Question

$100mL$ of $0.02M$ benzoic acid $\left( p{{K}_{a}}=4.2 \right)$ is titrated using $0.02M$ $NaOH.$ pH after $50mi$ , and $100mL$ of $NaOH$ have been added are
(A) $3.50,7$
(B) $4.2,7$
(C) $4.2,8.1$
(D) $4.2,8.25$

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 in each other’s presence.

Complete Step By Step Answer:
The overall reaction when benzoic acid is reacting with sodium hydroxide is shown as:
$C6H5COO{{H}_{s}}+NaOH\to C6H5COONa_{_{aq}}^{+}+{{H}_{2}}{{O}_{aq}}$
Therefore, moles of are $0.02$ moles and moles of is $0.02$ moles. It shows that they are added in $1:1$ mole ratios in the equation. Therefore, no moles are formed of moles. No of moles of used will be $0.02$ moles as $0.02$ mole of reacts with $0.02$ mole of to give $0.02$ mole. Since, it is a buffer solution, so we will apply Henderson HasselBalch equation written below:
$pH=p{{K}_{a}}+\log \dfrac{salt}{acid}$
$pH=4.2+\log \left( \dfrac{0.001}{0.001} \right)$
$\Rightarrow pH=4.2$
Thus now we have $pH=7+\dfrac{1}{2}\left[ pKa+\log C \right]$
$=7+\dfrac{1}{2}\left[ 4.2+\log 0.01 \right]$
On further simplifying;
$=7+\dfrac{1}{2}\left[ \dfrac{4.2}{2} \right]$
$=7+\dfrac{1}{2}\left[ 2.2 \right]$
$\Rightarrow 8.1$
Therefore the correct answer is option C.

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.