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Question: A solution has \[0.05\text{ }M M{{g}^{(2+)}}\]and \[0.05\text{ }M\text{ N}{{\text{H}}_{3}}\]. Calcul...

A solution has 0.05 MMg(2+)0.05\text{ }M M{{g}^{(2+)}}and 0.05 M NH30.05\text{ }M\text{ N}{{\text{H}}_{3}}. Calculate the concentration of NH4ClN{{H}_{4}}Cl required to prevent the formation of Mg(OH)2Mg{{\left( OH \right)}_{2}} in this solution. Ksp{{K}_{sp}} of Mg(OH)2=9.0×10−12Mg{{\left( OH \right)}_{2}}=9.0\times {{10}^{-12}} and ionization constant of NH3=1.8×10−5N{{H}_{3}}=1.8\times {{10}^{-5}}.

Explanation

Solution

Hint: The following formula can be used to solve these type of questions:
Ksp=[Mg2+][OH−]2{{K}_{sp}}=\left[ M{{g}^{2+}} \right]{{\left[ O{{H}^{-}} \right]}^{2}}
pOH=− log[OH−]1pOH=-~\text{log}{{\left[ O{{H}^{-}} \right]}^{1}}
pOH=pKb+log(NH4ClNH3)pOH=p{{K}_{b}}+\text{log}\left( \dfrac{N{{H}_{4}}Cl}{N{{H}_{3}}} \right)

Complete answer:
This solution is a buffer of a weak base and its conjugate acidic salt.
Mg(OH)2⇌ Mg2++2OH−                                                     s                    2s\begin{aligned} & Mg{{\left( OH \right)}_{2}}\rightleftharpoons ~M{{g}^{2+}}+2O{{H}^{-}} \\\ & \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; s \;\;\;\;\;\;\;\;\;\; 2s \end{aligned}
Ksp=s×(2s)2{{K}_{sp}}=s\times {{\left( 2s \right)}^{2}}
Ksp=[Mg2+][OH−]2{{K}_{sp}}=\left[ M{{g}^{2+}} \right]{{\left[ O{{H}^{-}} \right]}^{2}}so now we have to find the concentration of the hydroxyl ions,
[OH−]=(Ksp[Mg2+])12\left[ O{{H}^{-}} \right]={{\left( \dfrac{{{K}_{sp}}}{\left[ M{{g}^{2+}} \right]} \right)}^{\dfrac{1}{2}}}.
[OH−]=(9.0×10−120.05)12\left[ O{{H}^{-}} \right]={{\left( \dfrac{9.0\times {{10}^{-12}}}{0.05} \right)}^{\dfrac{1}{2}}}.
pOH=− log[OH−]1pOH=-~\text{log}{{\left[ O{{H}^{-}} \right]}^{1}}Now, we know for a buffer solution, pOH=pKb+log(NH4ClNH3)pOH=p{{K}_{b}}+\text{log}\left( \dfrac{N{{H}_{4}}Cl}{N{{H}_{3}}} \right)
4.8495=−5.2553+log(NH4Cl0.05)4.8495=-5.2553+\text{log}\left( \dfrac{N{{H}_{4}}Cl}{0.05} \right)
−0.4058 =log(NH4Cl0.05)-0.4058~=\text{log}\left( \dfrac{N{{H}_{4}}Cl}{0.05} \right)
[NH4Cl] =0.67×0.05\left[ N{{H}_{4}}Cl \right]~=0.67\times 0.05
[NH4Cl] =0.0335≃0.03M\left[ N{{H}_{4}}Cl \right]~=0.0335\simeq 0.03M
Hence, the correct answer is 0.03M0.03M.

Note: Solubility equilibrium is defined as a type of dynamic equilibrium that exists when a chemical compound in the solid state is in chemical equilibrium with a solution of that compound which here is ammonium chloride and ammonium hydroxide. The solid may dissolve unchanged, but with dissociation or with chemical reaction with other constituents of the solution which can be acid or alkali.