Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\underset{2}{\mathrm{NaOH}}+\underset{2}{\mathrm{HA}}\to \underset{}{\mathrm{Na}}A+\underset{}{{\mathrm{H}}_{2}O}$
At end point$\equiv 0.1\times 20=2$
$\because 20 \text{mL}\text{NaOH}$is required for the complete neutralization of$\text{HA}$
$\text{NaA}$is a salt of strong base and weak acid
Thus, will undergo hydrolysis and solution will becomes basic
$\text{C}\text{=}\left[\text{NaA}\right]\text{=}\frac{\text{2}}{\text{20}\text{+}\text{20}}\text{=}\text{0}\text{.05 M}$
And${\text{pK}}_{\text{a}}\text{=}-\text{log}\left(\text{6}\text{×}{\text{10}}^{-\text{6}}\right)\text{=}\text{5}\text{.2}$
$\text{pH}$at the end point$\text{=}\text{7}\text{+}\frac{\text{1}}{\text{2}}\left({\text{pK}}_{\text{a}}\text{+}\text{log}\text{C}\right)$
$\text{7}\text{+}\frac{\text{1}}{\text{2}}\left(\text{5}\text{.2}\text{+}\text{log}\text{0}\text{.05}\right)\text{=}\text{8}\text{.95}$