Chemical and Ionic Equilibrium - Result Question 56 — JEE PYQ Chapterwise

Solution:

In pure water, solubility $=\frac{9.57}{58} \times 10^{-3} \mathrm{M} $

$=1.65 \times 10^{-4} \mathrm{M} $

$ K_{\text {sp }}=4 S^3=4\left(1.65 \times 10^{-4}\right)^3=1.8 \times 10^{-11} $

$\text { In } 0.02 \hspace{1mm} \mathrm{M} \hspace{1mm}\mathrm{Mg}\left(\mathrm{NO}_3\right)_2 \text {; } $

solubility of water $\mathrm{Mg}(\mathrm{OH})2 =\sqrt{\frac{K{\mathrm{sp}}}{\left[\mathrm{Mg}^{2+}\right]}} \times 2$

$=1.5 \times 10^{-5} \mathrm{~mol} \mathrm{~L}^{-1} $

$=1.5 \times 10^{-5} \times 58 \mathrm{~g} \mathrm{~L}^{-1} $

$=8.7 \times 10^{-4} \mathrm{~g} \mathrm{~L}^{-1}$