Acid-base titration: Calculate pKa with only three values given There's an unkown acid, diluted with an unknown amount of water and titrated with $\ce{NaOH}$.
  
  
  After adding $\pu{10.00 mL}$ $\ce{NaOH}$, a $\mathrm{pH}$ value of $4.65$ is measured.  
  After adding another $\pu{12.22 mL}$ ($\pu{22.22 mL}$ $\ce{NaOH}$ in total), the equivalence point is reached.
  
  
  Calculate the $\mathrm{p}K_\mathrm{a}$ of the acid!


I'm currently facing this uncommon task regarding acid-base titration. There's quite few information in this text, only three values are given:

\begin{align}
V_1(\ce{NaOH}) &amp;= \pu{10.00 mL}: &amp;\quad &amp;\mathrm{pH = 4.65} \\
V_2(\ce{NaOH}) &amp;= \pu{22.22 mL}: &amp;\quad &amp;\text{equivalence point reached}
\end{align}

What I've got so far:

Before adding any $\ce{NaOH}$, the reaction should look like this:

$$\ce{HA + H2O &lt;=&gt; A- + H3O+}$$

I expect the neutralisation reaction to be:

$$\ce{A- + H3O+ + Na+ + OH- &lt;=&gt; NaA + 2 H2O}$$

In general $n = c V$. At the equivalence point $n(\ce{NaOH}) = n(\ce{HA})$, so:

$$C(\ce{NaOH}) \cdot V_2(\ce{NaOH}) = n(\ce{HA})$$

$$C(\ce{NaOH}) = \frac{n(\ce{HA})}{\pu{22.22 mL}}$$

Henderson-Hasselbalch:

\begin{align}
\mathrm{pH} &amp;= \mathrm{p}K_\mathrm{a} + \log{\frac{C(\ce{A-})}{C(\ce{HA})}} \\
\mathrm{p}K_\mathrm{a} &amp;= \mathrm{pH} - \log{\frac{C(\ce{A-})}{C(\ce{HA})}} \\
\mathrm{p}K_\mathrm{a} &amp;= 4.65 - \log{\frac{C(\ce{A-})}{C(\ce{HA})}}
\end{align}

I'm unsure how to express this equation as a function of $V_1(\ce{NaOH})$. I may be on the wrong track as well. Any help is appreciated!
First, we note that this is not a strong acid, since when it is only partially neutralized, the $\mathrm{pH}$ has already risen to almost $5$.

At that point, we know that $\frac{10}{22.2}$ of the total acid has been neutralized. Therefore, the ratio 

$$\frac{\ce{[A-]}}{\ce{[HA]}} = \frac{10}{22.2} \cdot \left(1 - \frac{10}{22.2}\right)^{-1} = \frac{1}{1.22}$$

Substitution into Henderson-Hasselbalch gives that the $\mathrm{p}K_\mathrm{a} = 4.73$.

Note that I assumed that the solution wasn't super dilute. If that were the case, I couldn't plug directly into the Henderson-Hasselbalch equation because the acid-base equilibrium would need to shift, but the change in the concentration of hydronium would be dominated by auto-ionization of water.

And in general, the total volume of solution does matter. You could consider a solution of hydrochloric acid that were dilute enough that even the first assumption of a weak acid were wrong. It's pretty easy to construct the solution such that the partially neutralized solution had a $\mathrm{pH}$ of $4.65$. I'll leave that as a follow-up exercise for you to explore.