Question

The solubility product of $BaS{{O}_{4}}$ is $1.5\times {{10}^{-9}}$. The precipitation in a 0.01 M $B{{a}^{2+}}$ ions solution will start on adding ${{H}_{2}}S{{O}_{4}}$ ​ of concentration?
(A) ${{10}^{-9}}M$
(B) ${{10}^{-8}}M$
(C) $1.5\times {{10}^{-7}}M$
(D) ${{10}^{-6}}M$

Answer 1

Hint: Here, a concept of “common ion effect” can be seen. An ionic precipitates solubility is reduced when a soluble compound with the precipitate common ion is added to the solution. This means when a salt is added to the solvent, which contains a common ion, then the equilibrium will shift in a backward direction. This is known as the common ion effect.

Complete Step by Step Solution:
Let the solubility of $BaS{{O}_{4}}$ be s.
$BaS{{O}_{4}}$can be dissociated into ions as

Given ${{K}_{sp}}$ of $BaS{{O}_{4}}$ =$1.5\times {{10}^{-9}}$
The $B{{a}^{2+}}$ ions are common in both the solute and the solvent. So, its solubility is taken to be $s+0.01$.
The solubility product (${{K}_{sp}}$) of $BaS{{O}_{4}}$can be calculated as ${{K}_{sp}}=\left[ B{{a}^{2+}} \right]\left[ S{{O}_{4}}^{2-} \right]$
${{K}_{sp}}=\left( s+0.01 \right)\times s$
${{K}_{sp}}={{s}^{2}}+0.01s$
As ${{s}^{2}}$is very small, therefore it can be neglected.
${{K}_{sp}}$ is thus given as ${{K}_{sp}}=0.01s$
Comparing the calculated value of ${{K}_{sp}}$ with the given value.
$0.01s=1.5\times {{10}^{-9}}$
$s=\frac{1.5\times {{10}^{-9}}}{0.01}$
$s=1.5\times {{10}^{-7}}M$
So, for precipitation of $B{{a}^{2+}}$, the concentration of ${{H}_{2}}S{{O}_{4}}$should be greater than $s=1.5\times {{10}^{-7}}M$.
Correct Option: (C) $1.5\times {{10}^{-7}}M$.

Additional Information: The solubility product depends upon temperature. With an increase in temperature, the solubility of a substance is increased. Hence, the solubility product increases. When a strong electrolyte, which dissociates to a greater extent, is added to a weak electrolyte, which dissociates to a lesser extent, and both have an ion in them, the solubility of the weak electrolyte is further decreased.

Note: The common ion effect has many applications in the processes involving water treatment; regulation of buffer solutions; production of sodium bicarbonate; salting out of soap; and purification of salts.