Question

Determine the molar solubility (S) of Ag2CO3 in a buffered solution with a pH of 5.868 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

Ksp of Ag2CO3 = [Ag+]^2[CO3^2-]

Ag2CO3 <==> 2Ag+ + CO3^2- Ksp

with S amount of salt in solution

Ksp = (2S)^2.(S)

CO3^2- + H+ <==> HCO3- 1/Ka2

S = [CO3^2-] + [HCO3-]

pH = -log[H+] = 5.868

[H+] = 1.36 x 10^-6 M

Ka2 = [H+][CO3^2-]/[HCO3-]

[HCO3-] = 1.36 x 10^-6[CO3^2-]/4.69 x 10^-11

              = 2.9 x 10^4[CO3^2-]

feed in above equation,

S = [CO3^2-] + 2.9 x 10^4[CO3^2-]

[CO3^2-] = 3.45 x 10^-5(S)

Feed in Ksp equation,

8.46 x 10^-12 = (2S)^2.(3.45 x 10^-5S)

So,

molar solubility S would be,

S = 3.94 x 10^-3 M