# The molar solubility of Ag_2CO_3 is 1.3xx10^-4*mol*L^-1. What is K_"sp" for this salt?

May 25, 2017

${K}_{\text{sp}} = 8.80 \times {10}^{-} 12$..........

#### Explanation:

We interrogate the equilibrium:

$A {g}_{2} C {O}_{3} \left(s\right) r i g h t \le f t h a r p \infty n s 2 A {g}^{+} + C {O}_{3}^{2 -}$

And thus ${K}_{\text{sp}} = {\left[A {g}^{+}\right]}^{2} \left[C {O}_{3}^{2 -}\right]$

But from the stoichiometry, $\left[C {O}_{3}^{2 -}\right] = {S}_{A {g}_{2} C {O}_{3}}$, and $\left[A {g}^{+}\right] = 2 {S}_{A {g}_{2} C {O}_{3}}$........and so

${K}_{\text{sp}} = S \times {\left(2 S\right)}^{2} = 4 {S}^{3} = 4 \times {\left(1.3 \times {10}^{-} 4\right)}^{3} = 8.80 \times {10}^{-} 12. \ldots .$

This site reports that ${K}_{\text{sp}}$, $A {g}_{2} C {O}_{3}$, is $8.46 \times {10}^{-} 12$, so our estimate is kosher.