In class, I performed an weak acid-strong base titration using commercial white vinegar and sodium hydroxide with the aim of finding the concentration of of ethanoic acid in the vinegar, and I used stoichiometric calculations as well as the $\mathrm{p}K_\mathrm{a}$ equation to find the concentration from the data I collected. However, these two methods are giving me different answers and I have no idea if I did something wrong in my calculations, or used the wrong method outright.

As a bit of background info, the label on the vinegar says it has a $\ce{CH3COOH}$ concentration of $5\%$, which according to my calculations (I measured the density of the vinegar to figure this out) is around $0.9\mathrm{~mol~dm^{-3}}$.

Here's the data from the titration:

Volume of $\ce{CH3COOH}$: $3.1\mathrm{~cm^3}$ (diluted with water to the $50\mathrm{~cm^3}$ mark for experimental purposes)

Concentration of $\ce{NaOH}$: $0.1 \mathrm{~M}$

Volume of $\ce{NaOH}$ added: $33.1\mathrm{~cm^3}$

Initial $\mathrm{pH}$ of $\ce{CH3COOH}$: $3.0$

Half-equivalence $\mathrm{pH}$: $4.6$

So, using the stoichiometric method, we have the formula

$$M_\mathrm{acid} V_\mathrm{acid} = M_\mathrm{base} V_\mathrm{base}$$


$$M_\mathrm{acid}(0.0031) = (0.1)(0.0331)$$

$$M_\mathrm{acid} = \frac{(0.1)(0.0331)}{0.0031} = 1.067 \mathrm{~mol~dm^{-3}}$$

This answer is fairly close to the literature value, so I'm assuming it's accurate. (The percentage concentration works out to be about $6.1\%$.)

On the other hand, the $\mathrm{p}K_\mathrm{a}$ equation is $K_\mathrm{a} = \frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}$

Since $\mathrm{p}K_\mathrm{a} = \mathrm{pH}$ at the half-equivalence point, the $K_\mathrm{a}$ is $10^{-4.6} = 2.5 \times 10^{-5}$; similarly the initial $\mathrm{pH}$ is $3$, so the initial $[\ce{H+}]$ is $10^{-3}$.


$$2.5 \times 10^{-5} = \frac{(10^{-3})^2}{[\ce{HA}]-0.001}$$

Entering this into a calculator returns the value $[\ce{HA}] = 0.041 \mathrm{~mol~dm^{-3}}$.

So, why is the second answer so much smaller than the first? Have I used one of the equations incorrectly or made an error in calculation? Does one of the methods not work for the acid-base titration in question (and why would that be?)

I sincerely apologize for the messy formatting; I literally had to learn LaTeX from scratch just to post this question. Thank you so much in advance.

  • $\begingroup$ Not sure what your convention is but $dm^3$ and $cm^3$ are weird units to me. I would have used liters ($l$) and milliners ($ml$). $\endgroup$
    – MaxW
    Commented Oct 29, 2015 at 18:12
  • $\begingroup$ I reside outside of the U.S. where our education system is predisposed to the S.I. unit system rather than the metric system which uses liters and milliliters. $\endgroup$
    – Razorlance
    Commented Oct 29, 2015 at 18:13

1 Answer 1


Part of the reason why there is such a large discrepancy is because in your second method of calculating $[\ce{HA}]$, you failed to take into account the fact that your vinegar was diluted from $3.1\mathrm{~cm^3}$ to $50\mathrm{~cm^3}$.

Effectively, if the original concentration of $\ce{HA}$ was $x$, then your dilution would have brought it to a concentration of $\frac{3.1}{50}x$. And you plugged the $\mathrm{pH}$ of this diluted solution into the the $K_\mathrm{a}$ equation. Let's adjust for that dilution:

$$\begin{align} \frac{3.1}{50}x &= 0.041\mathrm{~mol~dm^{-3}} \\ x &= 0.661\mathrm{~mol~dm^{-3}} \end{align}$$

As you can see, there is still some difference between the two values. However, you can calculate the theoretical $\mathrm{pH}$ of a $1.067\mathrm{~mol~dm^{-3}}$ solution of $\ce{CH3COOH}$, using the fact that the $\mathrm{p}K_\mathrm{a}$ is $4.6$. Here, I will make the simplifying assumptions that the final and initial concentrations of $\ce{HA}$ are the same, i.e. $[\ce{HA}] = c_{\ce{HA}}$, and also that $[\ce{H+}] = [\ce{A-}]$:

$$\begin{align} \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} &= 10^{-4.6} \\ &= 2.511\times 10^{-5}\mathrm{~mol~dm^{-3}} \\ [\ce{H+}] &= \sqrt{(2.511\times 10^{-5})(1.067)}\mathrm{~mol~dm^{-3}} \\ &= 0.00518\mathrm{~mol~dm^{-3}} \\ \mathrm{pH} &= 2.28 \end{align}$$

You could go all the way and do mass-balance and charge-balance equations in order to avoid making that assumption, but I doubt it alters the result significantly. So, if you really had a concentration of $1.067\mathrm{~mol~dm^{-3}}$, your observed initial $\mathrm{pH}$ would not have been $3.0$. There is likely some experimental error present, or perhaps the $\mathrm{pH}$ meter that you used to measure the initial $\mathrm{pH}$ is not calibrated well.

  • $\begingroup$ For the first point: LOL derp - I can't believe I missed that! I see, that makes perfect sense now. $\endgroup$
    – Razorlance
    Commented Oct 30, 2015 at 0:01
  • $\begingroup$ As for the pH meter being calibrated incorrectly - this is certainly a plausible reason for the discrepancy, but I also performed the titration using phenolphthalein indicator and obtained a very close equivalence volume of $32 \mathrm{cm^3}$. Is this to be expected even if there is an experimental error with the pH meter? $\endgroup$
    – Razorlance
    Commented Oct 30, 2015 at 0:05
  • $\begingroup$ Also, due to the fact that the vinegar was titrated about $16$ times, your equation gives me a theoretical pH of about $2.8$ to $2.9$. Although, yes I believe this could be due to a slight calibration error in the pH meter. $\endgroup$
    – Razorlance
    Commented Nov 3, 2015 at 2:38
  • $\begingroup$ "Also, due to the fact that the vinegar was titrated about $16$ times, your equation gives me a theoretical pH of about $2.8$ to $2.9$." What do you mean? As for your earlier question, I don't know what else might have gone wrong. It might just worth repeating if you could do so. However I would guess the most likely case is the pH meter is off. $\endgroup$ Commented Nov 3, 2015 at 9:23
  • $\begingroup$ Sorry, I meant that I think in calculating the theoretical pH the concentration you should have used was $0.669$ (that's $1.067/16$), which corrects for the dilution, unless I'm very much mistaken there. Oh, and I meant 'diluted', not 'titrated'. That was a typo. $\endgroup$
    – Razorlance
    Commented Nov 3, 2015 at 16:47

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