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}) &= \pu{10.00 mL}: &\quad &\mathrm{pH = 4.65} \\ V_2(\ce{NaOH}) &= \pu{22.22 mL}: &\quad &\text{equivalence point reached} \end{align}
What I've got so far:
Before adding any $\ce{NaOH}$, the reaction should look like this:
$$\ce{HA + H2O <=> A- + H3O+}$$
I expect the neutralisation reaction to be:
$$\ce{A- + H3O+ + Na+ + OH- <=> NaA + 2 H2O}$$
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} &= \mathrm{p}K_\mathrm{a} + \log{\frac{C(\ce{A-})}{C(\ce{HA})}} \\ \mathrm{p}K_\mathrm{a} &= \mathrm{pH} - \log{\frac{C(\ce{A-})}{C(\ce{HA})}} \\ \mathrm{p}K_\mathrm{a} &= 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!