Question

The equivalence point in a titration of \[40.0mL\] of a solution of a weak monoprotic acid occurs when \[35.0mL\] of a \[0.10M\] \[NaOH\] solution has been added. The pH of the solution is \[5.75\] after the addition of \[20.0mL\] of \[NaOH\] solution. What is the dissociation constant of the acid?

Answer 1

Hint: The titration of a weak acid and a strong base give rise to a buffer at the intermediate stage. The dissociation constant is easily determined with the help of Henderson equation for buffer solutions.

Complete step by step answer:
The given titration is the reaction of a weak acid and a strong base. This is an example of a neutralization reaction. The acid used is a weak monoprotic acid. The acid on dissociation gives only one proton. Thus it is represented as \[HA\].
The base used is sodium hydroxide which is indicated as \[NaOH\]. The neutralization reaction of the acid and the base generates a salt and water molecule. The reaction is written as
$HA + NaOH \to NaA + {H_2}O$
The salt produced contains the cation of the base (\[N{a^ + }\]) and the anion of the acid (\[{A^ - }\]). Also the reaction indicates that one mole of acid reacts with one mole of base to produce one mole of salt and one mole of water.
Given that the equivalence point is reached when \[35.0mL\] of \[0.10M\] \[NaOH\] reacts with \[40.0mL\] of \[HA\]. Also at equivalence point the titration is complete and the whole solution is neutralized. At this point the mole of base is equal to the mole of acid.
Thus mole of base =$0.10M \times 35mL = 3.5mmoles$.
So the moles of acid =\[3.5mmoles\].
Also, given that the pH of the solution is \[5.75\] when \[20.0mL\] of \[NaOH\] is added. Thus the mole of \[NaOH\] is =$0.1M \times 20mL = 2mmoles.$
Then when \[20.0mL\]of \[NaOH\] is added the \[2mmoles\] of base reacts with \[2mmoles\] of acid. But actually we have \[3.5mmoles\] of acid present. So the solution contains an excess moles of acid = $3.5 - 2 = 1.5mmoles.$
Also the solution contains a salt \[NaA\] which has the concentration of \[2mmoles\]. So the solution is regarded as an acid buffer containing an acid and its salt. Applying Henderson equation the dissociation constant is determined as
$[H + ] = \dfrac{{Ka[acid]}}{{[salt]}}$
$pH = 5.75,$ so $pH = - \log [{H^ + }]$
$[{H^ + }] = {10^{5.75}}$
Inserting in the Henderson equation,
${10^{5.75}} = \dfrac{{Ka \times 1.5mmoles}}{{2mmoles}}$
$Ka = \dfrac{{2 \times {{10}^{5.75}}}}{{1.5}}$
$Ka = 2.37 \times {10^{ - 6}}$.

Note: The dissociation constant measures the strength of an acid. The strength is the ability to produce hydrogen ions in a solution. The pH of a buffer solution is easily determined with the help of Henderson equation and the dissociation constant of the acid or base.