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Question: The solubility product constant for \( Mg{(OH)_2} \) is \( 1.8 \times {10^{ - 11}} \) . What would b...

The solubility product constant for Mg(OH)2Mg{(OH)_2} is 1.8×10−111.8 \times {10^{ - 11}} . What would be the solubility of Mg(OH)2Mg{(OH)_2} in 0.345 M NaOH?

Explanation

Solution

In this problem we are given the solubility constant Ksp{K_{sp}} . The value of Ksp{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)2Mg{(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)2Mg{(OH)_2} and NaOH. Here the common ion effect will come into play. The common ion in both the solutions is OH−O{H^ - } . Hence the concentration of OH−O{H^ - } will increase in the resultant solution.
We are given the value of Ksp{K_{sp}} of Mg(OH)2Mg{(OH)_2} as 1.8×10−111.8 \times {10^{ - 11}} and the concentration of NaOH is 0.345M. Let us consider the solubility of Mg(OH)2Mg{(OH)_2} to be ‘x’. The dissociation of NaOH will be complete as it is a strong base. The dissociation can be given as:
NaOH ⇌ Na+ + OH−{\text{NaOH }} \rightleftharpoons {\text{ N}}{{\text{a}}^ + }{\text{ }} + {\text{ O}}{{\text{H}}^ - }

T=00.345--
T=equilibrium0.3450.345M0.345M

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

T=0aa--
T=equilibriuma−xa - xxx2x+0.3452x + 0.345

Now, the Ksp{K_{sp}} of Mg(OH)2Mg{(OH)_2} as 1.8×10−111.8 \times {10^{ - 11}} . Mathematically can be given as: Ksp=[Mg+2][OH−]2=1.8×10−11{K_{sp}} = [M{g^{ + 2}}]{[O{H^ - }]^2} = 1.8 \times {10^{ - 11}}
[Mg+2][OH−]2=(x)(2x+0.345)2=1.8×10−11[M{g^{ + 2}}]{[O{H^ - }]^2} = (x){(2x + 0.345)^2} = 1.8 \times {10^{ - 11}}
We know that 2x<<<0.3452x < < < 0.345 , therefore we can ignore 2x. The equation will hence become:
(x)(0.345)2=1.8×10−11(x){(0.345)^2} = 1.8 \times {10^{ - 11}}
0.119x=1.8×10−110.119x = 1.8 \times {10^{ - 11}}
x=1.8×10−110.119=1.5×10−10x = \dfrac{{1.8 \times {{10}^{ - 11}}}}{{0.119}} = 1.5 \times {10^{ - 10}}
This is the solubility of the salt Mg(OH)2Mg{(OH)_2} which is equal to 1.5×10−10mol/L1.5 \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.