Question
The concentration of hydroxyl ion in a solution left after mixing $$100\,mL$$ of $$0.1\,M\,MgC{l_2}$$ and $$100\,mL$$ of $$0.2\,M\,NaOH$$
$$\left( {{K_{sp}}\,{\text{of}}\,Mg{{\left( {OH} \right)}_2} = 1.2 \times {{10}^{ - 11}}} \right)$$ is
A.
$$2.8 \times {10^{ - 4}}$$
B.
$$2.8 \times {10^{ - 3}}$$
C.
$$2.8 \times {10^{ - 2}}$$
D.
$$2.8 \times {10^{ - 5}}$$
Answer :
$$2.8 \times {10^{ - 4}}$$
Solution :
\[\begin{align}
& \underset{\begin{smallmatrix}
100\times 0.1 \\
0
\end{smallmatrix}}{\mathop{MgC{{l}_{2}}}}\,+\underset{\begin{smallmatrix}
100\times 0.2 \\
\,0
\end{smallmatrix}}{\mathop{2NaOH}}\,\to \underset{\begin{smallmatrix}
0 \\
10
\end{smallmatrix}}{\mathop{Mg}}\,\underset{\begin{smallmatrix}
0 \\
20
\end{smallmatrix}}{\mathop{{{\left( OH \right)}_{2}}}}\,+\underset{\begin{smallmatrix}
\text{Initial moles} \\
\text{After mixing}
\end{smallmatrix}}{\mathop{2NaCl}}\, \\
& \therefore \,\,{{K}_{sp}}=\left[ \frac{10}{200} \right]{{\left[ \frac{20}{200} \right]}^{2}}=5\times {{10}^{-4}} \\
& \text{But actually}\,{{K}_{sp}}\,\text{of} \\
& Mg{{\left( OH \right)}_{2}}=1.2\times {{10}^{-11}} \\
& \therefore \,\,4{{s}^{3}}=1.2\times {{10}^{-11}} \\
& \text{Find }S\text{ then}\left[ O{{H}^{-}} \right]=2S\,\,\text{which is}\,\,2.8\times {{10}^{-4}} \\
\end{align}\]