Question

Determine the molar solubility (S) of Ag2CO3 in a buffered solution with a pH of 3.573 using the systematic treatment of equilibrium. Ksp(Ag2CO3) = 8.46 × 10–12; Ka1(H2CO3) = 4.45 × 10–7; Ka2(H2CO3) = 4.69 × 10–11. Please show step as to how you got the answer, i would really appreciate it.

Answer 1

Ag2CO3(s) <==> 2Ag+ + CO3^2- ------ Ksp

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

adding borth equations,

Ag2CO3(s) + H+ <===> 2Ag+ + HCO3-

K = 8.46 x 10^-12/4.69 x 10^-11 = 0.180

pH = -log[H+] = 3.573

[H+] = 2.7 x 10^-4 M

ICE chart

              Ag2CO3(s)   +   H+          <===>     2Ag+     +    HCO3-

I                    -             2.7 x 10^-4                      -                     -

C                  -                   -x                               +x                +2x        

E                  -           2.7 x 10^-4 - x                     x                   2x

So,

K = [HCO3-][Ag+]^2/[H+]

0.180 = (x)(2x)^2/(2.7 x 10^-4 - x)

let x be a small amount then,

0.180 = (x)(2x)^2/(2.7 x 10^-4)

4x^3 = 4.86 x 10^-5

x = 0.03 M

so molar solubility of Ag2CO3 was found to be 0.03 M