Question

Consider a sparingly soluble salt $\left(AB\right)$ with concentration $C_1$ and highly soluble salt $\left(AX\right)$ with concentration $C_2$. Then which of the following relation holds true?
(Assume $K_{sp}\left(AB\right)<<{10}^{-3}$)

• A. $K_{sp}=\frac{s}{C_1}$
• B. $K_{sp}=\frac{s}{C_2}$
• C. $s=\frac{K_{sp}}{C_1}$
• D. $s=\frac{K_{sp}}{C_2}$

Answer 1

The correct option is D $s=\frac{K_{sp}}{C_2}$

Consider sparingly soluble salt $\left(AB\right)$ with concentration $C_1$ and highly soluble salt $\left(AX\right)$ with concentration $C_2$
$AX\left(aq\right)\to A^{+}\left(aq\right)+X^{-}\left(aq\right)$
$C_2--$
$-C_2+s{C}_2$
$AB\left(s\right)\rightleftharpoons A^{+}\left(aq\right)+B^{-}\left(aq\right)$
$C_1--$
$C_1-ss+C_{2}s$

Solubility of $AB$ is $s$ because out of $C_1$, $s$ moles has dissociated only into ions.

Here $A^{+}$ is the common ion in above reactions.

$K_{sp}=\left[A^{+}\right]\left[B^{-}\right]$
$K_{sp}=(s+C_2)\left(s\right)$
Since $AX$ is highly soluble ,
$\therefore s+C_2\approx C_2$ because of the sparingly soluble salt where $K_{sp}<<{10}^{-3}$
$K_{sp}=\left(C_2\right)\left(s\right)$
$s=\frac{K_{sp}}{C_2}$


Theory:
Solubility in presence of common ion:
Consider an aqueous solution of $PbCr{O}_4\left(s\right)$

The equilibrium between the undissolved solid and the ions in a saturated solution can be represented by the equation:
$PbCr{O}_4\left(s\right)\rightleftharpoons P{b}^{2+}\left(aq\right)+Cr{O}_4^{2-}$

$PbCr{O}_4$ is sparingly soluble salt in an aqueous solution. If in the solution, $K_{2}Cr{O}_4$ is added. The common ion formed is $Cr{O}_4^{2-}$.
The backward reaction is favoured when $\left[Cr{O}_4^{2-}\right]$ increases, and thus the solubility is further reduced.