Question

A saturated solution of silver benzoate AgOCOC6H5 has a pH of 8.03. The Ka for benzoic acid is 6.5x10-5. What is the Ksp for silver benzoate based on this data?

Answer 1

The pH of the solution is 8.03; therefore, pOH = 14.00 – 8.03 = 5.97 (pH + pOH = 14).

[OH^-] = antilog (-pOH) = antilog (-5.97) = 1.0715*10^-6 M.

Write down the reaction of C₆H₅COO^- with water, H₂O as below.

C₆H₅COO^- (aq) + H₂O (l) <======> C₆H₅COOH (aq) + OH^- (aq)

Since OH^- is formed, we will work with K_b.

Given K_a = 6.5*10^-5; K_b = K_w/K_a = (1.0*10^-14)/(6.5*10^-5) = 1.5385*10^-10

K_b = [C₆H₅COOH][OH^-]/[C₆H₅COO^-] = (1.0715*10^-6)*(1.0715*10^-6)/[C₆H₅COO^-] ([C₆H₅COOH] = [OH^-] due to the 1:1 nature of dissociation; also we assume small dissociation and hence assume [C₆H₅COO^-] to remain constant).

====> 1.5385*10^-10 = (1.0715*10^-6)²/[C₆H₅COO^-]

====> [C₆H₅COO^-] = (1.0715*10^-6)²/(1.5385*10^-10) = 7.4625*10^-3 M

Write down the ionization of silver benzoate, AgOCOC₆H₅.

AgOCOC₆H₅ (s) <======> Ag⁺ (aq) + C₆H₅COO^- (aq)

K_sp = [Ag⁺][C₆H₅COO^-] = (7.4625*10^-3).(7.4625*10^-3) ([Ag⁺] = [C₆H₅COO^-] due to the 1:1 nature of ionization)

====> K_sp = (7.4625*10^-3)² = 5.5689*10^-5 ≈ 5.57*10^-5 (ans).