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Question: The solubility constant \({{K}_{sp}}\) of \(Mg{{(OH)}_{2}}\) is \(9.0\text{ x 1}{{\text{0}}^{-12}}\)...

The solubility constant Ksp{{K}_{sp}} of Mg(OH)2Mg{{(OH)}_{2}} is 9.0 x 10−129.0\text{ x 1}{{\text{0}}^{-12}}. If a solution is 0.010 with respect to Mg2+M{{g}^{2+}} ion, what is the maximum hydroxide ion concentration which could be present without causing precipitation of Mg(OH)2Mg{{(OH)}_{2}} ?

Explanation

Solution

Write the solubility product reaction for Mg(OH)2Mg{{(OH)}_{2}}. Now write the reaction constant at equilibrium. The solubility constant is given to us as well as the concentration of Mg2+M{{g}^{2+}}. Substitute both these values in the reaction constant at equilibrium to obtain the concentration of hydroxide ions.

Complete step-by-step answer:
Solubility is defined as the property of a substance (solute) to get dissolved in a solvent in order to form a solution.
The solubility product constant is the equilibrium constant for the dissolution of a solid solute in a solvent to form a solution. It is denoted by the symbol Ksp{{K}_{sp}}. We can form an idea about the dissolution of an ionic compound by the value of its solubility product.
The concentration of Mg2+\text{M}{{\text{g}}^{\text{2+}}} = 0.010 M
Reaction:
Mg(OH)2 → Mg2+ + 2OH−\text{Mg(OH}{{\text{)}}_{\text{2}}}\text{ }\to \text{ M}{{\text{g}}^{\text{2+}}}\text{ + 2O}{{\text{H}}^{-}}
Ksp= !![!! Mg2+ !!]!! !![!! OH− !!]!! 2{{\text{K}}_{\text{sp}}}\text{= }\\!\\![\\!\\!\text{ M}{{\text{g}}^{\text{2+}}}\text{ }\\!\\!]\\!\\!\text{ }\\!\\![\\!\\!\text{ O}{{\text{H}}^{-}}{{\text{ }\\!\\!]\\!\\!\text{ }}^{\text{2}}}
 !![!! OH− !!]!! = 9.0 x 10-120.010\text{ }\\!\\![\\!\\!\text{ O}{{\text{H}}^{-}}\text{ }\\!\\!]\\!\\!\text{ = }\sqrt{\dfrac{\text{9}\text{.0 x 1}{{\text{0}}^{\text{-12}}}}{\text{0}\text{.010}}}
[OH−] = 3 x 10−5[\text{O}{{\text{H}}^{-}}]\text{ = 3 x 1}{{\text{0}}^{-5}}
The concentration of hydride ions is 3 x 10−53\text{ x 1}{{\text{0}}^{-5}}.

Therefore, the correct answer is option (D).

Note: The common ion effect describes the effect of adding a common ion on the ​equilibrium of the new solution. The common ion effect generally decreases ​solubility of a solute. The equilibrium shifts to the left to relieve off the excess product formed.
When a strong electrolyte is added to a solution of weak electrolyte, the solubility of weak electrolyte decreases leading to the formation of precipitate at the bottom of the testing apparatus.