Question

Calculate the ${{K}_{h}}$ and pH of 0.05 N $N{{H}_{4}}Cl$ solution where ${{K}_{b}}=1.8\text{ x 1}{{\text{0}}^{-5}}$?

Answer 1

Hint: ${{K}_{h}}$ of a base can be calculated by the formula ${{K}_{h}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}$, where ${{K}_{w}}$ is the ionic product of the water, and ${{K}_{b}}$ is the dissociation constant of the base. The pH of a salt which is formed by the weak base and strong salt can be calculated by the formula $pH=7-\dfrac{1}{2}[p{{K}_{b}}+\log c]$, where c is the concentration of the salt.

Complete step by step answer:
- $N{{H}_{4}}Cl$ is an inorganic compound whose name is ammonium chloride and it is a salt of a weak base and a strong acid. The weak base is ammonium hydroxide whose formula is $N{{H}_{4}}OH$ and the strong acid is hydrochloric acid whose formula is $HCl$.
- ${{K}_{h}}$ of a base can be calculated by the formula ${{K}_{h}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}$, where ${{K}_{w}}$ is the ionic product of the water, and ${{K}_{b}}$ is the dissociation constant of the base. ${{K}_{h}}$ is the hydrolysis constant of the salt, ${{K}_{w}}$ is taken as ${{10}^{-14}}$ and the value of ${{K}_{b}}$ is given in the question, i.e., $1.8\text{ x 1}{{\text{0}}^{-5}}$ . so, by [utting the values, we get:
${{K}_{h}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}=\dfrac{{{10}^{-14}}}{1.8\text{ x 1}{{\text{0}}^{-5}}}=5.56\text{ x 1}{{\text{0}}^{-10}}$
- Since $N{{H}_{4}}Cl$ is a salt of a weak base and a strong acid, we can calculate the pH of the solution of the salt by the formula $pH=7-\dfrac{1}{2}[p{{K}_{b}}+\log c]$, where c is the concentration of the salt.
- We are given the concentration of the salt is 0.05 N and ${{K}_{b}}$ is given in the question, i.e., $1.8\text{ x 1}{{\text{0}}^{-5}}$ so, the $p{{K}_{b}}$ will be $-\log (1.8\text{ x 1}{{\text{0}}^{-5}})$. Now, putting all the values, we get:
$pH=7-\dfrac{1}{2}[p{{K}_{b}}+\log C]=7-\dfrac{1}{2}\{[-\log (1.8\text{ x 1}{{\text{0}}^{-5}})]+\log 0.05\}=5.28$
So, the pH of the solution will be 5.28.

Note: If the salt is made up of a strong base and a weak acid then the formula will be ${{K}_{h}}=\dfrac{{{K}_{w}}}{{{K}_{a}}}$, where ${{K}_{a}}$ is the dissociation constant of the acid, and the formula for calculating the pH will be $pH=7-\dfrac{1}{2}[p{{K}_{a}}+\log c]$.