Question

The solubility of $BaS{O}_4$ in water is $2.42\times {10}^{-3}g{L}^{-1}$ at $298K$. The value of its solubility product $\left(K_{sp}\right)$ will be:
(Given molar mass of $BaS{O}_4=233gmo{l}^{-1}$)

• A. $1.08\times {10}^{-14}mo{l}^2{L}^{-2}$
• B. $1.08\times {10}^{-10}mo{l}^2{L}^{-2}$
• C. $1.08\times {10}^{-8}mo{l}^2{L}^{-2}$
• D. $1.08\times {10}^{-12}mo{l}^2{L}^{-2}$

Answer 1

The correct option is B $1.08\times {10}^{-10}mo{l}^2{L}^{-2}$

$\text{Solubility, S}=2.42\times {10}^{-3}g{L}^{-1}$
$S=\frac{2.42\times {10}^{-3}g{L}^{-1}}{233gmo{l}^{-1}}=1.04\times {10}^{-5}mol{L}^{-1}$
$K_{sp}=\left[B{a}^{2+}\right]\left[S{O}_4^{2-}\right]$
$K_{sp}=S\times S$
$K_{sp}=1.04\times {10}^{-5}\times 1.04\times {10}^{-5}{K}_{sp}=1.08\times {10}^{-10}mo{l}^2{L}^{-2}$
Hence, option (b) is correct answer.