The solubility product \(\left( K _{ sp }\right)\) of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. Thus, for
\(Ag _{2} CO _{3}\), we have
$$g _{2} CO _{3(x)} \rightleftharpoons 2 Ag _{\left(a_{q}\right)}^{+}+ CO _{3\left(a_{q}\right)}^{2-}
$$
\(K_{s p}=\left[ Ag ^{+}\right]^{2}\left[ CO _{3}^{2-}\right]\)
If \(S\) is the solubility of \(Ag _{2} CO _{3}\), then,
we have
\begin{aligned}
& Ag _{2} CO _{3(s)} \rightleftharpoons 2 Ag _{\text {(maq) }}^{+}+ CO _{3(\text { arp })}^{2-} \\
&\quad S \quad 2 S \quad S \\
&\text { i.c. }\left[ Ag ^{+}\right]=2 S \&\left[ CO _{3(\text { aq) }}^{2-}\right]= S \\
&\text { Thus, } K_{\text {sp }}=(2 s )^{2}( s )=4 s ^{3} \\
&=4\left(1.30 \times 10^{-5} moldm ^{-3}\right) \\
&=8.788 \times 10^{-15} mol ^{3} dm ^{-9} \\
&=8.79 \times 10^{-15} mol ^{3} dm ^{9}
\end{aligned}