Question

The solubility product of $Ba{SO}_4$ is $1.5\times {10}^{-9}$ at ${18}^{o}C$. Its solubility in water at ${18}^{o}C$ is:

• A. $1.5\times {10}^{-9}mole$ ${litre}^{-1}$
• B. $1.5\times {10}^{-5}mole$ ${litre}^{-1}$
• C. $3.9\times {10}^{-9}mole$ ${litre}^{-1}$
• D. $3.9\times {10}^{-5}mole$ ${litre}^{-1}$

Answer 1

The correct option is D $3.9\times {10}^{-5}mole$ ${litre}^{-1}$

Solubility in water:

$BaS{O}_4\left(s\right)\rightleftharpoons B{a}^{2+}S{O}_4^{2-}$

$Smol{L}^{-1}$

Applying the law of solubility product:

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

or $S^2=K_{sp}$

or $S^2=K_{sp}$

$\sqrt{K_{sp}=\sqrt{1.5\times 10^{-9}}}$

$3.78X10^{-5}mol{L}^{-1}$