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

Hint: 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.
\[\begin{align}
  & pH=7-\dfrac{1}{2}[{{p}{{{K}_{b}}}}+\log C] \\
 & \text{ = 7}-\dfrac{1}{2}[\log {{10}^{-5}}+\log 0.02] \\
 & \text{ = 7}-\dfrac{5}{2}-\dfrac{1}{2}(\log 2\times {{10}^{-2}}) \\
 & \text{ = 5}\text{.35} \\
\end{align}\]
-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 (\[N{{H}_{4}}OH\]).