Equilibrium Question 904
Question: The solubility product of $ Mg{{(OH)}_2} $ is $ {10^{-14}} $ . The solubility of $ Mg{{(OH)}_2} $ in a buffer solution of $ pH=8 $ is
Options:
A) $ {10^{-8}} $
B) $ {10^{-6}} $
C) $ {10^{-2}} $
D) $ {10^{-4}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ Mg{{(OH)}_2},M{g^{2+}},+2O{H^{-}} $
$ K _{sp}=[M{g^{2+}}]{{[O{H^{-}}]}^{2}} $
$ 1\times {10^{-14}}=[M{g^{2+}}],{{[{10^{-6}}]}^{2}} $
$ \because $ $ pH=8 $
$ \therefore $ $[{H^{+}}]={10^{-8}},mol,{L^{-1}} $ . and $ [O{H^{-}}]={10^{-6}}mol{L^{-1}}] $
$ [M{g^{2+}}]=\frac{{10^{-14}}}{{10^{-12}}}={10^{-2}}mol,{L^{-1}} $
Solubility $ [M{g^{2+}}]={10^{-2}},mol{L^{-1}} $
