Question

A weak acid of dissociation constant ${10^{ - 5}}$ is being titrated with aqueous NaOH solution. The pH at the point of one-third neutralisation of the acid will be:
A. 5 + log 2 — log 3
B. 5 — log 2
C. 5 — log 3
D. 5 — log 6

Answer 1

Hint- In this question we have to find the pH of the acid base solution when the acid has been neutralized to one third of its initial concentration. It is given that the acid is weak, therefore, the solution will form an acidic buffer on one third neutralization. We will use the Henderson-Hasselbalch Equation to find the required answer.

Complete step-by-step answer:
Buffer solutions are solutions containing a weak acid or base and along with its respective salt. The pH values of these solutions do not change much when a little bit of acid or base is added.
Given, dissociation constant of acid ${K_a} = {10^{ - 5}}$
$
  p{K_a} = - \log ({K_a}) \\
  p{K_a} = - \log ({10^{ - 5}}) \\
  p{K_a} = 5 \\
$
Let the initial concentration of weak acid be x. Then, at one third neutralization:
Conc. of acid= $\dfrac{2}{3}x$
Conc. of salt= $\dfrac{1}{3}x$ (Acid+Base will form the respective salt)
To solve this question we will be using the Henderson-Hasselbalch Equation:
\[pH = p{K_a} + \log \dfrac{{[salt]}}{{[acid]}}\]
Putting the values in the given equation:
\[
  pH = 5 + \log (\dfrac{{\dfrac{1}{3}x}}{{\dfrac{2}{3}x}}) = 5 + \log \dfrac{1}{2} \\
  pH = 5 - \log 2 \\
\]
The pH of the given solution at one third neutralization is pH= 5- log2.
Hence the correct option is B.

Note- Titration is a chemical technique where a solution of known concentration is used to determine the concentration of an unknown solution. The known solution is taken in a flask and the unknown solution is titrated or added dropwise into it. The volume of the unknown solution added is used to find the concentration.