Question

The solubility product of $BaS{O}_4$ is $4\times {10}^{-10}$. The solubility product of $BaS{O}_4$ in presence of $0.02N{H}_{2}S{O}_4$ will be:

• A. $4\times {10}^{-8}M$
• B. $2\times {10}^{-7}M$
• C. $2\times {10}^{-5}M$
• D. $2\times {10}^{-4}M$

Answer 1

The correct option is B $2\times {10}^{-7}M$

$2\times$ Molarity of $H_{2}S{O}_4=$ Normality

$\therefore$ Molarity $=\frac{N}{2}$

$=\frac{0.02}{2}=0.01M$

Now $BaS{O}_4\to B{a}^{2+}+S{O}_4^{2-}$

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

$=5\times (5+0.01)=0.01$

$4\times {10}^{-10}=5\times 0.01$

$5=\frac{4\times {10}^{-10}}{0.01}$

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

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