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

Answer: (B)

$1.08\times {10}^{-10}mo{l}^2{L}^{-2}$

The solubility $S=2.42\times {10}^{-3}g{L}^{-1}$

$S=\frac{2.42\times {10}^{-3}g{L}^{-1}}{233gmo{l}^{-1}}=1.04\times {10}^{-5}mol{L}^{-1}$

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

$K_{sp}=S\times S$

$K_{sp}=1.04\times {10}^{-5}mol{L}^{-1}\times 1.04\times {10}^{-5}mol{L}^{-1}=1.08\times {10}^{-10}mo{l}^2{L}^{-2}$