# Why do these two calculations give me different answers for the same acid-base titration?

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

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

so,

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

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

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 $10^{-3}$.

So,

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

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.

• 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 ($ml$).
– MaxW
Oct 29 '15 at 18:12
• 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. Oct 29 '15 at 18:13

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.

• For the first point: LOL derp - I can't believe I missed that! I see, that makes perfect sense now. Oct 30 '15 at 0:01
• 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? Oct 30 '15 at 0:05
• 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. Nov 3 '15 at 2:38
• "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. Nov 3 '15 at 9:23
• 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. Nov 3 '15 at 16:47