Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

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

seo-qna
Last updated date: 20th Apr 2024
Total views: 399.9k
Views today: 8.99k
Answer
VerifiedVerified
399.9k+ views
Hint: The following formula can be used to solve these type of questions:
${{K}_{sp}}=\left[ M{{g}^{2+}} \right]{{\left[ O{{H}^{-}} \right]}^{2}}$
$pOH=-~\text{log}{{\left[ O{{H}^{-}} \right]}^{1}}$
$pOH=p{{K}_{b}}+\text{log}\left( \dfrac{N{{H}_{4}}Cl}{N{{H}_{3}}} \right)$

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

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.
Recently Updated Pages