Question

The values of $K_{sp}$ for sparingly soluble $AB$ and $M{B}_2$ are each equal to $4.0\times {10}^{-18}$. Which salt is more soluble?

Answer 1

For $AB\rightleftharpoons A^{+}+B^{-}$

$K_{sp}=\left[A^{+}\right]\left[B^{-}\right]=\left(S\right)\left(S\right)=S^2=4\times {10}^{-18}$

$S_{AB}=2\times {10}^{-9}$

For $M{B}_2\rightleftharpoons M^{+}+2B^{-}$

$K_{sp}=\left[A^{+}\right][B^{-}{]}^2=(S\left)\right(2S{)}^2=4S^3=4\times {10}^{-18}$

$S_{M{B}_2}={10}^{-6}$

So $M{B}_2$ salt is more soluble than $AB$.