Question

The solubility product constant for $ Mg{(OH)_2} $ is $ 1.8 \times {10^{ - 11}} $ . What would be the solubility of $ Mg{(OH)_2} $ in 0.345 M NaOH?

Answer 1

Hint: In this problem we are given the solubility constant $ {K_{sp}} $ . The value of $ {K_{sp}} $ is given as the product of the solubilities of the respective ions present in the salt, raised to their respective stoichiometries. The salts given to us are Magnesium Hydroxide and Sodium Hydroxide.

Complete answer:
The given salts are magnesium hydroxide ( $ Mg{(OH)_2} $ ) and Sodium Hydroxide (NaOH). We can see that in both the compound hydroxide ions are common. In this question we are given $ Mg{(OH)_2} $ and NaOH. Here the common ion effect will come into play. The common ion in both the solutions is $ O{H^ - } $ . Hence the concentration of $ O{H^ - } $ will increase in the resultant solution.
We are given the value of $ {K_{sp}} $ of $ Mg{(OH)_2} $ as $ 1.8 \times {10^{ - 11}} $ and the concentration of NaOH is 0.345M. Let us consider the solubility of $ Mg{(OH)_2} $ to be ‘x’. The dissociation of NaOH will be complete as it is a strong base. The dissociation can be given as:
 $ {\text{NaOH }} \rightleftharpoons {\text{ N}}{{\text{a}}^ + }{\text{ }} + {\text{ O}}{{\text{H}}^ - } $

T=0	0.345	-	-
T=equilibrium	0.345	0.345M	0.345M

 $ {\text{Mg(OH}}{{\text{)}}_2}{\text{ }} \rightleftharpoons {\text{ M}}{{\text{g}}^{ + 2}}{\text{ }} + {\text{ 2O}}{{\text{H}}^ - } $

T=0	$ a $	-	-
T=equilibrium	$ a - x $	$ x $	$ 2x + 0.345 $

Now, the $ {K_{sp}} $ of $ Mg{(OH)_2} $ as $ 1.8 \times {10^{ - 11}} $ . Mathematically can be given as: $ {K_{sp}} = [M{g^{ + 2}}]{[O{H^ - }]^2} = 1.8 \times {10^{ - 11}} $
 $ [M{g^{ + 2}}]{[O{H^ - }]^2} = (x){(2x + 0.345)^2} = 1.8 \times {10^{ - 11}} $
We know that $ 2x < < < 0.345 $ , therefore we can ignore 2x. The equation will hence become:
 $ (x){(0.345)^2} = 1.8 \times {10^{ - 11}} $
 $ 0.119x = 1.8 \times {10^{ - 11}} $
 $ x = \dfrac{{1.8 \times {{10}^{ - 11}}}}{{0.119}} = 1.5 \times {10^{ - 10}} $
This is the solubility of the salt $ Mg{(OH)_2} $ which is equal to $ 1.5 \times {10^{ - 10}}mol/L $ . This is the required answer.

Note:
If we are given two or more solutions having at least one common ion, always the common ion effect will come into play, and an increase in the concentration of that common ion is observed. If concentration of one ion is increased, the system will try to decrease the concentration and the equilibrium will move forward.