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The solubility product of $Mg(OH_2)$ is $1 \times 10^{-12}$. At what $pH$ precipitation will occur in $0.01\, M$ $MgCl_2$ solution?
$8$
$9$
$10$
$12$
Solution
The solubility product expression is:
$\mathrm{K}_{\mathrm{SP}}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}$
Rearrange the above expression:
$\left[\mathrm{OH}^{-}\right]=\sqrt{\frac{\mathrm{K}_{\mathrm{sp}}}{\left[\mathrm{Mg}^{2+}\right]}}$
Substitute values in the above expression and calculate hydroxide ion concentration-
$\left[\mathrm{OH}^{-}\right]=\sqrt{\frac{1 \times 10^{-12}}{0.01}}$
$\left[\mathrm{OH}^{-}\right]=1 \times 10^{-5} \mathrm{M}$
Calculate pOH from the hydroxide ion concentration:
$\mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log \left(1 \times 10^{-5}\right)$
$\mathrm{pOH}=5$
Calculate pH from pOH-
$\mathrm{pH}=14-\mathrm{pOH}$
$\mathrm{pH}=14-5$
$\mathrm{pH}=9$
