Question

The pH of a 0.02 M $N{{H}_{4}}Cl$ solution will be: [given ${{K}_{b}}(N{{H}_{4}}OH)={{10}^{-5}}$and log 2 = 0.301].
A. 4.65
B. 5.35
C. 4.35
D. 2.65

Answer 1

Ammonium chloride ($N{{H}_{4}}Cl$) is a salt formed by the reaction of a strong acid (HCl) and a weak base ($N{{H}_{4}}OH$).
The chemical reaction between HCl and $N{{H}_{4}}OH$is as follows.
$HCl+N{{H}_{4}}OH\to N{{H}_{4}}Cl+{{H}_{2}}O$

Complete step by step answer:
-In the question it is given that the concentration of $N{{H}_{4}}Cl$is 0.02 M and base dissociation constant of $N{{H}_{4}}OH$is${{10}^{-5}}$.
- We have to find the pH of the 0.02 M $N{{H}_{4}}Cl$ solution from the given data.
-To calculate the pH of the $N{{H}_{4}}Cl$solution (formed by strong acid and weak base) there is a formula. The formula is as follows.
$pH=7-\dfrac{1}{2}[{{p}{{{K}_{b}}}}+\log C]$
Here pH = pH of the salt solution
${{K}_{b}}$ = base dissociation constant (${{p}{{{K}_{b}}}}=-\log {{k}_{b}}$)
C = concentration of the salt
- Now we have to substitute all the known values in the above formula to get the pH of the solution.

& pH=7-\dfrac{1}{2}[{{p}{{{K}_{b}}}}+\log C] \\\ & \text{ = 7}-\dfrac{1}{2}[\log {{10}^{-5}}+\log 0.02] \\\ & \text{ = 7}-\dfrac{5}{2}-\dfrac{1}{2}(\log 2\times {{10}^{-2}}) \\\ & \text{ = 5}\text{.35} \\\ \end{aligned}$$ -Therefore the pH of the ammonium chloride solution is 5.35. **-So, the correct option is B.** **Note:** Strong acid is an acid which dissolves in an aqueous solution and dissociates completely in less time. Example for a strong acid is Hydrochloric acid (HCl). Weak acid is an acid which does not dissociate completely in an aqueous solution. Example for weak acid is acetic acid ($$C{{H}_{3}}COOH$$). Strong base is a base which dissolves in an aqueous solution and dissociates completely in less time. Example for a strong base is sodium hydroxide (NaOH). Weak base is a base which does not dissociate completely in an aqueous solution. Example for a weak base is ammonium hydroxide ($$N{{H}_{4}}OH$$).