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Question: \(Mg{\left( {OH} \right)_2}\)is precipitated when NaOH is added to a solution of \(M{g^{2 + }}\). If...

Mg(OH)2Mg{\left( {OH} \right)_2}is precipitated when NaOH is added to a solution of Mg2+M{g^{2 + }}. If the final concentration of Mg2+M{g^{2 + }}is 10−10{10^{ - 10}}M, the concentration of OH−(M)O{H^ - }\left( M \right)in the solution is [Solubility product forMg(OH)2Mg{\left( {OH} \right)_2} =5.6×10−12 = 5.6 \times {10^{ - 12}}].
(A) 0.0560.056
(B) 0.120.12
(C) 0.240.24
(D) 0.0250.025

Explanation

Solution

As we learnt previously that solubility is the ability to be dissolved, especially in water. Talking about solubility products, it is a kind of equilibrium constant and its value depends on temperature solubility product can be written as Ksp. So, it usually increases with an increase in temperature due to which solubility increases.

Complete step by step answer:
As we have magnesium dihydroxide given in the question. So, if we make the equation of Mg(OH)2Mg{\left( {OH} \right)_2}. Therefore, Mg(OH)2→Mg2++(OH−)2Mg{\left( {OH} \right)_2} \to M{g^{2 + }} + {\left( {O{H^ - }} \right)_2}……… (i) solubility product: KsP=[Mg2+][OH−]K_sP = \left[ {M{g^{2 + }}} \right]\left[ {O{H^ - }} \right]
A solubility product mathematically is the product of dissolved ion concentrations raised to the power of their stoichiometric coefficient.
KsP=[Mg2+]1[OH−]2K_sP = {\left[ {M{g^{2 + }}} \right]^1}{\left[ {O{H^ - }} \right]^2}, From equation (i) we write the powers and it is given in the equation, the solubility product of Mg(OH)2Mg{\left( {OH} \right)_2} =5.6×10−12 = 5.6 \times {10^{ - 12}}
5.6×10−12=5.6 \times {10^{ - 12}} = [Mg2+]1[OH−]2{\left[ {M{g^{2 + }}} \right]^1}{\left[ {O{H^ - }} \right]^2}
5.6×10−12=5.6 \times {10^{ - 12}} = =10−10×[OH−]2 = {10^{ - 10}} \times {\left[ {O{H^ - }} \right]^2}
5.6×10−1210−10=[OH−]2\dfrac{{5.6 \times {{10}^{ - 12}}}}{{{{10}^{ - 10}}}} = {\left[ {O{H^ - }} \right]^2}
By solving this, we get
[OH−]2=5.6×10−2{\left[ {O{H^ - }} \right]^2} = 5.6 \times {10^{ - 2}}
[OH−]=5.6×10−2\left[ {O{H^ - }} \right] = \sqrt {5.6 \times {{10}^{ - 2}}}
[OH−]=0.24M\left[ {O{H^ - }} \right] = 0.24M

So, the correct answer is Option C.

Note:
As we know that concentration of a solution is that amount of solute which has been dissolved in given amount of solution and a concentrated solution has a relatively large amount of the dissolved solute whereas a dilute solution has relatively small amount of dissolved solute.