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6-2.Equilibrium-II (Ionic Equilibrium)
medium
Solution of $0.01$ $M$ $MgCl_2$ will form precipitate of $Mg(OH)_2$ at limiting $pH$ of Given : $K_{sp}$ of $Mg(OH)_2$ is $1 × 10^{-12}$
A
$9$
B
$5$
C
$12$
D
$2$
Solution
$\left[\mathrm{Mg}^{2+}\right]=10^{-2}\, \mathrm{M}$
$\mathrm{Ksp}=\left[\mathrm{Mg}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}$
$10^{-12}=0.01 \times\left[\mathrm{OH}^{-2}\right]$
$\left[\mathrm{OH}^{-}\right]=10^{-5} \quad, \mathrm{pOH}=5, \mathrm{pH}=9$
Standard 11
Chemistry
