In short, the problem states that:

  • We have a molecule that has two ionizable groups
  • $\mathrm{p}K_\mathrm{a,1}$ is between $5.0$ and $9.0$; $\mathrm{p}K_\mathrm{a,2}$ is outside of this range.
  • $100\ \mathrm{mL}$ of a $0.1\ \mathrm{M}$ solution of molecule, all in the unprotonated form
  • added $12\ \mathrm{mL}$ of $1\ \mathrm{M}\ \ce{HCl}$
  • final $\mathrm{pH}$ is $7.4$

The goal is to find $\mathrm{p}K_\mathrm{a,1}$. This is how I've approached it:

  • First, I converted everything to moles:

    $0.012\ \mathrm{mol}$ of $\ce{HCl}$ added $\ce{->} 0.012\ \mathrm{mol}\ [\ce{H+}]$

  • $0.01\ \mathrm{mol}$ of molecule in solution $\ce{->} 0.02\ \mathrm{mol}$ of $[\ce{A-}]$ since the problem states that everything is in unprotonated form I think this is were I am wrong in my thinking
  • Next, I reason that some $\ce{A-}$ is going to be neutralized by $\ce{H+ ->} 0.02-0.012$. Leaves me with $0.008\ \mathrm{mol}$ of $\ce{A-}$
  • Then, I have $0.012~\mathrm{mol}$ of $\ce{HA}$ that was formed.
  • Im plugging this in to the Henderson-Hasselbalch equation $\ce{->} \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log ([\ce{A-}]/[\ce{HA}]) \ce{->}$
  • leads to the incorrect answer.

I've tried looking at similar problems but I am just having a hard time with this. I feel that the problem lies with this the neutralization. Also, I know that upper boundary of $\mathrm{p}K_\mathrm{a,1}$ is $9$, which is above the final $\mathrm{pH}$, and I know that at $\mathrm{pH}=\mathrm{p}K_\mathrm{a} 50~\%$ would be protonated/unprotonated, however, I don't know how to account for this information in my calculation.


For starters, I’m going to define your molecule as being $\ce{HA}$ and its two ionised forms to be $\ce{A-}$ and $\ce{H2A+}$. This means, we are observing the following equilibrium:

$$\ce{H2A+ <=>[H+] HA <=>[H+] A-}$$

As you correctly calculated, you have $10~\mathrm{mmol}$ of $\ce{H_nA^m+}$ and $12~\mathrm{mmol}\ \ce{HCl}$ that you add to your solution.

$12~\mathrm{mmol} > 10~\mathrm{mmol}$. Thus, there are two possibilities:

  1. we started with $10~\mathrm{mmol}\ \ce{HA}$. We ended up with $10~\mathrm{mmol}\ \ce{H2A+}$ and $2~\mathrm{mmol}\ \ce{HCl}$

  2. we started with $10~\mathrm{mmol}\ \ce{A-}$. We ended up with $8~\mathrm{mmol}\ \ce{HA}$ and $2~\mathrm{mmol}\ \ce{H2A+}$

$\mathrm{pH}~7.4$ is only consistent with possibility 2; in case of 1 it would mean $2~\mathrm{mmol}\ \ce{HCl}$ controlling the $\mathrm{pH}$ which would give us $\mathrm{pH} < 3$. Thus, we are well within the operating scheme of $\mathrm{p}K_\mathrm{a,1}$ and we can easily use the Henderson–Hasselbalch equation:

$$\begin{align}\mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{HA}]}{[\ce{H2A+}]}\\ 7.4 &= \mathrm{p}K_\mathrm{a} + \log \frac{\frac{8~\mathrm{mmol}}{V}}{\frac{2~\mathrm{mmol}}{V}}\\ 7.4 &= \mathrm{p}K_\mathrm{a} + \log 4\\ 7.4 &= \mathrm{p}K_\mathrm{a} + 0.60\\ 6.8 &= \mathrm{p}K_\mathrm{a}\end{align}$$

You did indeed identify the spot where you went wrong. Just because a compound is dibasic does not mean you should double its concentration. As I showed above, $10~\mathrm{mmol}$ of original molecule, whether it was entered as $\ce{H2A+}, \ce{HA}$ or $\ce{A-}$ will remain $10~\mathrm{mmol}$; only the number of attached protons will differ.

  • $\begingroup$ Thanks. This reinforces the first answer. I've managed to work my way through similar problems but still in need of more practice. $\endgroup$
    – J Vazquez
    Oct 3 '16 at 1:34
  • $\begingroup$ @JVazquez You know when the teacher/professor tells you to write out the equation, and no one does it because it's an extra step? Try writing out the equations, at least until you have enough practice that you can see the stoichiometric relationships in your head. You probably could have avoided your mistake if you had written out the equation. $\endgroup$
    – Zhe
    Oct 4 '16 at 1:31

I think you did go wrong in that step. Why are there 0.02 moles of A?

You should consider writing out your mystery substance as H2A. Then the conjugate base is HA. You don't really care about the second ionization, but I think the diprotic nature of this acid is causing you confusion. You only create one mole of the monoprotic form from one mole of the diprotic form. There's no extra 2 in there.


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