# 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!

• Who marked this off-topic as homework? There is clearly an attempt to solve the poblem...
– Zhe
Dec 8 '17 at 20:05
• @Zhe It's not a mandatory homework but a voluntary exercise. My fellow students also had problems with it, so I decided to ask for help here. Dec 9 '17 at 9:36
• That's not what I meant. Someone voted to close as homework, but there's no reason to do that because you showed a lot of work towards solving the problem.
– Zhe
Dec 9 '17 at 14:39

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 $\mathrm{p}K_\mathrm{a}$:

$$\mathrm{p}K_\mathrm{a} = 4.65 - \log{\frac{V_1}{V_2 - V_1}}$$

• I brushed up the formatting a little. \ce{...} from mhchem is used for chemical equations and chemistry-related notations. Also, generally I wouldn't recommend putting chemicals in super- or subscripts; this would make indices in chemical formulas hard to read and to parse (compare $X_\ce{H2O}$ vs $X(\ce{H2O})$). Aside from that, looks great:) Dec 9 '17 at 0:06
• Actually, I missed the connection between $n\ce{(HA)}$, $n(\ce{A-})$ and $n\ce{(NaOH)}$ expressed in your first two equations. From my neutralisation reaction, the mole ratio is pretty obvious, but I didn't know that it's allowed to use it to equate $n(\ce{A-})$ and $n\ce{(NaOH)}$ this way. Thank you @Gert! Dec 9 '17 at 9:08
• @andselisk: personally I prefer the subscript notation but each to their own. Thanks!
– Gert
Dec 9 '17 at 14:02
• @Candor: yes, that's where you went wrong. Easy when you know it, I guess!
– Gert
Dec 9 '17 at 14:04

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.

• My calculation looks like this: $\ce{{pKa} = 4.65 - log(\frac{1}{1.22}) = 4.73}$. Eventually you used addition instead of substraction? Anyways, thanks for your explanation! Dec 9 '17 at 9:24
• @Candor Yes I did that in my head and didn't add a minus sign for the log.
– Zhe
Dec 9 '17 at 14:38