Question

The solubility of $ Mg(OH)_2$ in pure water is $9.57 \times 10^{ - 3} \, g/L$. Calculate its solubility (in $g/L$) in $0.02\, M \,Mg(N_3 )_2 $ solution

Answer 1

In pure water, solubility = $ \frac{ 9.57}{ 58} \times 10^{ - 3 } $ M
= $1.65 \times 10^{ - 3}\, M$
$K_{ sp} = 4S^3 = 4 ( 1.65 \times 10^{ - 4 })^3 = 1.8 \times 10^{ - 11} $
In $0.02 \,M \,Mg(NO_3)_2$
solubility of $ Mg(OH)_2 = \sqrt{ \frac{ K_{ sp}}{ [ Mg^{ 2 + } ] }} \times \frac{1}{2} $
$ = 1.5 \times 10^{ - 5} $ mol $ L^{ - 1} $
$ = 1.5 \times 10^{ - 5} \times 58 \, g \, L^{ - 1} $
$ = 8.7 \times 10^{ - 4} g \, L^{ - 1}$