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Q.
A weak acid of dissociation constant $10^{-5}$ is being titrated with aqueous $NaOH$ solution. The $pH$ at the point of one-third neutralisation of the acid will be
$K_{a}=10^{-5} \Rightarrow p K_{a}=-\log K_{a}=-\log 10^{-5}=5$
(Assumed weak acid to be monoprotic, since only one dissociation constant value is provided.)
Final solution acts as an acidic buffer.
$pH = p K_{a}+\log \frac{[\text { salt }]}{[\text { acid] }}$
or $pH =5+\log \frac{\frac{1}{3}}{\frac{2}{3}}$
$=5+\log \frac{1}{2}$
$\therefore pH =5-\log 2$
