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Q.
$100 \,mL$ of $0.02 \,M$ benzoic acid $\left( p K_{a}=4.2\right)$ is titrated using $0.02\, M \,NaOH$. $pH$ after $50\, mL$ and $100\, mL$ of $NaOH$ have been added are
Equilibrium
Solution:
When $50\, mL$ of $0.02 \,M \,NaOH$ is added inside, we will have [salt] = [acid] at half equivalence point,
$pH = p K_{a}+\log \frac{[\text { Salt }]}{[\text { Acid }]}=4.2$
When $100\, mL$ of $NaOH$ is added we will have equivalence point so $pH$ will be calculated according to hydrolysis of salt of weak acid and strong base.
$pH =\frac{1}{2}\left( p K_{w}+ p K_{a}+\log C\right)$