pKa and equivalence point
$99.9\%$ of the acid $\ce{HA}$ is titrated at  $\mathrm{pH}=7.9$ resp.$ 7.5$.     

$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+\log \frac{[\ce{A-}]}{[\ce{HA}]}=\mathrm{p}K_\mathrm{a}+3$$

Considering the final concentration 0.05M,
titrating 0.1M acid by 0.1M hydroxide,
100% of the acid is titrated at:

$$\begin{align}
\mathrm{pH}&amp;=7+\frac12 \cdot ( \mathrm{p}K_\mathrm{a} + \log  c_\mathrm{acid, total}) \\
&amp;=9.45(9.25) + \frac12 \cdot (-1.3) \\
&amp;=8.8(8.6)\\
\end{align}$$

Titration with $100.1\%$ of used hydroxide we can consider
as $0.05 \cdot 0.001=5\cdot 10^{-5}\ \mathrm{M} \ \ce{NaOH}$
with $\mathrm{pH}=-\log ( 5\cdot 10^{-5})=9.7$

The indicator  phenolphthalein with range $8.2-10.0$ works well in both cases.