Diluting a sample of weak base before titration. What happens to equivalence point pH of diluted vs undiluted?
At the point of equivalence, when the added molar amount of $\ce{HCl}$ is just matching the present molar amount of $\ce{CH3NH2}$, there is solution of the weak acid $\ce{CH3NH3+}(+\ce{Cl-})$, that I will denote as $\ce{HA}$.

If we neglect $\ce{H+}$ from water dissociation and consider $c_\ce{A-}\ll c_\ce{HA}$, then $c_\ce{A-} = c_\ce{H+}$
Therefore

$$K_\mathrm{a} = \frac{c_\ce{H+} c_\ce{A-}}{c_\ce{HA}} = \frac{c_\ce{H+} c_\ce{H+}}{c_\ce{HA} + c_\ce{A-}}$$

$$\begin{align}
\mathrm{pH} &amp;= \frac12(\mathrm{p}K_\mathrm{a} - \log c) \\
  &amp;= \frac12(14 - \mathrm{p}K_\mathrm{b} - \log c) \\
  &amp;= 7 - \frac12(\mathrm{p}K_\mathrm{b} + \log c) \\
\end{align}$$

where $\mathrm{p}K_\mathrm{b}$ is the methylamine basicity constant;
$c$ is summary concentration of methylamine and it's a conjugate acid.

Therefore, if there is final volume 30 instead of 20 ml, the final

$$\Delta\mathrm{pH} = -0.5\log \left(\frac{\pu{20 ml}}{\pu{30 ml}}\right) = +0.088$$