0.1MNaOH is titrated with 0.1 M, 20 ml HA till the end point, K _ a...

$\mathop{NaOH}_{2} +\mathop{HA}_{2} \toNa A + {H}_{2}O$
At end point $≡ 0.1 \times 20 = 2$
$∵ 20\text{mL}\text{NaOH}$ 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
$\text{C}\text{=}\left[\text{NaA}\right]\text{=}\dfrac{2}{\text{20}\text{+}\text{20}} \text{=}\text{0}\text{.05 M}$
And {\text{pK}_{a} \text{=} - \text\log\left(\text{6}\text{\times}{\text{10}^{- \text{6}}\right)\text{=}\text{5}\text{.2}
$\text{pH}$ at the end point $\text{=}\text{7}\text{+}$\dfrac{1}{2} \left({\text{pK}_{a} \text{+}\text\log\text{C}\right)$<br />$\text{7}\text{+}\dfrac{1}{2} \left(\text{5}\text{.2}\text{+}\text\log\text{0}\text{.05}\right)\text{=}\text{8}\text{.95}