Equilibrium Question 815
Question: The 0.001M solution of $ Mg{{(NO_3)}_2} $ is adjusted to pH 9, $ K _{sp} $ of $ Mg{{(OH)}_2} $ is $ 8.9\times {10^{-12}} $ . At this pH
Options:
A) $ Mg{{(OH)}_2} $ will be precipitated
B) $ Mg{{(OH)}_2} $ is not precipitated
C) $ Mg{{( OH )}_3} $ will be precipitated
D) $ Mg{{( OH )}_3} $ is not precipitated
Show Answer
Answer:
Correct Answer: B
Solution:
$ pH=9;[ {H^{+}} ]={10^{-9}};[ O{H^{-}} ]={10^{-5}}; $
$ [M{g^{2+}}]=1\times {10^{-3}};[M{g^{2+}}]{{[O{H^{-}}]}^{2}}=1\times {10^{-13}} $ given $ K _{sp} $ of $ Mg{{( OH )}_2}=8.9\times {10^{-12}} $ which is more than $ 1\times {10^{-13}} $ . Hence $ Mg{{(OH)}_2} $ will not precipitate
BETA
