
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 $N{{H}_{3}}=1.8\times {{10}^{-5}}$.
Answer
584.7k+ 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.
${{K}_{sp}}=\left[ M{{g}^{2+}} \right]{{\left[ O{{H}^{-}} \right]}^{2}}$
$pOH=-~\text{log}{{\left[ O{{H}^{-}} \right]}^{1}}$
$pOH=p{{K}_{b}}+\text{log}\left( \dfrac{N{{H}_{4}}Cl}{N{{H}_{3}}} \right)$
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
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

What is a periderm How does periderm formation take class 11 biology CBSE

What are porins class 11 biology CBSE

Why is steel more elastic than rubber class 11 physics CBSE

Differentiate between red algae and brown algae class 11 biology CBSE

What is boron A Nonmetal B Metal C Metalloid D All class 11 chemistry CBSE

