Question:
At $25^{\circ} \mathrm{C}$, the solubility producct of $\mathrm{Mg}(\mathrm{OH})_{2}$ is $1.0 \times 10^{-11}$. At which $\mathrm{pH}$, will $\mathrm{Mg}^{2+}$ ions start precipitating in the form of $\mathrm{Mg}(\mathrm{OH})_{2}$ from a solution of $0.001 \mathrm{M} \mathrm{Mg}^{2+}$ ions?
Correct Option: , 3
Solution:
$10^{-11}=\left[\mathrm{Mg}^{+2}\right]\left[\mathrm{OH}^{-}\right]^{2}$
$10^{-11}=\left(10^{-3}\right)\left[\mathrm{OH}^{-}\right]^{2}$
${\left[\mathrm{OH}^{-}\right]=10^{-4} \quad \mathrm{pOH}=4 \quad \mathrm{pH}=11 }$
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