At $25^{\circ} \mathrm{C}$, the solubility product of $\mathrm{MCl}$ is $1 \times 10^{-10}$. What is its…
At $25^{\circ} \mathrm{C}$, the solubility product of $\mathrm{MCl}$ is $1 \times 10^{-10}$. What is its molar solubility in $0.1 \mathrm{M}$ $\mathrm{NaCl}$ solution at same temperature?
$0.1$
$0.05$
$10^{-9}$
$10^{-5}$
Solution
$
K_{\text {sp }} \text { of } M C l=1 \times 10^{-10} \text {. }
$
Let the molar solubility of $\mathrm{MCl}$ in $0.1 \mathrm{M} \mathrm{NaCl}$ be ' $S$ ' $\mathrm{mol} \mathrm{L}^{-1}$.
$
\mathrm{MCl} \longrightarrow \mathrm{M}^{+}+\mathrm{Cl}^{-}
$
The concentration of $\mathrm{Cl}^{-}$will be $(S+0.1) \mathrm{mol} \mathrm{L}^{-1}$, as $0.1 \mathrm{~mol} \mathrm{~L}^{-1}$ are provided by $0.1 \mathrm{M} \mathrm{NaCl}$.
$
\begin{aligned}
& K_{\text {sp }}=\left[M^{+}\right]\left[\mathrm{Cl}^{-}\right] \\
& K_{\text {sp }}=S \times(S+0.1)=1 \times 10^{-10}
\end{aligned}
$
As the $K_{\text {sp }}$ is very small, the $S \ll 0.1$ and therefore can be ignored.
$
S \times 0.1=1 \times 10^{-10} \Rightarrow S=10^{-9} \mathrm{~mol} \mathrm{~L}^{-1}
$