# 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?

• What other formulas related to acids and bases do you know? Can you make a graph of a titration of a weak acid with a strong base when all concentrations and the pKa is known?
– Karsten
Commented Feb 28, 2022 at 18:48
• The answer is the same no matter what the concentration of the NaOH solution is. Does it help if I say the concentration is 0.1 mol/L? Once you have an answer, you can also try with 0.2 mol/L and check you get the same answer. Then, you can think about why the concentration of NaOH does not matter in terms of what the solution to the question is.
– Karsten
Commented Feb 28, 2022 at 18:55
• Did you discuss this with your faculty advisor or team coach?
– Karsten
Commented Feb 28, 2022 at 18:56
• The main formulas I know are that pH + pOH = 14, pKa + pKb = 14. I also understand how to calculate each of these quantities... At my school, there isn't really an advisor or a coach. They select the students w/ the top 5 grades and we are supposed to self study the material, so I don't have someone to go to. I'll try using 0.1 mol/L and will update if I make progress/get stuck. Commented Feb 28, 2022 at 18:59
• Thank you so much! I have a solution that I think is relatively similar to Poutnik's. I put it as an answer: how does it look? Commented Feb 28, 2022 at 23:47

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, $$\mathrm{p}K_\mathrm{a} = -\log_{10} \left(10^{-4.15} \cdot \frac{2}{3} \right) = 4.33.$$

• This method would work for all points in the titration where weak acid and conjugate base are major species, i.e. from 2 mL in to 48 mL or so.
– Karsten
Commented Mar 1, 2022 at 1:13
• You solved it without having to make an assumption about the concentration of NaOH. That shows a nice ability to reason in the abstract.
– Karsten
Commented Mar 1, 2022 at 1:21
• Great! Thank you so much for all of your help! Commented Mar 1, 2022 at 4:06