.6 =) 8.4. [H + ] = 10 pH, [OH ] = 10 pOH [H + ] = 10 5.2 = 6.31 10 6 = 6 10 6 M; [H + ] = 10 5.6 = 2.51 10 6 M. 10 9 = 4 10 9 M. 10 6 = 3 10 6 M
The range of [H + ] is 6 [OH ] = 10 8.8 = 1.58
10 6 M to 3 10 9 = 2
10 9 M; [OH ] = 10 8.4 = 3.98 10 9 M.
The range of [OH ] is 2
10 9 M to 4
(The pH has one decimal place, so concentrations are reported to one sig fig.)
Strong Acids and Bases
16.42 (a) (b) A strong base is completely dissociated in aqueous solution; a strong base is a strong electrolyte. Sr(OH) 2 is a soluble strong base. Sr(OH) 2 (aq) Sr 2 +(aq) + 2OH (aq)
0.045 M Sr(OH) 2 (aq) = 0.090 M OH (c) Base strength should not be confused with solubility. Base strength describes the tendency of a dissolved molecule (formula unit for ionic compounds such as Mg(OH) 2 ) to dissociate into cations and hydroxide ions. Mg(OH) 2 is a strong base because each Mg(OH) 2 unit that dissolves also dissociates into Mg 2 +(aq) and OH (aq). Mg(OH) 2 is not very soluble, so relatively few Mg(OH) 2 units dissolve when the solid compound is added to water.
245
16 Acid-Base Equilibria
16.44 (a) (b)
0.525 g HClO 4 2.00 L soln 1 mol HClO 4 100.5 g HClO 4
Solutions to Black Exercises
For a strong acid, which is completely ionized, [H + ] = the initial acid concentration. 0.00835 M HNO 3 = 0.00835 M H + ; pH = log (0.00835) = 2.08
2.612 10
3
2.61
10
3
M HClO 4
[H + ] = 2.61 (c) Mc
10 3 M; pH = log (2.612 V d ; 0.500 L = 500 mL
10 3) = 2.583
Vc = Md
1.00 M HCl
Md HCl
5.00 mL HCl = M d HCl
1.00
500 mL HCl
10
2
1.00 M 5.00 mL 500 mL
M HCl 1.00
10
2
MH
pH = log (1.00 (d)
[H ]tota l [H ]tota l
10 2) = 2.000
M L
mol H from HCl mol H from HI ; mol total L solution (0.020 M HCl 0.0500 L) (0.010 M HI 0.200 L
0.150 L)
[H ]total
1.0
10
3
mol H 1.50 0.200 L
10
3
mol H
0.0125
0.013 M
pH = log (0.0125) = 1.90 16.46 For a strong base, which is completely dissociated, [OH ] = the initial base concentration. Then, pOH = log [OH ] and pH = 14 pOH. (a) (b) 0.012 M KOH = 0.012 M OH ; pOH = log (0.012) = 1.92; pH = 14 1.92 = 12.08
1.565 g KOH 0.5000 L 1 mol KOH 56.106 g KOH 0.055787 0.05579 M [OH ]
pOH = log (0.055787) = 1.2535; pH = 14 pOH = 12.7465 (c) Mc Vc = Md Vd 10.0 mL = M d Ca(OH) 2
10.0 mL
0.250 M Ca(OH) 2
Md Ca(OH)2
500 mL
2.10 10
4
0.0105 M Ca(OH)2 500.0 mL
M Ca(OH)2
Ca(OH) 2 (aq)
Ca 2 +(aq) + 2OH (aq) 10 4 M) = 4.20 10 4 M
[OH ] = 2[Ca(OH) 2 ] = 2(2.10 pOH = log (4.20 (d)
[OH ]tota l
10 4) = 3.377; pH = 14 pOH = 10.623
mol OH from NaOH mol OH from Ba(OH)2 total L solution
3
(7.5
10
M
3.00
0.0400 L) 2(0.015 M 0.0500 L
10
4
0.0100 L)
10
4
[OH ]total
mol OH 3.0 0.0500 L
mol OH
0.01200
0.012 M OH
pOH = log (0.0120) = 1.92; pH = 14 pOH = 12.08
246
16 Acid-Base Equilibria
16.48 pOH = 14 pH = 14.00 12.05 = 1.95
Solutions to Black Exercises
10 2 M 10 3 M
pOH = 1.95 = log[OH ]; [OH ] = 10 1.95 = 0.01122 = 1.1
[OH ] = 2[Ca(OH) 2 ]; [Ca(OH) 2 ] = [OH ] / 2 = 0.01122/2 = 5.6 16.50
Upon dissolving, Li 2 O dissociates to form Li + and O 2 . According to Equation 16.22, O 2 is completely protonated in aqueous solution. Li 2 O(s) + H 2 O(l) 2Li + (aq) + 2OH (aq)
Thus, initial [Li 2 O] = [O 2 ]; [OH ] = 2[O 2 ] = 2[Li 2 O]
[Li 2 O] mol Li 2 O L solution 2.50 g Li 2 O 1 mol Li 2 O 29.88 g Li 2 O 1 1.500 L 0.0558 0.0558 M
[OH ] = 0.11156 = 0.112 M; pOH = 0.9525 = 0.958 pH = 14.00 pOH = 13.0475 = 13.048
Weak Acids
16.52 (a) HC6H5O (aq) H (aq) C 6 H 5 O (aq); K a
[H ] [C 6 H 5 O ] [HC6 H 5 O]
[ H 3 O ] [C 6 H 5 O ] [HC 6 H 5 O]
HC6H5O (aq) + H2O (l) H 3 O (aq) C 6 H 5 O (aq); K a (b) HCO3 (aq) H (aq) CO 3 2 (aq);
[H ] [CO 3 2 ] [HCO3 ]
Ka
HCO3 (aq) + H2O (aq) H 3 O (aq) CO 3 2 (aq); K a HC8H7O2 (aq) H (aq) C 8 H 7 O 2 (aq); K a [H + ] = [C 8 H 7 O 2 ] = 10 2.68 = 2.09 [HC 8 H 7 O 2 ] = 0.085 2.09
Ka ( 2.09 10 3 ) 2 0.0829 5.3
[H 3 O ] [CO 3 2 ] [HCO3 ]
16.54
[ H ] [C 8 H 7 O 2 ] [HC8 H 7 O 2 ]
10 3 = 2.1
10 3 M
10 3 = 0.0829 = 0.083 M
10
5
16.56
[H + ] = 0.132
[BrCH 2 COOH] i nitial = 0.0132 M
BrCH 2 COOH(aq) initial equil. 0.100 M 0.087
[H ] [ BrCH2 COO ] [ BrCH2 COOH]
H + (aq) 0 0.0132 M
+
BrCH 2 COO (aq) 0 0.0132 M
Ka
(0.0132) 2 0.087
2.0
10
3
247
16 Acid-Base Equilibria
16.58 [H + ] = 10 pH = 10 3.25 = 5.623 10 4 = 5.62
Solutions to Black Exercises
10 4 M
4
Ka
6.8 6.8
6.8
10
4
[H ][F ] [HF]
(5.623 x 5.623
10 4 ) 2 10
10 4(x 5.623 10 4 x = 3.824 10 3 = 1.0 L = 1.027
10 4) = (5.623 10 7 + 3.162 10 3 M HF 10 3 M
10 4) 2 ; 10 7 = 6.986 10 7
x = 1.027 mol = M 16.60 initial equil.
Ka
0.200 L = 2.055
10 4 = 2.1 +
10 4 mol HF ClO (aq) 0 xM
HClO(aq) 0.0075 M (0.0075 x) M
x2 (0.0075 x)
H + (aq) 0 xM
[H ] [ClO ] [HClO]
x2 0.0075
3.0
10
5
x 2 = 0.0075 (3.0 [HClO] = 7.5
Check .
10 8); x = 1.5
10 5 M = [H + ] = [H 3 O + ] = [ClO ] 10 3 = 7.5 10 3 M
10 3 1.5
10 6 = 7.485
100
4.7 10 5 MH 0.0075 M HClO
0.20% ionization; the assumption is valid
16.62
(a) initial equil
Ka
HOCl(aq) 0.125 M (0.125 x) M
x2 (0.125 x)
H + (aq) 0 xM
+
OCl (aq) 0 xM
[H ][OCl ] [HOCl]
x2 0.125
3.0
10
8
x 2 = 0.125 (3.0
Check .
10 8); x = [H + ] = 6.1
100
10 5 M, pH = 4.21
6.1 10 5 M H 0.125 M HOCl
0.049% ionization; the assumption is valid
(b)
Ka
[H ][C 6 H 5 O ] [C 6 H 5 OH]
x2 (0.0085 x)
x2 0.0085
1.3
10
10
x 2 = 0.0085 (1.3
10 10); x = [H + ] = 1.1
10 6 M, pH = 5.98 10 3 M C 6 H 5 OH, and + OH (aq) 0 xM
Check. Clearly 1.1 10 6 M H + is small compared to 8.5 the assumption is valid. (c) initial equil HONH 2 (aq) + H 2 O(l) 0.095 M (0.095 x) M
HONH 3 + (aq) 0 xM
248
16 Acid-Base Equilibria
Kb [HONH3 ] [OH ] [HONH2 ] x2 (0.095 x)
Solutions to Black Exercises
x2 0.095 1.1 10
8
x 2 = 0.095 (1.1
Check .
10 8); x = [OH ] = 3.2
100
10 5 M, pH = 9.51
3.2 10 5 M OH 0.095 M HONH2
0.034% ionization; the assumption is valid
16.64
Calculate the initial concentration of HC 9 H 7 O 4 .
2 tablets 500 mg tablet 1g 1000 mg 0.02220 1 mol HC9 H 7 O 4 180.2 g HC9 H 7 O 4 0.005549 0.00555 mol HC9 H 7 O 4
0.005549 mol HC9 H 7 O 4 0.250 L
0.0222 M HC9 H 7 O 4
HC9H7O4 (aq) Initial equil
Ka 3.3 10
C9H7O4 + 0M xM
x2 (0.0222 x)
H+ (aq ) 0M xM
0.0222 M (0.0222 x)
4
[H ][C 7 H 9 O 4 ] [HC7 H 9 O 4 ]
Assuming x is small compared to 0.0222, x 2 = 0.0222 (3.3 10 4); x = [H + ] = 2.7
100
10 3 M
2.7 10 3 M H 0.0222 M HC9 H 7 O 4
12% ionization; the assumption is not valid
Using the quadratic formula, x 2 + 3.3
x 3.3 10
4
10 4 x 7.325
10 6 = 0
3.3 10
4
( 3.3
10 4 ) 2 2(1)
4(1) ( 7.325
10 6 )
2.941 2
10
5
x = 2.547 16.66 (a)
10 3 = 2.5
10 3 M H + ; pH = log(2.547
10 3) = 2.594 = 2.59
HC 3 H 5 O 2 (aq) H + (aq) + C 3 H 5 O 2 (aq)
Ka 1.3 10
5
[H ][C 3 H 5 O 2 ] [HC 3 H 5 O 2 ]
x2 0.250 x
x 2 = 0.250 (1.3
% ionization
10 5); x = 1.803
10 3 = 1.8
100
3
10 3 M H +
0.721%
3
1.803 10 3 M H 0.250 M HC3 H 5 O 2
(b)
x2 ~ 1.3 0.0800
% ionization
10
5
;x
1.020
10
1.0
10
MH
1.020 10 3 M H 0.0800 M HC3 H 5 O 2
100
1.27%
249
16 Acid-Base Equilibria
(c)
x2 0.0200 1.3 10 5 ; x 5.099 10
4
Solutions to Black Exercises
5.1 10
4
MH
% ionization
5.099 10 4 M H 0.0200 M HC 3 H 5 O 2
100
2.55%
16.68
HX(aq) H (aq)
X (aq); K a
[H ] [X ] [HX]
[ H ]2 ; [H ] K 1 / 2 [HX]1 / 2 a [HX]
[H ] [X ]; assume the % ionization is small; K a
pH
1/2 log K a [HX]1/2
1/2 log K a
log [HX]1/2 ; pH = 1/2 log K a 1/2 log [HX]
This is the equation of a straight line, where the intercept is 1/2 log K a , the slope is 1/2, and the independent variable is log [HX]. 16.70 H 2 C 4 H 4 O 6 (aq) HC 4 H 4 O 6 (aq
H + (aq) H + (aq)
+ +
HC 4 H 4 O 6 (aq) C 4 H 4 O 6 2 (aq)
K a 1 = 1.0 K a 2 = 4.6
10 3 10 5
Begin by calculating the [H + ] from the first ionization. The equilibrium concentrations are [H + ] = [HC 4 H 4 O 6 ] = x, [H 2 C 4 H 4 O 6 ] = 0.25 x.
K a1 [H ] [HC 4 H 4 O 6 ] [H 2 C 4 H 4 O 8 ] x2 ; x 2 1.0 0.25 x 10
3
x 2.5
10
4
0
Using the quadratic formula, x = 1.532 10 2 = 0.015 M H + from the first ionization. Next calculate the H + contribution from the second ionization. HC 4 H 4 O 6 (aq) initial equil.
K a2
H + (aq) 0.015 (0.015 + y)
+
C 4 H 4 O 6 2 (aq) 0 y
0.015 (0.015 y)
(0.015 y) (y) (0.015 y) 4.6
10 5 ; assuming y is small compared to 0.015,
y = 4.6
10 5 M HC 4 H 4 O 6 2 (aq)
This assumption is reasonable, since 4.6 10 5 is only 0.3% of 0.015. [H + ] = 0.015 M (first ionization) + 4.6 10 5 (second ionization). Since 4.6 10 5 is 0.3% of 0.015 M, it can be safely ignored when calculating total [H + ]. Thus, pH = log(0.01532) = 1.18148 = 1.181. Assumptions: 1) 2) The ionization can be treated as a series of steps (valid by Hess' law). The extent of ionization in the second step (y) is small relative to that from the first step (valid for this acid and initial concentration). This assumption was used twice, to calculate the value of y from K a 2 and to calculate total [H + ] and pH.
250
16 Acid-Base Equilibria
Weak Bases
16.72
Solutions to Black Exercises
Organic amines (neutral molecules with nonbonded pairs on N atoms) and anions that are the conjugate bases of weak acids function as weak bases. (a) C3H7NH2(aq) + H2O(l) C 3 H 7 NH 3 (aq) OH (aq); K b HPO42 (aq) + H2O(l) H 2 PO4 (aq) OH (aq); K b
[C 3 H 7 NH 3 ] [OH ] [C 3 H 7 NH 2 ]
[HPO4 2 ]
16.74
(b)
[H 2 PO4 ] [OH ]
(c) 16.76 initial equil.
Kb
C6H5CO2 (aq) + H2O(l) C 6 H 5 CO 2 H(aq) OH (aq); K b BrO (aq) + H 2 O(l) 1.15 M (1.15 x) M
[HOBr] [OH ] [ BrO ] x2 1.15 x x2 1.15 4.0
[C 6 H 5 CO 2 H] [OH ] [C 6 H 5 CO 2 ]
HOBr(aq) 0 xM
10
6
+
OH (aq) 0 xM
x 2 = 1.15 (4.0
10 6); x = [OH ] = 2.14
3
10 3 = 2.1
10 3 M; pH = 11.33
Check .
16.78 (a)
2.1
10
M OH
1.15 M C 3 H 5 O 2
100
0.19% hydrolysis; the assumption is valid
10 6 = 8.9 10 5 M + OH-(aq) 0 8.9 10 5 M
pOH = 14.00 9.95 = 4.05; [OH ] = 10 4.05 = 8.91 C18H21NO3(aq) + H2O(1) initial equil. 0.0050 M (0.0050 8.9 10 5) 8.9
C18H21NO3H+(aq) 0 10 5 M
5
Kb
(b)
[C18 H 21NO3 H ] [OH ] [C18 H 21NO3 ]
(8.91
10 5 ) 2 10 )
(0.0050 8.91
10 5) = 5.79
1.62
10
5
1.6
10
6
pK b = log (K b ) = log (1.62
The a K K b Relationship; Acid-Base Properties of Salts
16.80 (a) (b) (c) We need K a for the conjugate acid of CO 3 2 , K a for HCO 3 . K a for HCO 3 is K a 2. K b = K w /K a = 1.0 K b for CO 3 2 (1.8 10 14/5.6 10 11 = 1.8 10 4 10 5).
10 4) > K b for NH 3 (1.8
CO 3 2 is a stronger base than NH 3 . 16.82 (a) (b) Ammonia is the stronger base because it has the larger K b value. Hydroxylammonium is the stronger acid because the weaker base, hydroxylamine, has the stronger conjugate acid.
251
16 Acid-Base Equilibria
(c) K a for NH 4 + = K w /K b for NH 3 = 1.0
Solutions to Black Exercises
10 14/1.8 10 5 = 5.6 10 10 10 7 10 14/1.1 10 8 = 9.1
K a for HONH 3 + = K w /K b for HONH 2 = 1.0
Note that K a for HONH 3 + is larger than K a for NH 4 + . 16.84 (a) Proceeding as in Solution 16.83(a): F (aq) + H 2 O(l) initial equil 0.085 M (0.085 x) M
HF(aq) 0M xM
+
OH (aq) 0M xM
K b for F
1.5 10
11
[HF] [OH ] [F ]
Kw K a for HF
1.0 6.8
10 10
14 4
1.47
10
11
1.5
10
11
( x) (x) ; assume the amount of F that hydrolyzes is small (0.085 x)
x 2 = 0.085(1.47
10 11); x = [OH ] = 1.118
10 6 = 1.1
10 6 M
pOH = 5.95; pH = 14 5.95 = 8.05 (b) Na 2 S(aq) S 2 (aq) + 2Na + (aq)
S 2 (aq) + H 2 O(l)
HS (aq) + OH (aq)
Kw K a for HS 1.0 1 10 10
14 19
As in part (a) above, [OH ] = [HS ] = x; [S 2 ] = 0.055 M
Kb
[HS ] [OH ] [S 2 ]
1
10 5
Since K b >> 1, the equilibrium above lies far to the right and [OH ] = [S 2 ] = 0.055 M. K b for HS = 1.05 10 7; [OH ] produced by further hydrolysis of HS amounts to 7.6 10 5 M. The second hydrolysis step does not make a significant contribution to the total [OH ] and pH. [OH ] = 0.055 M; pOH = 1.26, pH = 12.74 (c) As in Solution 16.83(c), calculate total [C 2 H 3 O 2 ]. [C 2 H 3 O 2 ] t = [C 2 H 3 O 2 ] from NaC 2 H 3 O 2 + [C 2 H 3 O 2 ] from Ba(C 2 H 3 O 2 ) 2 [C 2 H 3 O 2 ] t = 0.045 M + 2(0.055 M) = 0.155 M The hydrolysis equilibrium is C 2 H 3 O 2 (aq) + H 2 O(l) HC 2 H 3 O 2 (aq) + OH (aq)
Kb [HC 2 H 3 O 2 ] [OH ] [C 2 H 3 O 2 ] Kw K a for HC 2 H 3 O 2 1.0 1.8 10 10
14 5
5.56 5.6
10 10
10 10
[OH ] = [HC 2 H 3 O 2 ] = x, [C 2 H 3 O 2 ] = 0.155 x
Kb 5.56 10
10
x2 ; assume x is small compared to 0.155 M (0.155 x)
x 2 = 0.155 (5.56
10 10); x = [OH ] = 9.280 10 6) = 8.97
10 6 = 9.3
10 6
pH = 14 + log (9.280
252
16 Acid-Base Equilibria
16.86 (a) (b) (c) (d) (e) 16.88 basic;
Solutions to Black Exercises
acidic; Cr 3 + is a highly charged metal cation and a Lewis acid; Br is negligible. neutral; both Li + and I are negligible. PO 4 3 is the conjugate base of HPO 4 2 ; K + is negligible.
acidic; CH 3 NH 3 + is the conjugate acid of CH 3 NH 2 ; Cl is negligible. acidic; HSO 4 is a negligible base, but a fairly strong acid (K a = 1.2 K + is negligible. 10 2).
Plan. Estimate pH of salt solution by evaluating the ions in the salts. Calculate to confirm if necessary. Solve. KBr: salt of strong acid and strong base, neutral solution. The unknown is probably KBr. Check the others to be sure.
NH 4 Cl: salt of a weak base and a strong acid, acidic solution KCN: salt of a strong base and a weak acid, basic solution
K 2 CO 3 : salt of a strong base and a weak acid (HCO 3 ), basic solution Only KBr fits the acid-base properties of the unknown. 16.90 The solution is basic because of the hydrolysis of PO 4 3 . The molarity of PO 4 3 is
45.0 g Na 3 PO 4 1.00 L soln 1 mol Na 3 PO 4 163.9 g Na 3 PO 4 0.2746 0.275 M PO 4 3
PO 4 3 (aq) + H 2 O(l) HPO 4 2 (aq) + OH (aq)
Kb
[HPO4 2 ] [OH ] [PO4 3 ]
Kw K a for HPO4 2
1.0 4.2
10 10
14 13
0.0238
2.4
10
2
Ignoring the further hydrolysis of HPO 4 2 , [OH ] = [HPO 4 2 ] = x, [PO 4 3 ] = 0.275 x
2.4 10
2
x2 ; x2 (0.275 x)
2.4
10
2
x 0.0065
0
Since K b is relatively large, we will not assume x is small compared to 0.275.
x 0.024 (0.024) 2 2(1) 4(1) ( 0.0065) 0.024 0.0267 2(1)
x = 0.070 M OH ; pH = 14 + log (0.070) = 12.84
Acid-Base Character and Chemical Structure
16.92 (a) (b) Acid strength increases as the polarity of the XH bond increases and decreases as the strength of the XH bond increases. Assuming the element, X, is more electronegative than H, as the electronegativity of X increases, the XH bond becomes more polar and the strength of the acid increases. This trend holds true as electronegativity increases across a row of the periodic chart. However, as electronegativity decreases going down a family, acid strength increases because the strength of the HX bond decreases, even though the HX bond becomes less polar.
253
16 Acid-Base Equilibria
16.94 (a) (b) (c) (d) (e)
Solutions to Black Exercises
For binary hydrides, acid strength increases going across a row, so HCl is a stronger acid than H 2 S. For oxyacids, the more electronegative the central atom, the stronger the acid, so H 3 PO 4 is a stronger acid than H 3 AsO 4 . HBrO 3 has one more nonprotonated oxygen and a higher oxidation number on Br, so it is a stronger acid than HBrO 2 . The first dissociation of a polyprotic acid is always stronger because H + is more tightly held by an anion, so H 2 C 2 O 4 is a stronger acid than HC 2 O 4 . The conjugate base of benzoic acid, C 7 H 5 O 2 , is stabilized by resonance, while the conjugate base of phenol, C 6 H 5 O , is not. HC 7 H 5 O 2 has greater tendency to form its conjugate base and is the stronger acid. NO 2 PO 4 3 (HNO 3 is the stronger acid because it has more nonprotonated O atoms, so NO 2 is the stronger base.) (K a for HAsO 4 2 is greater than K a for HPO 4 2 , so K b for PO 4 3 is greater and PO 4 3 is the stronger base. Note that P is more electronegative than As and H 3 PO 4 is a stronger acid than H 3 AsO 4 , which could lead to the conclusion that AsO 4 3 is the stronger base. As in all cases, the measurement of base strength, K b , supercedes the prediction. Chemistry is an experimental science. (The more negative the anion, the stronger the attraction for H + .)
16.96
(a) (b)
(c) 16.98 (a) (b) (c)
CO 3 2 True.
False. For oxyacids with the same structure but different central atom, the acid strength increases as the electronegativity of the central atom increases. False. HF is a weak acid, weaker than the other hydrogen halides, primarily because the HF bond energy is exceptionally high.
Lewis Acids and Bases
16.100 No. If a substance is a Lewis acid, it is not necessarily a Brnsted or an Arrhenius acid. The Lewis definition of an acid, an electron pair acceptor, is most general. A Lewis acid does not necessarily fit the more narrow description of a Brnsted or Arrhenius acid. An electron pair acceptor isn't necessarily an H + donor, nor must it produce H + in aqueous solution. An example is Al 3 +, which is a Lewis acid, but has no ionizable hydrogen. Lewis Acid (a) (b) (c) (d) 16.104 (a) (b) (c) HNO 2 (or H + ) FeBr 3 (Fe 3 +) Zn 2 + SO 2 Cu(NO 3 ) 2 , higher cation charge NiBr 2 , smaller cation radius, same charge Lewis Base OH Br NH 3 H2O
16.102
ZnBr 2 , smaller cation radius, same charge
254
16 Acid-Base Equilibria
Additional Exercises
16.106 (a) (b) (c) (d) Correct.
Solutions to Black Exercises
Incorrect. A Brnsted acid must have ionizable hydrogen. Lewis acids are electron pair acceptors, but need not have ionizable hydrogen. Correct. Incorrect. K + is a negligible Lewis acid because it is the conjugate of strong base KOH. Its relatively large ionic radius and low positive charge render it a poor attractor of electron pairs. Correct. 10 6 M, we must
(e) 16.108
Assume T = 25 C. For acid or base solute concentrations less than 1 consider the autoionization of water as a source of [OH ] and [H + ]. H2O(l) initial equil K w = 1.0 C C
[H+] 0 x
+ 2.5
[OH-] 10 9 M 10 9) M 10 9 x 1.0
4( 1
(x + 2.5 10 9); x 2 + 2.5
9
10 14 = [H + ][OH ] = (x)(x + 2.5
2.5 10
10 14 = 0
10
14
From the quadratic formula, x
( 2.5
10 9 ) 2 2
)
9.876
10
6
9.9
10
6
MH
[H + ] = 9.9
10 8 M; [OH ] = (9.9
10 8 + 2.5
10 9) = 1.013
10 7 = 1.0
10 7 M
pH = 7.0054 = 7.01 Check: [9.876 10 8][1.013 10 7] = 1.0 10 14. Our answer makes sense. The very small concentration of OH from the solute raises the solution pH to slightly more than 7. 16.109 The solution with the higher pH has the lower [H + ]. (a) (b) (c) 16.111 (a) For solutions with equal concentrations, the weaker acid will have a lower [H + ] and higher pH. The acid with K a = 8 10 5 is the weaker acid, so it has the higher pH.
The base with pK b = 4.5 is the stronger base, has greater [OH ] and smaller [H + ], so higher pH. H2X H + + HX
Assuming HX does not ionize, [H + ] = 0.050 M, pH = 1.30 (b) (c) H2X 2H + + X ; 0.050 M H 2 X = 0.10 M H + ; pH = 1.00
The observed pH of a 0.050 M solution of H 2 X is only slightly less than 1.30, the pH assuming no ionization of HX . HX is not completely ionized; H 2 X, which is completely ionized, is a stronger acid than HX .
255
16 Acid-Base Equilibria
(d) 16.112
Solutions to Black Exercises
Since H 2 X is a strong acid, HX has no tendency to act like a base. HX does act like a weak acid, so a solution of NaHX would be acidic.
Analyze/Plan. Evaluate the acid-base properties of the cation and anion to determine whether a solution of the salt will be acidic, basic, or neutral. Solve. (i) (ii) (iii) NH 4 NO 3 : NH 4 + , weak conjugate acid of NH 3 ; NO 3 , negligible conjugate base of HNO 3 ; acidic solution. NaNO 3 : Na + , negligible conjugate acid of NaOH; NO 3 , negligible conjugate base of HNO 3 ; neutral solution. NH 4 C 2 H 3 O 2 : NH 4 + , weak conjugate acid of NH 3 , K a = K w /1.8 10 5 = 5.6 10 10; C 2 H 3 O 2 , weak conjugate base of HC 2 H 3 O 2 , K b = K w /1.8 10 5 = 5.6 10 10; neutral solution (K a for the cation and K b for the anion are accidentally equal, producing a neutral solution). NaF: Na + , negligible conjugate acid of NaOH; F , weak conjugate base of HF, K b = K w /6.8 10 4 = 1.5 10 11; basic solution. NaC 2 H 3 O 2 : Na + , negligible; C 2 H 3 O 2 , weak base, K b = 5.6 solution. 10 10, basic
(iv) (v)
In order of increasing acidity and decreasing pH: 0.1 M NaC 2 H 3 O 2 > 0.1 M NaF > 0.1 M NH 4 C 2 H 3 O 2 = 0.1 M NaNO 3 > 0.1 M NH 4 Cl; (v) > (iv) > (iii) ~ (ii) > (i) (iv) and (v) are both bases, and (v) has the greater K b value and higher pH. (ii) and (iii) are both neutral and (i) is acidic. 16.114 Then, Q(aq) + H 2 O(l) QH + (aq) + OH . K b = [QH + ] [OH ]/[Q] The ratio in question is [QH + ]/[Q], which equals K b /[OH ] for each compound. At pH = 2.5, pOH = 11.5, [OH ] = antilog (11.5) = 3.16 10 12 = 3 10 12 M. Now calculate K b /[OH ] for each compound:
Nicotine [QH ] [Q] [QH ] [Q] [QH ] [Q] [QH ] [Q] 7 10
7
Call each compound in the neutral form Q.
/ 3.16
10
12
2
10 5
Caffeine
4
10
14
/ 3.16
10
12
1
10
2
Strychnine
1
10
6
/ 3.16
10
12
3
10 5
Quinine
1.1
10
6
/ 3.16
10
12
3.5
10 5
For all the compounds except caffeine the protonated form has a much higher concentration than the neutral form. However, for caffeine, a very weak base, the neutral form dominates.
256
16 Acid-Base Equilibria
16.115 (a)
NH2 CH 2 COOH H 2 O
NH2 CH 2 COOH H 2 O
Solutions to Black Exercises
Consider the formation of the zwitterion as a series of steps (Hess' law).
NH2
10
CH 2 COO
H3O
Ka Kb
1/K
NH3 CH 2 COOH OH
H3O
OH
2H 2 O
NH3 CH 2 COO
NH2 CH 2 COOH
Ka K b Kw
K
Ka Kw
Kb
(4.3
10 3 ) (6.0 1.0
14
10 5 )
2.6
10 7
The large value of K indicates that formation of the zwitterion is favorable. The assumption is that the same NH 2 CH 2 COOH molecule is acting like an acid (COOH H + + COO ) and a base (NH 2 + H + NH 3 + ), simultaneously. Glycine is both a stronger acid and a stronger base than water, so the H + transfer should be intramolecular, as long as there are no other acids or bases in the solution. (b) Since glycine exists as the zwitterion in aqueous solution, the pH is determined by the equilibrium below.
NH
3
CH 2 COO
H 2O
NH 2
CH 2 COO
H 3O
14 5
Ka
[NH2 CH 2 COO ] [H 3 O ] [ NH3 CH 2 COO ]
Kw Kb
1.0 6.0
1.67
10 10
10
1.67
10
10
1.7
10
10
x [H 3 O ] [NH2 CH 2 COO ]; K a
10
( x)(x) (0.050 x)
x2 0.050
x = [H 3 O + ] = 2.89 (c)
10 6 = 2.9
10 6 M; pH = 5.54
In a strongly acidic solution the CO 2 function would be protonated, so glycine would exist as + H 3 NCH 2 COOH. In strongly basic solution the NH 3 + group would be deprotonated, so glycine would be in the form H 2 NCH 2 CO 2 .
16.116
The general Lewis structures for these acids and their conjugate bases are shown below.
where X = H or Cl. Replacement of H on the acid by the more electronegative chlorine atoms causes the central carbon to become more positively charged, thus withdrawing more electrons from the attached COOH group, in turn causing the OH bond to be more polar, so that H + is more readily ionized.
257
16 Acid-Base Equilibria
Solutions to Black Exercises
For the conjugate base (two resonance structures), the electronegative X atoms delocalize negative charge and stabilize these forms relative to the unsubstituted anions. This favors products in the ionization equilibrium and increases the value of K a . To calculate pH proceed as usual, except that the full quadratic formula must be used for all but acetic acid. Acid acetic chloroacetic dichloroacetic trichloroacetic pH 3.37 2.51 2.09 2.0
Integrative Exercises
16.118 Analyze. Given mass % and density of concentrated HCl, calculate volume of concentrated solution required to produce 10.0 L of HCl with pH = 2.05. Plan. Calculate molarity of concentrated solution from density and mass %. Calculate molarity of dilute solution from pH. Use the dilution formula to calculate volume (mL) concentrated solution required. Solve.
1.18 g conc. soln. mL conc. soln. 36.0 g HCl 100 g conc. soln. 1000 mL 1L 1 mol HCl 36.46 g HCl 11.651 mol HCl/L
11.7 M HCl/L
For the dilute HCl solution, [H + ] = 10 pH = 10 2.05 = 8.913 MC
LC
10 3 = 8.9
10 3 M HCl
LC = MD
7.650
M D ; 11.651
10
L C = 8.913
3
10 3 M
10.0 L;
10 3 ; 7.650
L
1000 mL 1L
7.65 7.7 mL conc. HCl
16.119
[H + ] = 10 pH = 10 2 = 1 HCl(aq) + HCO 3 (aq)
4 10
3
10 2 M H + ; 1
10 2 M
0.400 L = 4.0
10 3 = 4
10 3 mol H +
Cl (aq) + H 2 O(l) + CO 2 (g)
10
3
mol H
4
mol HCO 3
84.01 g NaHCO3 1 mol HCO 3
0.336
0.3 g NaHCO3
16.121
(a)
24 valence e , 12 e pairs
The formal charges on all atoms are zero. Structures with multiple bonds lead to nonzero formal charges. There are three electron domains about Al. The electrondomain geometry and molecular structure are trigonal planar.
258
16 Acid-Base Equilibria
(b) (c)
Solutions to Black Exercises
The Al atom in AlCl 3 has an incomplete octet and is electron deficient. It "needs" to accept another electron pair, to act like a Lewis acid.
Both the Al and N atoms in the product have tetrahedral geometry. (d) The Lewis theory is most appropriate. H + and AlCl 3 are both electron pair acceptors, Lewis acids.
16.122
Plan. Use acid ionization equilibrium to calculate the total moles of particles in solution. Use density to calculate kg solvent. From the molality (m) of the solution, calculate T b and T b . Solve. HSO 4 (aq) Initial equil. 0.10 M 0.10 x M 0.071 M
Ka 1.2 10
2
H + (aq) 0 xM 0.029 M
+
SO 4 2 (aq) 0 xM 0.029 M
[H ] [SO4 2 ] [HSO4 ]
x2 ; K a is relatively large, so use the quadratic. 0.10 x
x2
0.012 x 0.0012
0; x
0.012
(0.012) 2 2
4(1)( 0.0012)
;x
0.029 M H , SO4 2
Total ion concentration = 0.10 M Na + + 0.071 M HSO 4 + 0.029 M H + + 0.029 M SO 4 2 = 0.229 = 0.23 M. Assume 100.0 mL of solution. 1.002 g/mL
0.10 M NaHSO4 0.1000 L
100.0 mL = 100.2 g solution.
120.1 g NaHSO4 mol NaHSO4
0.010 mol NaHSO4
1.201 12 g NaHSO
4
100.2 g soln 1.201 g NaHSO 4 = 99.0 g = 0.099 kg H 2 O
m mol ions kg H 2 O 0.229 M 0.1000 L 0.0990 kg 0.231 0.23 m ions
T b = K b (m) = 0.52 C/m 16.124 (a) (b)
(0.23 m) = +0.12 C: T b = 100.0 + 0.12 = 100.1 C
rate = k[IO 3 ][SO 3 2 ][H + ] pH = pH 2 pH 1 = 3.50 5.00 = 1.50 pH = log [H + ] 2 (log [H + ] 1 ); pH = log [H + ] 2 log [H + ] 1 pH = log [H + ] 2 / [H + ] 1 ; [H + ] 2 / [H + ] 1 = 10
pH
[H + ] 2 /[H + ] 1 = 10 1 .50 = 31.6 = 32. The rate will increase by a factor of 32 if [H + ] increases by a factor of 32. The reaction goes faster at lower pH.
259
16 Acid-Base Equilibria
(c)
Solutions to Black Exercises
Since H + does not appear in the overall reaction, it is either a catalyst or an intermediate. An intermediate is produced and then consumed during a reaction, so its contribution to the rate law can usually be written in terms of concentrations of other reactants (Sample Exercise 14.15). A catalyst is present at the beginning and end of a reaction and can appear in the rate law if it participates in the rate-determining step (Solution 14.72). This reaction is pH dependent because H + is a homogeneous catalyst that participates in the ratedetermining step. The structures of the two acids are similar, but lactic acid has an OH group on the C atom adjacent to the COOH group. This electronegative substituent withdraws electron density from the COOH group, and stabilizes its conjugate base, increasing the strength of lactic acid relative to propionic acid. The stronger the acid, the larger the K a value and the smaller the pK a . pKa = 3.85, Ka = 103.85 = 1.4
Ka 1.4 10
4
16.126
(a)
(b)
104; [H+] = [C3H5O3] = x; [HC3H5O3] = 0.050 x
x2 0.050 x x2 0.050
[H ]1[C 3 H 5 O 3 ] [HC3 H 5 O 3 ]
x 2 = 0.050(1.4
10 4) = 2.646
10 3 = 2.6
10 3 M C 3 H 5 O 3
(This represents 5.3% dissociation; solution by the quadratic yields essentially the same result.) (c) Strategy: Assume a 100 g sample. Calculate mol Cu in sample. Use mole ratios from formula to calculate mass of O and H not due to H 2 O. Subtract masses of Cu, C, H, and O from 100 g to get mass of H 2 O. Calculate mol H 2 O and X. Assume a 100 g sample, 22.9 g Cu.
22.9 g Cu 1 mol Cu 63.55 g Cu 0.3603 0.360 mol Cu
mole ratios of Cu, O, and H (not due to H 2 O): 1 Cu:6 O:10 H
gO 0.3603 mol Cu 6 mol O 1 mol Cu 10 mol H 1 mol Cu 16.00 g O 1 mol O 1.008 g H 1 mol H 34.59 34.6 g O
gH
0.3603 mol Cu
3.632
3.63 g H
g H 2 O = 100 g sample [22.9 g Cu + 26.0 g C + 34.6 g O + 3.63 g H] = 12.87 = 12.9 g H 2 O
12.87 g H 2 O 1 mol H 2 O 18.02 g H 2 O 0.7142 mol H 2 O/0.3604 1.98 2
X = 2 mol H 2 O in the hydrate.
260
16 Acid-Base Equilibria
(d)
Solutions to Black Exercises
Compare K a for Cu 2 +(aq) and K b for C 3 H 5 O 3 (aq). The ion with the larger K value will undergo hydrolysis to the greater extend and will determine the pH of the solution. pK b for C 3 H 5 O 3 = 14 pK a for HC 3 H 5 O 3 = 14.00 3.85 = 10.15 K b = 10 10.15 = 7.1 10 11. 10 11) the
Since K a for Cu 2 + (1.0 10 8) is greater than K b for C 3 H 5 O 3 (7.1 solution will be slightly acidic.
261