Diluting a sample of weak base before titration. What happens to equivalence point pH of diluted vs undiluted? A student titrated a 10.00 mL of 0.20 M methylamine ($\ce{CH3NH2}$) and determined that 10.00 mL of $\ce{HCl}$ was needed to reach the equivalence point.
If you diluted your original sample of methylamine by adding 10.00 mL of water to your sample before you started adding $\ce{HCl}$, which of the following statements would be true?
a) You now only need to add 5.00 mL of $\ce{HCl}$ to reach equivalence.
b) The equivalence point pH would be the same as in the undiluted titration.
c) Both (a) and (b) are true.
d) Neither (a) nor (b) are true.

The answer is d) Neither (a) nor (b) are true.
I always thought that both the $\mathrm{p}K_\mathrm{a}$ and equivalence $\mathrm{pH}$ of an acid stay the same no matter what the concentration. Except where it is so dilute, ionization from water would have an effect. But I don't understand why or how it would change. I can't find anything in the textbook or online, can someone please explain this?
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

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