Equilibrium Question 543
Question: The $ K _{sp} $ of $ Mg{{(OH)}_2} $ is $ 1\times {10^{-12}},,0.01M,Mg{{(OH)}_2} $ will precipitate at the limiting pH [DPMT 2005]
Options:
A) 3
B) 9
C) 5
D) 8
Show Answer
Answer:
Correct Answer: B
Solution:
$ Mg{{(OH)}_2} $ ⇌ $ M{g^{2+}}+2O{H^{-}} $
$ K _{sp}=[M{g^{2+}}]{{[O{H^{-}}]}^{2}} $
$ 1\times {10^{-12}}=0.01,{{[O{H^{-}}]}^{2}} $
$ {{[O{H^{-}}]}^{2}}=1\times {10^{-10}} $
$ \Rightarrow [O{H^{-}}]={10^{-5}} $
$ [{H^{+}}]={10^{-14}}/{10^{-5}}=10^{9} $
$ pH=-\log [{H^{+}}]=-\log [{10^{-9}}]=9 $
