Acid-base titration: Calculate pKa with only three values given


> 
> There's an unkown acid, diluted with an unknown amount of water and titrated with $\ce{NaOH}$.
> 
> 
> 1. After adding $\pu{10.00 mL}$ $\ce{NaOH}$, a $\mathrm{pH}$ value of $4.65$ is measured.
> 2. After adding another $\pu{12.22 mL}$ ($\pu{22.22 mL}$ $\ce{NaOH}$ in total), the equivalence point is reached.
> 
> 
> Calculate the $\mathrm{p}K\_\mathrm{a}$ of the acid!
> 
> 
> 


I'm currently facing this uncommon task regarding acid-base titration. There's quite few information in this text, only three values are given:


\begin{align}
V\_1(\ce{NaOH}) &= \pu{10.00 mL}: &\quad &\mathrm{pH = 4.65} \\
V\_2(\ce{NaOH}) &= \pu{22.22 mL}: &\quad &\text{equivalence point reached}
\end{align}


What I've got so far:


Before adding any $\ce{NaOH}$, the reaction should look like this:


$$\ce{HA + H2O <=> A- + H3O+}$$


I expect the neutralisation reaction to be:


$$\ce{A- + H3O+ + Na+ + OH- <=> NaA + 2 H2O}$$


In general $n = c V$. At the equivalence point $n(\ce{NaOH}) = n(\ce{HA})$, so:


$$C(\ce{NaOH}) \cdot V\_2(\ce{NaOH}) = n(\ce{HA})$$


$$C(\ce{NaOH}) = \frac{n(\ce{HA})}{\pu{22.22 mL}}$$


Henderson-Hasselbalch:


\begin{align}
\mathrm{pH} &= \mathrm{p}K\_\mathrm{a} + \log{\frac{C(\ce{A-})}{C(\ce{HA})}} \\
\mathrm{p}K\_\mathrm{a} &= \mathrm{pH} - \log{\frac{C(\ce{A-})}{C(\ce{HA})}} \\
\mathrm{p}K\_\mathrm{a} &= 4.65 - \log{\frac{C(\ce{A-})}{C(\ce{HA})}}
\end{align}


I'm unsure how to express this equation as a function of $V\_1(\ce{NaOH})$. I may be on the wrong track as well. Any help is appreciated!




You've done quite well there. You correctly calculated the number of moles of acid in the original solution as:


$$n(\ce{HA}) = C(\ce{NaOH}) \times V\_2$$


After the first addition of base, the number of moles of conjugated base is:


$$n(\ce{A}) = C(\ce{NaOH}) \times V\_1$$


And by stoichiometry, the remaining number of moles of acid:


\begin{align}
n'(\ce{HA}) &= n(\ce{HA}) - C(\ce{NaOH}) \times V\_1 \\ 
&= C(\ce{NaOH}) \times V\_2 -C(\ce{NaOH}) \times V\_1 \\
&= C(\ce{NaOH})(V\_2 - V\_1)
\end{align}


Because the volumes the remaning acid and conjugated base are contained are the same:


$$\mathrm{p}K\_\mathrm{a} = 4.65 - \log{\frac{C(\ce{A-})}{C(\ce{HA})}} = 4.65 - \log{\frac{n(\ce{A})}{n'(\ce{HA})}}$$


The $C(\ce{NaOH})$ drop out and you have your $\mathrm{p}K\_\mathrm{a}$:


$$\mathrm{p}K\_\mathrm{a} = 4.65 - \log{\frac{V\_1}{V\_2 - V\_1}}$$


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First, we note that this is not a strong acid, since when it is only partially neutralized, the $\mathrm{pH}$ has already risen to almost $5$.


At that point, we know that $\frac{10}{22.2}$ of the total acid has been neutralized. Therefore, the ratio 


$$\frac{\ce{[A-]}}{\ce{[HA]}} = \frac{10}{22.2} \cdot \left(1 - \frac{10}{22.2}\right)^{-1} = \frac{1}{1.22}$$


Substitution into Henderson-Hasselbalch gives that the $\mathrm{p}K\_\mathrm{a} = 4.73$.


Note that I assumed that the solution wasn't super dilute. If that were the case, I couldn't plug directly into the Henderson-Hasselbalch equation because the acid-base equilibrium would need to shift, but the change in the concentration of hydronium would be dominated by auto-ionization of water.


And in general, the total volume of solution does matter. You could consider a solution of hydrochloric acid that were dilute enough that even the first assumption of a weak acid were wrong. It's pretty easy to construct the solution such that the partially neutralized solution had a $\mathrm{pH}$ of $4.65$. I'll leave that as a follow-up exercise for you to explore.


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