Question

The $K_{sp}$ of $Mg(OH{)}_2$ is $1\times 10^{-12}.0.001Mg(OH{)}_2$ will precipitate at the limited $pH$ :

• A. 3
• B. 9
• C. 5
• D. 8

Answer 1

Answer: (B)

First find $\left[O{H}^{-}\right]$from $K_{sp}.$
Then find $\left[H^{+}\right]$ by using formula
$\left[H^{+}\right]\left[O{H}^{-}\right]=10^{-14}$.
Finally calculate $pH$ by using following formula.
$pH=-\log \left[H^{+}\right]$ Given,
$K_{sp}Mg(OH{)}_2=1\times 10^{-12}$ Concentration
$Mg(OH{)}_2=0.01MMg(OH{)}_{2}M{g}^{2+}+2O{H}^{-}$
$K_{sp}=[M{g}^{2+}+2O{H}^{-}$
or $1\times 10^{-12}=0.01\times {\left[O{H}^{-}\right]}^2$
or ${\left[O{H}^{-}\right]}^2=1\times 10^{-10}$
or $\left[O{H}^{-}\right]=1\times 10^{-5}$
$\left[H^{+}\right]\left[O{H}^{-}\right]=10^{-14}$
or $\left[H^{+}\right]\left[1\times 10^{-5}\right]=10^{-14}$
or $\left[H^{+}\right]=\frac{10^{-14}}{10^{-5}}=10^{-9}pH=-\log \left[H^{+}\right]$
$=-\log 10^{-9}=9$
$\therefore Mg(OH{)}_2$ will precipitate at limited pH of $9$.