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Question: Solubility product of \[Mg{{\left( OH \right)}_{2}}\] is \[1\times {{10}^{-11}}.\] At what pH, preci...

Solubility product of Mg(OH)2Mg{{\left( OH \right)}_{2}} is 1×10−11.1\times {{10}^{-11}}. At what pH, precipitation of Mg(OH)2Mg{{\left( OH \right)}_{2}}will begin from 0.1M Mg2+0.1M\text{ }M{{g}^{2+}} solution:
A.99
B.55
C.33
D.77

Explanation

Solution

We know that the solubility product constant is a simplified equilibrium constant denoted as Ksp{{K}_{sp}} which is defined for equilibrium between a solid and its respective ions in a given solution. Its value shows the degree to which a compound can dissociate in water.

Complete answer:
As we know that the Solubility products are always dependent upon the solubility equation and again, the solubility equation depends upon the concentration of the dissociated ions and their numbers. If the concentration of the solutes is small enough that solute-solute interactions are negligible, then the solution can be considered to be ideal and we can approximate the activities of solutes by the concentration terms. Each concentration is raised to a power which is equal to the coefficient of that ion in a balanced equation to get the solubility equilibrium.
The solubility product constants are used for describing the saturated solutions of ionic compounds of relatively low solubility. A saturated solution is in a dynamic equilibrium state between the dissolved and dissociated ionic compound and the undissolved solid. In this article, we will learn about what is the solubility product, what is Ksp{{K}_{sp}} the solubility product definition, and some solubility product numerical. Here the reaction is given out by; Mg(OH)2→Mg+2(0.1) +2OH−(2S) Mg{{\left( OH \right)}_{2}}\to \underset{\left( 0.1 \right)}{\mathop{M{{g}^{+2}}}}\,+2\underset{\left( 2S \right)}{\mathop{O{{H}^{-}}}}\,
Thus, Ksp=[Mg+2][OH−]2Ksp=\left[ M{{g}^{+2}} \right]{{\left[ O{{H}^{-}} \right]}^{2}} here [OH−]=10−5.\left[ O{{H}^{-}} \right]={{10}^{-5}}.
The pOH is carried out by pOH=−log⁡[OH−]=−log⁡[10−5]=5.pOH=-\log \left[ O{{H}^{-}} \right]=-\log \left[ {{10}^{-5}} \right]=5.
As we know that the pH=14−pOH=14−5=9.pH=14-pOH=14-5=9.
Therefore, the correct answer is option A i.e., at pH = 9,pH\text{ }=\text{ }9, precipitation will occur.

Note:
Remember that solubility is defined as the property of a substance known as solute to get dissolved in a solvent for the formation of a solution. The solubility of ionic compounds that dissociate and form cations and anions in water vary to a great extent.