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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?

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  • $\begingroup$ 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? $\endgroup$ Feb 28 at 18:48
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    $\begingroup$ 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. $\endgroup$ Feb 28 at 18:55
  • $\begingroup$ Did you discuss this with your faculty advisor or team coach? $\endgroup$ Feb 28 at 18:56
  • $\begingroup$ 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. $\endgroup$
    – sfs2007
    Feb 28 at 18:59
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    $\begingroup$ 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? $\endgroup$
    – sfs2007
    Feb 28 at 23:47

1 Answer 1

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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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  • $\begingroup$ 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. $\endgroup$ Mar 1 at 1:13
  • $\begingroup$ You solved it without having to make an assumption about the concentration of NaOH. That shows a nice ability to reason in the abstract. $\endgroup$ Mar 1 at 1:21
  • $\begingroup$ Great! Thank you so much for all of your help! $\endgroup$
    – sfs2007
    Mar 1 at 4:06

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