Question

The solubility product of $BaS{O}_4$ is $1.5\times {10}^{-9}$. The precipitation in a $0.01M$ $B{a}^{2+}$ solution will start on adding $H_{2}S{O}_4$ of concentration:

• A. $1\times {10}^{-9}M$
• B. $1.5\times {10}^{-7}M$
• C. $2\times {10}^{-7}M$
• D. $1\times {10}^{-6}M$

Answer 1

The correct option is B $1.5\times {10}^{-7}M$

The solubility product of $BaS{O}_4$ is $1.5\times {10}^{-9}$.

$K_{sp}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]$

Substitute values in the above expression.

$1.5\times {10}^{-9}=0.01\times \left[S{O}_4^{2-}\right]$

Hence, $\left[S{O}_4^{2-}\right]=1.5\times {10}^{-7}M$.

Thus, the sulfuric acid solution of the concentration $1.5\times {10}^{-7}M$ should be added to start precipitation of barium sulphate.

Option B is correct.