Question

How to calculate the pH of the buffer system $\text{0}\text{.15M N}{\text{H}}_{\text{3}}\text{ /0}\text{.35M N}{\text{H}}_{\text{4}}\text{C}$ ?

Answer 1

Hint: In the above question, we are asked to find the pH of the buffer system when the concentration of base and acid of ammonia is given. First we have to use the base dissociation constant of ammonia to get the acid dissociation constant so that we can put the values in Henderson- Hasselbach equation to get the pH value.

Formula used:
Henderson-Hasselbach equation states that-
 $\text{pH = p}{\text{K}}_{\text{a}}\text{ + log}\left({\displaystyle \frac{\text{[base]}}{\text{[acid]}}}\right)$
Where ${\text{K}}_{\text{a}}$ is the acid dissociation constant.
 $\text{p}{\text{K}}_{\text{a}}\text{ = - log}{\text{K}}_{\text{a}}$
[base]= concentration of base
[acid]= concentration of acid.

Complete step by step solution:
We know that buffer system is the system which contains acid or base or else it consists of conjugate acid and base.
According to Henderson-Hasselbach equation-
 $\text{pH = p}{\text{K}}_{\text{a}}\text{ + log}\left({\displaystyle \frac{\text{[base]}}{\text{[acid]}}}\right)$ .........................(1)
So, let us first find the value of ${\text{K}}_{\text{a}}$ .
We know that ${\text{K}}_{\text{a}}{\text{K}}_{\text{b}}\text{ = }{\text{K}}_{\text{w}}$ …………...(2)
Where ${\text{K}}_{\text{w}}$ is a constant and is equal to $\text{1}{\text{0}}^{\text{ - 14}}$
 ${\text{K}}_{\text{b}}$ for ammonia is equal to $\text{1}.8\times 1{\text{0}}^{\text{ - 5}}$
So, rearranging equation 2, we get:
 ${\text{K}}_{\text{a}}\text{ = }{\displaystyle \frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}}$
Substituting the values, we get:
 ${\text{K}}_{\text{a}}\text{ = }{\displaystyle \frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{b}}}}\text{ = }{\displaystyle \frac{\text{1}{\text{0}}^{\text{ - 14}}}{\text{1}.8\times 1{\text{0}}^{\text{ - 5}}}}\text{ = 5}.6\times 1{\text{0}}^{\text{ - 10}}$
Now $\text{p}{\text{K}}_{\text{a}}\text{ = - log}{\text{K}}_{\text{a}}\text{ = - log(5}.6\times 1{\text{0}}^{\text{ - 10}}\text{) = 9}\text{.26}$
The concentration of $\text{N}{\text{H}}_{\text{3}}$ (base) is given as $\text{0}\text{.15}$ M and the concentration of $\text{N}{\text{H}}_{\text{4}}\text{C}$ (acid) is $\text{0}\text{.35}$ M.
So, substituting the values in equation 1, we get:
 $\text{pH = p}{\text{K}}_{\text{a}}\text{ + log}\left({\displaystyle \frac{\text{[base]}}{\text{[acid]}}}\right)\text{ = 9}\text{.26 + log}\left({\displaystyle \frac{\text{0}\text{.15}}{\text{0}\text{.35}}}\right)\text{ = 8}\text{.89}$
Hence, pH of the buffer system is $\text{8}\text{.89}$

Note:
The $\text{p}{\text{K}}_{\text{a}}$ is the pH value at which the chemical species taking part in an reaction will accept or donate a proton.
The higher the value of $\text{p}{\text{K}}_{\text{a}}$ , low is the acid strength and hence, it has probably less ability to donate a proton in an aqueous solution.
  $\text{p}{\text{K}}_{\text{a}}$ and pH values are equal when half of the acid has been dissociated. The buffering capacity of a species or its ability to maintain pH of a solution is highest when the $\text{p}{\text{K}}_{\text{a}}$ and pH values are close.