Question 22.6: Calculate the solubility of silver benzoate in an aqueous so......
Calculate the solubility of silver benzoate in an aqueous solution buffered at pH = 2.0 at 25°C. Compare your result with Figure 22.5.
Adding Equations 22.15 and 22.16 we have the overall reaction equation,
AgC_{6}H_{5}COO(s) ⇋ Ag^+(aq) + C_{6}H_{5}COO^−(aq) (22.15)
C_{6}H_{5}COO^−(aq) + H_{3}O^+(aq) ⇋ C_{6}H_{5}COOH(aq) + H_{2}O(l) (22.16)
AgC_{6}H_{5}COO(s) + H_{3}O^+(aq) ⇋ Ag^+(aq) + C_{6}H_{5}COOH(aq) + H_{2}O(l )with
K_{c} = K_{sp} \left(\frac{1}{K_{a}} \right) =(2.5 × 10^{–5}\ M^2)(1.6 × 10^4\ M^{–1}) = 0.40\ MFor every formula unit of AgC_{6}H_{5}COO(s) that dissolves, one Ag^+(aq) ion occurs in solution. Therefore, the solubility can be expressed as s = [Ag^+]. Because the value of K_{c} for this reaction is much greater than the value of K_{sp} for Equation 22.15 we can ignore the additional contribution to s from Equation 22.15. The pH is buffered at 2.0, so [H_{3}O^+] is essentially constant and may be taken as equal to 1.0 × 10^{-2}\ M throughout the reaction. In addition, according to the overall reaction stoichiometry,
[Ag^{+}] = [C_{6}H_{5}COOH]So at equilibrium s=[Ag^{+}] = [C_{6}H_{5}COOH]. Constructing a concentration table, we have the following:
Substituting these values into the equilibrium expression, we get
0.40\ M = \frac{[Ag^+][C_{6}H_{5}COOH]}{[H_{3}O^+]} =\frac{s^2}{1.0 × 10^{–2}\ M}Solving for s and taking the positive root, we find s = 0.063 M, in agreement with Figure 22.5.