Question

The values of $K_{sp}$ for the slightly soluble salts $MX$ and ${QX}_2$ are each equal to $4.0\times {10}^{-18}$. Which salt is more soluble? Explain your answer fully.

Answer 1

$MX\rightleftharpoons M^{+}+X^{-}$

$K_{sp}=\left[{Ag}^{+}\right]\left[X^{-}\right]=4\times {10}^{-18}$

Let $\left[M^{+}\right]=s_1\Rightarrow \left[X^{-}\right]=s_1$

$\therefore {s_1}^2=4\times {10}^{-18}$

$\Rightarrow s_1=\sqrt{4\times {10}^{-18}}=2\times {10}^{-9}.....\left(1\right)$

$Q{X}_2\rightleftharpoons Q^{+2}+2X^{-}$

$K_{sp}=\left[Q^{2+}\right]{\left[X^{-}\right]}^2$

Let $\left[Q^{+2}\right]=s_2\Rightarrow \left[X^{-}\right]=2s_2$

$\therefore 4{s_2}^3=4\times {10}^{-18}$

$\Rightarrow s_2=\sqrt[3]{3{10}^{-18}}={10}^{-6}.....\left(2\right)$

From comparing ${eq}^n\left(1\right)\&\left(2\right)$, we get

$s_2>s^1$

Hence the salt $Q{X}_2$ is more soluble than the salt $MX$.