Question

A certain buffer solution contains an equal concentration of ${{X}^{-}}$and HX. The ${{K}_{b}}$for${{X}^{-}}$is ${{10}^{-10}}$. The pH of the buffer is :
A.4
B.7
C.10
D.14

Answer 1

Before solving this question, we should first know about the Henderson equation: This equation gives a relationship of pH of acids that are in aqueous solutions and their $p{{K}_{a}}$ i.e Acid dissociation constant. This equation helps in the estimation of the pH of a buffer solution if the concentration of base and its corresponding conjugate acid is known or concentration of acid and its corresponding conjugate base is known.

Complete answer:
Henderson-Hasselbalch Equation-
We can write this equation as-
$pH=p{{K}_{a}}+{{\log }_{10}}\dfrac{({{A}^{-}})}{(H{{A}^{-}})}$
When 50% of the acid already faces dissociation, Then the value $\dfrac{(A)}{(HA)}$becomes unity i.e 1 which tells that the$p{{K}_{a}}$of acid = pH of the solution at this point. (pH=p{{K}_{a}}+\log 1=pK{{ & }_{a}}).
Whenever there is a change in the ratio of pH and$p{{K}_{a}}$ by 1, there would be a change in the ratio of associated acid to the dissociated acid by tenfold.
$pH\,and\,p{{K}_{a}}$are the factors responsible for the value of $\dfrac{(A)}{(HA)}$.
If $pH\,$pH,>p{{K}{a}}$, the value of$\dfrac{(A)}{(HA)}>1$We know,${{K}{b}}$=${{10}^{-10}}$${{K}{a}}\times {{K}{b}}={{10}^{-14}}$Therefore, we will put the value of${{K}{b}}$in the equation-${{K}{a}}\times {{10}^{-10}}={{10}^{-14}}$${{K}{a}}={{10}^{-4}}$$pH=-\log {{K}{a}}+{{\log }_{10}}\dfrac{salt}{acid}$=$-\log {{10}^{-4}}+\log \dfrac{1.0}{1.0}$
= 4

So, Option (A) 4 is correct.

Note:
Limitations of the Henderson Equation are-
We don’t get the accurate values of strong acid and strong base by the Henderson equation because it supposes the value of the concentration of the acid and its conjugate base will remain as it is as the formal concentration.
It also fails to deliver accurate values of the pH of very dilute buffer solutions.