Question

At ${25}^{\circ }C$, will a precipitate of $Mg(OH{)}_2$ form when a $0.0001M$ solution of $Mg(N{O}_3{)}_2$ is adjusted to a $pH$ of $9.0$? At what minimum value of $pH$ will precipitation start?
[Given that: $K_{sp}\left(Mg\right(OH{)}_2)={10}^{-11}{M}^3$]

• A. $No,pH=3.5$
• B. $No,pH=10.5$
• C. $No,pH=6.0$
• D. $Yes,pH=8.5$

Answer 1

The correct option is B $No,pH=10.5$

From data, it can be seen that,

$\left[M{g}^{2+}\right]=0.0001M$
If pH= 9 then pOH= 5 $\Rightarrow \left[O{H}^{-}\right]={10}^{-5}M$

So, $IP=\left[M{g}^{2+}\right][O{H}^{-}{]}^2={10}^{-14}<K_{sp}$ s

Hence, precipitation does not occur.

Precipitation will start after IP > Ksp.

$\left[M{g}^{2+}\right][O{H}^{-}{]}^2>{10}^{-11};[O{H}^{-}{]}^2={10}^{-7},\left[O{H}^{-}\right]=3.16\ast {10}^{-4}$

$pOH=3.5$ so $pH=14-pOH=10.5$