Question

Solubility product of $Ba{Cl}_2$ is $4\times {10}^{-9}$. Its solubility would be:

• A. $1\times {10}^{-27}$
• B. $1\times {10}^{-3}$
• C. $1\times {10}^{-7}$
• D. $1\times {10}^{-2}$

Answer 1

Answer: (B)

$1\times {10}^{-3}$

Let the solubility of $BaC{l}_2$ be $S$ $mol{litre}^{-1}$
$BaC{l}_2\rightleftharpoons {Ba}^{+}+{2Cl}^{-}$
$K_s=\left[{Ba}^{2+}\right]{\left[{Cl}^{-}\right]}^2=S\times {\left(2S\right)}^2$

$K_s=4\times S^3$

$S={10}^{-3}$