Find second acid dissociation constant at a given pH 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 $\ce{-&gt;} 0.012\ \mathrm{mol}\ [\ce{H+}]$
$0.01\ \mathrm{mol}$ of molecule in solution $\ce{-&gt;} 0.02\ \mathrm{mol}$ 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 $\ce{H+ -&gt;} 0.02-0.012$. 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 $\ce{-&gt;} \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log ([\ce{A-}]/[\ce{HA}]) \ce{-&gt;}$
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:

$$\ce{H2A+ &lt;=&gt;[H+] HA &lt;=&gt;[H+] A-}$$

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.

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


we started with $10~\mathrm{mmol}\ \ce{HA}$. We ended up with $10~\mathrm{mmol}\ \ce{H2A+}$ and $2~\mathrm{mmol}\ \ce{HCl}$
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 $\mathrm{pH} &lt; 3$. Thus, we are well within the operating scheme of $\mathrm{p}K_\mathrm{a,1}$ and we can easily use the Henderson–Hasselbalch equation:

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



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.