Question

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

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

Answer 1

The correct option is B $2\times {10}^{-8}$ M

In presence of $H_{2}S{O}_4,\left[S{O}_4^{2-}\right]=0.02M$

$BaS{O}_4⟶B{a}^{2+}+S{O}_4^{2-}$

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

$\Rightarrow 4\times {10}^{-10}=S\times 2\times {10}^{-2}$

$\therefore S=2\times {10}^{-8}M$

Hence, solubility of $BaS{O}_4=S=2\times {10}^{-8}M$