Question

The solubility product of $BaCl_2$ is $3.2 \times 10^{-9}$. What will be its solubility in $mol \,L^{-1}$?

• A. $4\times10^{-3}$
• B. $3.2 \times 10^{-9}$
• C. $1 \times 10^{-3}$
• D. $1 \times 10^{-9}$

Answer 1

Answer: (C)

$BaCl_2 \rightleftharpoons Ba^{2+} +2Cl^-$
$K_{sp}=[Ba^{2+}][Cl^-]^2$
$=x \times (2x)^2=4x^3$
$4x^3=3.2 \times 10^{-9}$
$\Rightarrow x=9.28\times10^{-4}$
$=0.928\times10^{-3}\approx1\times10^{-3}$