Question

The solubility product $\left(K_{s p}\right)$ of solid barium sulphate at $298 K$ is $1.1 \times 10^{-10}$. The molar solubility, $S$ of $\left[ Ba ^{2+}\right]$ and $\left[ SO _4^{2-}\right]$ is

• A. $1.05 \times 10^{-7}\,mol\,L ^{-1}$
• B. $1.05 \times 10^{-10}\, mol\,L ^{-1}$
• C. $1.05 \times 10^{-6} \,mol\,L ^{-1}$
• D. $1.05 \times 10^{-5} \,mol\,L ^{-1}$

Answer 1

Answer: (D)

$ BaSO _4(s) \rightleftharpoons sa ^{2+}\underset{s}{(a q)}+S O_4^{2-}\underset{s}{(a q)} $
$ K_{s p}=S \times S=S^2 $
$1.1 \times 10^{-10}=S^2 $
$S=1.05 \times 10^{-5} mol\,L ^{-1}$