Question

100ml of 0.1 M $NaOH$ is added to 100 ml of a 0.2 M \[C{{H}_{3}}COOH\] solution. The pH of the resulting solution will be (pKa=4.74).

Answer 1

Hint: A mixture of acetic acid and sodium acetate act as a buffer solution and since in this reaction sodium hydroxide is reacting with acetic acid producing sodium acetate.
Therefore, the overall solution is acting as a buffer solution as when acetic acid and sodium acetate are present in the same medium, they act as buffers in each other’s presence.

Complete step-by-step answer:
The overall reaction when $NaOH$ is reacting with \[C{{H}_{3}}COOH\] is shown as:
\[NaOH+C{{H}_{3}}COOH\to C{{H}_{3}}COONa+{{H}_{2}}O\]
The moles reacting of each species are given by:
${{n}_{O{{H}^{-}}}}=c\times v=0.1\times \dfrac{100}{1000}=0.01$
${{n}_{C{{H}_{3}}COOH}}=c\times v=0.2\times \dfrac{100}{1000}=0.02$

Therefore, moles of $NaOH$ are 0.01 moles and moles of \[C{{H}_{3}}COOH\] is 0.02 moles. It shows that they are added in 1:2 mole ratios in the equation. Therefore, no moles of \[C{{H}_{3}}COONa\] are formed of 0.01 moles. No of moles of \[C{{H}_{3}}COOH\] used will be 0.01 moles as 0.01 mole of reacts with 0.01 mole of \[C{{H}_{3}}COOH\] to give 0.01 mole of \[C{{H}_{3}}COONa\].
Since, it is a buffer solution, so we will apply Henderson HasselBalch equation written below:
$pH = p{{K}_{a}}+\log \dfrac{salt}{acid}$
$pH = 4.74+\log 1$
$pH = 4.74$
So, $4.74$ is the pH of the resulting solution. The solution is bit acidic as its pH is less than the 7.

Note: In the above problem the mixture is acting as a buffer solution. Buffer solution comprises weak acid and its conjugate base or weak base and its conjugate acid.
In the above question sodium acetate is a weak basic salt while acetic acid is a conjugate acid of it. So it acts as a good buffer for weak acidic solutions. This can be solved by the Henderson HasselBalch equation which is $pH=p{{K}_{a}}+\log \dfrac{salt}{acid}$.