Question

Determine the molar solubility (S) of Ag2CO3 in a buffered solution with a pH of 4.271 using the systematic treatment of equilibrium. Ksp(Ag2CO3) = 8.46 × 10–12; Ka1(H2CO3) = 4.45 × 10–7; Ka2(H2CO3) = 4.69 × 10–11.

Answer 1

pH = 4.271

pOH = 14 – 4.271 = 9.729

[OH-] = 10^-9.729 = 1.866 x 10^-10 M

Ag₂CO₃ --> 2Ag+ + CO₃^2-

C – S --> S + (S – x)

CO32- + H2O --> HCO3- + OH-

(S – x) + H2O --> x + OH-

Kb1 = Kw/Ka2 = 10^-14 / 4.69 x 10^-11

= 2.13 x 10^-4 = [OH-] * x / (S – x)

Since Kb1 >> Kw/Ka1 = Kb2, we can neglect second step

Or [OH-] / 2.13 x 10^-4 = (S – x) / x = 8.75 x 10^-7 .. eqn (1)

Ksp = 8.46 x 10^-12 = [Ag+]²[CO32-]

8.46 x 10^-12 = S² * (S – x).. eqn (2)

Solving equations (1) & (2) we get

x S² = 9.66 x 10^-6

S = 0.0213 M