Page 270 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 17] ACID-BASE THEORY 259

from the data of part a.(e) Using the equation of part d and the data of part b, calculate the concentration of Y at
equilibrium. ( f ) Explain why the answer to part e is the same as that to part b.
[Y][Z] (0.010) 2
Ans. (a) K = = = 2.0 × 10 −5
[W][X] (0.090)(55.39)
x 2 −5
(b) = 2.0 × 10
(0.200)(55.40)
2
x = 2.2 × 10 −4
x = 1.5 × 10 −2

0.185 55.38
(c) = 0.925 = 0.9996 (practically no change for X)
0.200 55.40
(0.010) 2

(d ) K = = 1.1 × 10 −3
0.090
x 2
(e) = 1.1 × 10 −3
0.200
2
x = 2.2 × 10 −4
x = 1.5 × 10 −2
( f ) The concentration of X is essentially constant throughout the reaction (see part c). The equilibrium
constant of part a is amended to include that constant concentration; the new constant is K . That

constant is just as effective in calculating the equilibrium concentration of Y as is the original constant
K. This is the same effect as using K a , not K and [H 2 O], for weak acid and weak base equilibrium
calculations.

17.33. (a) What ions are present in a solution of sodium acetate? (b) According to Le Chˆatelier’s principle, what effect
would the addition of sodium acetate have on a solution of 0.250 M acetic acid? (c) Compare the hydronium ion
concentrations of Example 17.7 and Problem 17.24 to check your answer to part b.
Ans. (a)Na and C 2 H 3 O 2 − (b) The acetate ion should repress the ionization of the acetic acid, so that there
+
would be less hydronium ion concentration in the presence of the sodium acetate. (c) The hydronium ion
concentration dropped from 2.12 × 10 −3 to 2.38 × 10 −5 as a result of the addition of 0.190 M sodium
acetate.

17.34. Consider the ionization of the general acid HA:
−→
+
HA + H 2 O ←− H 3 O + A −
Calculate the hydronium ion concentration of a 0.100 M solution of acid for (a) K a = 2.0 × 10 −10 and
(b) K a = 8.0 × 10 −2
+
Ans. (a) Let [H 3 O ] = x
+
−
[H 3 O ][A ]
K a = = 2.0 × 10 −10
[HA]
x 2 −10
= 2.0 × 10
0.10 − x
2
x = 2.0 × 10 −11
x = 4.5 × 10 −6