Question

The solubility product of $Ba{SO}_4$ is $4\times {10}^{-10}$. The solubility of $Ba{SO}_4$ in presence of $0.02N$ $H_2{SO}_4$ will be:

• A. $4\times {10}^{-8}M$
• B. $2\times {10}^{-8}M$
• C. $4\times {10}^{-5}M$
• D. $2\times {10}^{-4}M$

Answer 1

Answer: (A)

$4\times {10}^{-8}M$

Let the solubility of ${BaSO}_4$be $S$ mole per litre

${BaSO}_4\left(S\right)\rightleftharpoons {Ba}^{2+}+S{O}_4^{2-}SMSMSM$

Applying the law of solubility product,

$K_{SP}=\left[Ba{]}^{2+}\right[S{O}_4^{2-}]=S\times S$

$\Rightarrow S^2=K_{SP}$

$\Rightarrow S=\sqrt{K_{SP}}=\sqrt{4\times {10}^{-10}}=2\times {10}^{-5}$

Solubility in $0.02N{H}_2{SO}_4=0.01M{H}_{2}S{O}_4:$

$\left[{Ba}^{2+}\right]=S,\left[{SO}_4^{2-}\right]=0.01+S$

$K_{SP}=\left[Ba{]}^{2+}\right[S{O}_4^{2-}]$

$=\left(S\right)(0.01+S)=0.01S+S^2$

$S^2$ being the product of two small quantities is neglected

$\therefore K_{SP}=4\times 10^{-10}$

$\Rightarrow 0.01S=4\times {10}^{-10}$

$\Rightarrow S=\frac{4\times {10}^{-10}}{0.01}$ $={4\times 10}^{-8}M$