Question

When a solution of benzoic acid was titrated with \[NaOH\], the pH of the solution when half of the acid neutralized was \[4.2\]. Dissociation constant of the acid is:
A.\[6.31 \times {10^{ - 5}}\]
B.\[3.2 \times {10^{ - 5}}\]
C.\[8.7 \times {10^{ - 5}}\]
D.\[6.42 \times {10^{ - 4}}\]

Answer 1

Hint: Benzoic acid is a weak acid and \[NaOH\] is a strong base which when reacts, then a solution of weak acid and its salt is formed which is termed as acidic buffer. We need to use the relation between the pH and dissociation constant to calculate the desired values.

Complete answer:
When benzoic acid reacts with sodium hydroxide, then the formation of salt and water takes place and the reaction is known as neutralization reaction. The solution formed is mildly basic in nature whereas the medium of the solution is termed as acidic buffer.
The reaction proceeds as follows:
\[{C_6}{H_5}COOH + NaOH \to {C_6}{H_5}COONa + {H_2}O\]
For acidic buffers, the expression for pH of the solution can be written as:
\[pH = p{K_a} + \log \dfrac{{[salt]}}{{[acid]}}\,\, - (i)\]
Where, \[{K_a}\] is the dissociation constant, \[[salt]\] is the concentration of salt formed and \[[acid]\] is the concentration of acid reacted.
As per reaction, one mole of acid is titrated to form one mole of salt. But, according to the data given in question, pH value is given when half of the acid is neutralized. So,
moles of acid reacted \[ = 0.5\]
moles of salt formed \[ = 0.5\]
Therefore, we can consider that \[[salt] = [acid] = c\]
Substituting values in the equation \[(i)\]
\[4.2 = p{K_a} + \log \dfrac{c}{c}\]
\[ \Rightarrow p{K_a} = 4.2 - \log 1\]
\[ \Rightarrow {K_a} = {10^{ - 4.2}}\]
\[ \Rightarrow {K_a} = 6.3 \times {10^{ - 5}}\]
Hence the value of dissociation constant of the acid is \[6.3 \times {10^{ - 5}}\].
So, option (A) is the correct answer.

Note:
A basic buffer solution is a mixture of weak base and its salt in the presence of a strong acid. A mixture of Ammonium hydroxide and ammonium chloride is an example of a basic buffer. The expression for the relation of pOH and base dissociation constant is as follows:
\[pOH = p{K_b} + \log \dfrac{{[salt]}}{{[base]}}\]