# 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? Feb 28 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. Feb 28 at 18:55
• Did you discuss this with your faculty advisor or team coach? Feb 28 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. Feb 28 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? Feb 28 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. Mar 1 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. Mar 1 at 1:21
• Great! Thank you so much for all of your help! Mar 1 at 4:06