Question

Silver (I) carbonate (Ag₂CO_{3 (s)}) is a slightly soluble ionic compound with K_sp = 1.8 x 10^-12. Calculate the solubility of Ag₂CO_{3 (s)} in mol/L.

Answer 1

Solubility product constant (Ksp) is the product of the dissolved ion concentrations raised to the power of their stoichiometric coefficients.

Here, Ag2CO3 (s)= 2Ag+ (aq)+ CO32- (aq).

So, Ksp = [Ag+]2[CO32-]

The dissolved ion concentration is the molar solubility, denoted by s.

Thus, if ‘s’ Ag2CO3 dissolves, we get 2s amount of Ag+ and s amount of CO32-

So, [Ag+] = 2s, and [CO32-] = s

So, Ksp = (2s)2(s)

Putting the values:

1.8 x 10-12 = 4s3

Or, s = 7.66 x 10-3 mol/L