Question

Four solutions of $N{H_4}Cl$ are taken with concentrations $1M,{\text{ }}0.1M,{\text{ }}0.1M$ and $0.001M$. Their degree of hydrolysis are ${h_1},{\text{ }}{h_2},{\text{ }}{h_3},$ and ${h_4}$. What is the graduation degree of hydrolysis?
A. ${h_1} > {h_2} > {h_3} > {h_4}$
B. ${h_1} = {h_2} = {h_3} = {h_4}$
C. ${h_4} > {h_3} > {h_2} > {h_1}$
D. None of these

Answer 1

The degree of hydrolysis of a salt may be defined as the fraction of salt which is hydrolysed by water. It is the ratio of the number of salt hydrolysed to the total number of salt taken.

Complete step by step answer:
Salt hydrolysis is the reaction of salt with water to produce an acid and a base. Thus, it is the reverse of neutralisation. The nature of the solution after hydrolysis depends upon the relative strengths of the acid and base formed. Ammonium chloride is a slat of weak base and a strong acid, so its hydrolysis produces with $pH < 7$. So, the solution will be acidic in nature.
Dissociation is the process by which ionic compounds or molecules get separated into its constituent ions, atoms or molecules. The dissociation process is reverse in nature.
Given: The concentration of ammonium chloride solutions are $1M,{\text{ }}0.1M,{\text{ }}0.1M$ and $0.001M$ their degree of hydrolysis are ${h_1},{\text{ }}{h_2},{\text{ }}{h_3},$and ${h_4}$.
By hydrolysis of ammonium chloride $\left( {N{H_4}Cl} \right)$ salt,
After hydrolysis $C\left( {1{\text{ }}-h} \right)\;Ch\;\;\;\;\;\;\;\;\;Ch$
$\therefore$ Hydrolysis constant $\left( {{K_h}} \right) = \dfrac{{\left[ {Product} \right]}}{{\left[ {\operatorname{Re} ac\tan t} \right]}}$
$\Rightarrow$Hydrolysis constant $\left( {{K_h}} \right) = \dfrac{{[N{H_4}OH][HCl]}}{{[N{H_4}OH]}}$
By putting the value of each concentration of reactant and product
Then, ${K_h} = \dfrac{{Ch \times Ch}}{{C(1 - h)}}$
${K_h} = \dfrac{{Ch \times Ch}}{C}\,\,\,\,\,\,\,\,\,\,\,\,\,\,[\because \,h < < 1,\,\,\because 1 - h \approx 1]$
${K_h} = C{h^2}$
$\therefore h = \sqrt {\dfrac{{{K_h}}}{C}}$
Where h is the degree of hydrolysis, ${K_h}$ is the hydrolysis constant and h is the degree of hydrolysis.
$\therefore h \propto \dfrac{1}{{\sqrt C }}$
Higher the value of concentration lower its degree of hydrolysis, from the above given data four solutions of $N{H_4}Cl$ are taken with concentration $1M,{\text{ }}0.1M,{\text{ }}0.01M$ and $0.001M$.
Their degree of hydrolysis are ${h_1},{\text{ }}{h_2},{\text{ }}{h_3},{\text{ }}{h_4},$then the order of degree of hydrolysis is ${h_4} > {\text{ }}{h_2} > {\text{ }}{h_3} > {\text{ }}{h_1}.$

Hence, the correct option is (c).

Note: So, you can start by describing the various terms in the question involved like degree of hydrolysis, dissociation constant. Then use equations $\left( {{K_h}} \right) = \dfrac{{\left[ {Product} \right]}}{{\left[ {\operatorname{Re} ac\tan t} \right]}}$ and $\therefore h = \sqrt {\dfrac{{{K_h}}}{C}}$ to find out the dissociation constant and degree of hydrolysis. Every equation is salt hydrolysis is derived from Ostwald’s law of dilution. Students must be very careful while finding out the value of hydrolysis constant as it is further used to find out the degree of dissociation and pH of the resulting solution.