Question

The solubility product of $BaC{l}_2$ is $3.2\times 10^{-9}$. What will be its solubility in $mol{L}^{-1}$?

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

Answer 1

Answer: (C)

$BaC{l}_2\rightleftharpoons B{a}^{2+}+2C{l}^{-}$
$K_{sp}=[B{a}^{2+}][C{l}^{-}{]}^2$
$=x\times (2x{)}^2=4x^3$
$4x^3=3.2\times 10^{-9}$
$\Rightarrow x=9.28\times 10^{-4}$
$=0.928\times 10^{-3}\approx 1\times 10^{-3}$