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Question: The \( {K_{sp}} \) of \( Mg{(OH)_2} \) is \( 1 \times {10^{ - 12}} \) . \( 0.01 \) m \( Mg{(OH)_2} \...

The Ksp{K_{sp}} of Mg(OH)2Mg{(OH)_2} is 1×10−121 \times {10^{ - 12}} . 0.010.01 m Mg(OH)2Mg{(OH)_2} will precipitate at the limiting pHpH equal to:

Explanation

Solution

Ksp{K_{sp}} is called the solubility product because it is the product of the solubilities of the ions in moles per liter. Here the given compound is Mg(OH)2Mg{(OH)_2} which will dissociate into 11 . Hence, Solubility product expression becomes
Ksp{K_{sp}} = [Mg2+][OH−]2[M{g^{2 + }}]{[O{H^ - }]^2}
The above expression can be simplified to find the OH−O{H^ - } concentration.
[OH−][O{H^ - }] = Ksp[Mg2+]\sqrt {\dfrac{{{K_{sp}}}}{{[M{g^{2 + }}]}}}

Complete step by step solution:
Mg(OH)2Mg{(OH)_2} will release Mg2+M{g^{2 + }} and OH−O{H^ - } ions according to the reaction,
⇌\rightleftharpoons Mg(OH)2Mg{(OH)_2} Mg2+M{g^{2 + }} ++ 2OH−2O{H^ - }
Thus, the solubility product is
Ksp{K_{sp}} = [Mg2+][OH−]2[M{g^{2 + }}]{[O{H^ - }]^2}
Where Ksp{K_{sp}} = solubility product constant
[Mg2+][M{g^{2 + }}] =concentration of Mg2+M{g^{2 + }} ions
[OH−][O{H^ - }] =concentration of OH−O{H^ - } ions
In this question we are given the concentration of Mg2+M{g^{2 + }} ions ( 0.010.01 m) and the value of Ksp{K_{sp}} ( 1×10−121 \times {10^{ - 12}} ).
Using the simplified formula we can find the concentration of OH−O{H^ - } ions.
[OH−][O{H^ - }] = Ksp[Mg2+]\sqrt {\dfrac{{{K_{sp}}}}{{[M{g^{2 + }}]}}}
⇒\Rightarrow [OH−][O{H^ - }] = 1×10−120.01\sqrt {\dfrac{{1 \times {{10}^{ - 12}}}}{{0.01}}} m
⇒\Rightarrow [OH−][O{H^ - }] = 10−10\sqrt {{{10}^{ - 10}}} m
⇒\Rightarrow [OH−][O{H^ - }] = 10−5{10^{ - 5}} m
Now we got the concentration of OH−O{H^ - } ions. By using the formula mentioned below, we can find the pOHpOH when the concentration of OH−O{H^ - } ion is known.
pOH=−log⁡[OH−]pOH = - \log [O{H^ - }]
⇒\Rightarrow pOH=−log⁡[10−5]pOH = - \log [{10^{ - 5}}]
⇒pOH=−1×−5\Rightarrow pOH = - 1 \times - 5
⇒pOH=5\Rightarrow pOH = 5
Now the general equation in terms of pHpH and pOHpOH is,
pH+pOH=14pH + pOH = 14
⇒pH=14−pOH\Rightarrow pH = 14 - pOH
As we know pOH=5pOH = 5 , thus by substituting the value we get,
⇒pH=14−5\Rightarrow pH = 14 - 5
⇒pH=9\Rightarrow pH = 9
Therefore, 0.010.01 m Mg(OH)2Mg{(OH)_2} will precipitate at the limiting pHpH of 99 .

Additional information:
The solubility product is the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is denoted by the symbol Ksp{K_{sp}} . The value of Ksp{K_{sp}} depends on temperature and is different for every salt. Ksp{K_{sp}} value generally increases with the increase in temperature due to increased solubility. Some of the factors which affect the value of Ksp{K_{sp}} are:
∙\bullet Common-ion effect (the presence of a common ion lowers the value of Ksp{K_{sp}} .
∙\bullet The diverse ion effect (if the ions of solute are uncommon, the value of Ksp{K_{sp}} will be high).
∙\bullet Ion pair presence.

Note:
The idea here is that you need to use magnesium hydroxide’s solubility product constant to determine what concentration of OH−O{H^ - } ions would cause the solid to precipitate out of solution.