Question

The pH of a buffer solution prepared by mixing 50ml of 0.2M $C{{H}_{3}}COOH$and 25 ml of $C{{H}_{3}}COONa$is 4.8. What is the concentration of $C{{H}_{3}}COONa$? [pKa of $C{{H}_{3}}COOH$= 4.8]
(A) 0.2M
(B) 0.4M
(C) 0.5M
(D)0.8M

Answer 1

The Henderson- Hasselbalch equation can be given as-
$pH=p{{K}_{a}}+{{\log }_{10}}\left( \frac{[{{A}^{- }}]}{[HA]} \right)$
where $[{{A}^{- }}]$ is the molar concentration of the conjugate base (of the acid)
[HA] denotes the molar concentration of the weak acid
Therefore, the Henderson- Hasselbalch equation can also be given as:
$pH=p{{K}_{a}}+\log \frac{\text{ }\\!\\![\\!\\!\text{ conjugate base }\\!\\!]\\!\\!\text{ }}{[\text{acid }\\!\\!]\\!\\!\text{ }}$

Complete Step by step answer:
- A buffer is a solution which is a mixture of a weak acid and its conjugate base or vice versa and it can resist the change in pH upon the acidic of acidic or basic solutions up to a certain limit.
- Buffers are broadly classified into two types- Acidic buffer solutions or Alkaline buffer solutions.
- Acidic buffer solutions are the buffers that have a pH below 7 and contains a weak acid and one of its salt. Alkaline buffer solutions are the buffers which have a pH above 7 and contains a weak base and one of its salts.
- Here acetic acid is taken, which is an acidic buffer.
- The Henderson- Hasselbalch equation provides a relationship between the pH of acids (in aqueous solutions) and their pKa (acid dissociation constant) values. The pH of a buffer solution can be determined with the help of the following equation when the concentration of the acid and its conjugate base, or the base and its conjugate acid is known.
- Since we know that concentration is actually $\frac{\text{Number of moles}}{Volume}$
Therefore, the Handerson- Hasselbalch equation can be written as- $pH=p{{K}_{a}}+\log \frac{\text{moles of conjugate base}}{\text{moles of acid}}$
- According to question, pH = 4.8; pKa = 4.8
Calculating the moles of the conjugate base $=Concentration(C)\times Volume=C\times 25\times {{10}^{- 3}}l$
Calculating the moles of acid $=0.2\times 50\times {{10}^{- 3}}l$
- Inserting the values in the equation of Handerson- Hasselbalch equation, we will get
$pH=p{{K}_{a}}+\log \frac{\text{ }\\!\\![\\!\\!\text{ conjugate base }\\!\\!]\\!\\!\text{ }}{[\text{acid }\\!\\!]\\!\\!\text{ }}$
$\Rightarrow 4.8=4.8+\log \frac{C\times 25\times {{10}^{- 3}}}{0.2\times 50\times {{10}^{- 3}}}$
$\Rightarrow C=0.4M$
So, the correct answer is option B.

Note: The Henderson- Hasselbalch equation had some limitations too. It failed to predict accurate values for the strong acids and strong bases as it assumed that the concentration of the acid and its conjugate at chemical equilibrium will remain the same as the formal concentration, which neglected the binding of protons to the base. Another limitation was that it does not consider the self dissociation undergone by water and thus failed to offer accurate pH values for extremely dilute buffer solutions.