Finding Ka of an Acid from incomplete titration data

I'm studying for chem olympiad and have a question about a problem from a past year's test:


A pure sample of a monoprotic acid is dissolved in water.
The sample is titrated with sodium hydroxide solution.
At the point where $20.0$ mL of the NaOH solution has
been added, the pH is $4.15$. The phenolphthalein
endpoint of the titration is observed when $50.0$ mL of
NaOH have been added. What is the p$K\_{\mathrm{a}}$ of the acid?


Here are my thoughts:


Since pH = $4.15$, pOH = $9.85$ $\implies$ $\pu{[OH^-] = 10^{-9.85}M}$, which I think must also be the molarity of NaOH (this might be wrong).


I don't understand otherwise how to solve this question. I have $K\_a = \frac{ \mathrm{[H^+]^2}}{\mathrm{[HA]}}$ as concentration of protons equals concentration of conjugate base. I guess I have $\pu{[H^+] = 10^{-4.15}}$ from the pH but I'm not sure if that's right since we added $20$ mL of NaOH first, which would have changed the pH from what it originally was.


How would I solve the question?




The formula for $K\_\mathrm{a}$ is : $$K\_\mathrm{a} = [\ce{H+}] \frac{ [\ce{A^-}]}{[\ce{HA}]}.$$


We'll use the data from after 20mL of titration: the pH is 4.15, so $[\ce{H^+}]$ is $10^{-4.15}.$


Now, we need the fraction of concentrations. In the titration at 20mL, $\ce{NaOH}$ is clearly limiting, and at the titration at 50mL, the entire acid gets used up. This means that at 20mL, 2/5 of the acid is used (and thus, 3/5 is left).


the formula for titration is: $$\ce{HA + NaOH -> H\_2O + Na^+ + A^-}$$ since $\ce{NaA}$ is always soluble.


At 20mL titration, the concentration of $\ce{A^-}$ is 2/5 the original concentration of acid and the concentration of acid is 3/5 the original. So, the fraction is 2/3, which means we get: $$K\_\mathrm{a} = 10^{-4.15} \cdot \frac{2}{3}.$$


Thus, $$\mathrm{p}K\_\mathrm{a} = -\log\_{10} \left(10^{-4.15} \cdot \frac{2}{3} \right) = 4.33.$$


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