Equilibrium Question 847
Question: The concentration of hydroxyl ion in a solution left after mixing 100mL of 0.1M $ MgCl_2 $ and 100mL of 0.2M $ NaOH $
$ (K _{sp} of ,Mg{{(OH)}_2}=1.2\times {10^{-11}}) $ is
Options:
A) $ 2.8\times {10^{-4}} $
B) $ 2.8\times {10^{-3}} $
C) $ 2.8\times {10^{-2}} $
D) $ 2.8\times {10^{-5}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{0}{\mathop{\underset{100\times 0.1}{\mathop{MgCl_2}},}}+,\underset{0}{\mathop{\underset{100\times 0.2}{\mathop{2NaOH}},}},\to \underset{10}{\mathop{\underset{0}{\mathop{Mg}},}},\underset{20}{\mathop{\underset{0}{\mathop{{{(OH)}_2}}},}},+\underset{After,mixing}{\mathop{\underset{Initial,moles}{\mathop{2NaCl}},}}, $
$ \therefore K _{sp}=[ \frac{10}{200} ]{{[ \frac{20}{200} ]}^{2}}=5\times {10^{-4}} $ But actually $ K _{sp} $ of $ Mg{{( OH )}_2}=1.2\times {10^{-11}} $
$ \therefore 4S^{3}=1.2\times {10^{-11}} $ Find S then $ [ O{H^{-}} ]=2S $ which is $ 2.8\times {10^{-4}}. $
