Question

Solubility product of \[Mg{(OH)_2}\] is \[1 \times {10^{ - 11}}\]. At what pH, precipitation of \[Mg{(OH)_2}\]​ will begin from 0.1 M \[M{g^{2 + }}\] solution:
A.9
B.5
C.3
D.7

Answer 1

Hint: The solubility product constant is an equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol \[{K_{sp}}\]. It is usually represented as the product of the concentration of the two ions found in the ionic compound. The solubility product value depends on temperature. It usually increases with an increase in temperature due to increased solubility in the aqueous medium.

Complete answer:
We know that the solubility product of any compound can be expressed as the product of concentration of its constituent ions. This means that in the case of \[Mg{(OH)_2}\]
\[ \Rightarrow {K_{sp}} = [M{g^{2 + }}]{[O{H^ - }]^2}\]
Since it is given that \[{K_{sp}}\] = \[1 \times {10^{ - 11}}\],
Concentration of \[M{g^{2 + }}\] ions is given as 0.1 M
 By substituting in the equation, we get
\[ \Rightarrow 1 \times {10^{ - 11}} = 0.1 \times {[O{H^ - }]^2}\]
Thus we can say that the concentration of \[O{H^ - }\] can be found out as
\[ \Rightarrow [O{H^ - }] = \sqrt {\dfrac{{1 \times {{10}^{ - 11}}}}{{0.1}}} \]
\[ \Rightarrow [O{H^ - }] = \sqrt {1 \times {{10}^{ - 10}}} \]
\[ \Rightarrow [O{H^ - }] = 1 \times {10^{ - 5}}\] M
Now we know that the concentration of \[O{H^ - }\] in the given solution is \[1 \times {10^{ - 5}}M\]
We also know that \[pH\] is defined as the \[ - \log [{H^ + }]\]
We can redefine this and say that
\[ \Rightarrow pH = 14 - pOH\]
\[ \Rightarrow pH = 14 - ( - \log [O{H^ - }])\]
\[ \Rightarrow pH = 14 + \log [1 \times {10^{ - 5}}])\]
\[ \Rightarrow pH = 14 - 5\]
\[ \Rightarrow pH = 9\]
 Thus we can say that at \[pH\] of 4, the given solution of \[Mg{(OH)_2}\] will start to precipitate.
Thus the correct option for the given question is option (A).

Note:
\[{K_{sp}}\] is a very important tool in chemistry. It is used for determination of solubility of sparingly soluble salts, for predicting precipitations, to manipulate the preferential and simultaneous solubility of a mixture of salts.
Note that when the ionic product of a solution is greater than its solubility product, the solute will start to precipitate.