Question

If the solubility of $Mg{\left(OH\right)}_2$ in water is $1.4\times {10}^{-x}$. $K_{sp}=1.2\times {10}^{-11}$. Find out the value of $x$.

• A. $4$
• B. $5$
• C. $6$
• D. $8$

Answer 1

The correct option is A $4$

$Mg(OH{)}_2\to M{g}^{+2}+2O{H}^{-}$

$x$ $2x$

$K_{sp}=\left[M{g}^{+2}\right]\left[O{H}^{-}\right]=x\times (2x{)}^2=4x^3=1.2\times {10}^{-11}$

As given, $4\times (1.4\times {10}^{-x}{)}^3=1.2\times {10}^{-11}$

By comparison $x=4$