Why is a theoretical titration curve higher than a practical one?

I'm currently in high school and I did two titrations. The first one is a titration of $10.0~\mathrm{mL}$ $0.1~\mathrm{M}$ $\ce{HCl}$ with $x~\mathrm{mL}$ $\ce{NaOH}$. I obtained the following titration curve:

The blue curve is what I measured and the red curve is what the curve would look like in theory, with these data.

The second titration is a titration of $10.0~\mathrm{mL}$ $0.2~\mathrm{M}$ $\ce{CH3COOH}$ with $x~\mathrm{mL}$ $\ce{NaOH}$. I obtained the following titration curve:

Again, my measurements are in blue and the red curve is the theoretical model. I noticed that (looking at the second curve in specific) the measured pH value is most of the time lower than in the theoretical model, and close to the equivalence point, the graph is not as steep. Why is this? Not a valid reason would be that the pH sensor does not measure the right value instantly, because near the equivalence point, only one drop of $\ce{NaOH}$ was added about every two seconds. Do you have a more reasonable explanation?

• I suspect the (in)homogeneity of the mixture is playing an important role. Though acid-base reactions are often very fast, very small concentration gradients can be difficult to eliminate entirely due to the low speed of diffusion of substances in liquids, precluding the formation of true equibliria. How vigorously was the titrated solution being mixed? Another factor is that the familiar pH scale with $pK_w=14$ only works at 25°C. How well was the temperature controlled during the experiment? – Nicolau Saker Neto Jun 1 '13 at 20:11
• @NicolauSakerNeto Thanks for your comment. I mixed it by hand: that is, I kept shaking the beaker the titration was done in continuously throughout the experiment. I could almost go around the beaker with my thumb and my middle finger, so it wasn't very large. The temperature was approximately $21^\circ$C, so that could also make a difference. – Tim Vermeulen Jun 1 '13 at 20:29
• Usually titrations are performed inside erlenmeyers to allow better mixing by hand and to avoid spillage from droplets formed on impact. Another option is to use an automatic stirrer, either mechanical or magnetic. I believe mixing is then a large part of the answer. The deviation seems to be in general more severe for the second half of the titration, which is presumably due to the larger volume of liquid, making it even harder to homogenize. Also you get tired after a while! – Nicolau Saker Neto Jun 1 '13 at 20:35
• The volume has indeed almost tripled at the equivalence point, so that makes sense. One last thing bothers me: in the first titration, the measured value was higher than the expected value, while in the second titration, it was the other way around. Could this have anything to do with the fact I already started with $10.0$mL of $\ce{NaOH}$ added to the solution, to skip a boring part of the curve? That doesn't make sense to me though. – Tim Vermeulen Jun 1 '13 at 20:46
• $\ce{HCl}$ and $\ce{NaOH}$ solutions are a bit tough to prepare at an exact concentration, because $\ce{HCl}$ is volatile and $\ce{NaOH}$ is hygroscopic, making both usually slightly more dilute than expected. For proper analytical precision, the concentration is usually tested by a previous titration with a more reliably prepared acid/base solution (a primary standard), such as potassium hydrogen phthalate. It may be simply that you didn't prepare the solutions to quite the concentration you expected. – Nicolau Saker Neto Jun 1 '13 at 20:55

One thing that that has already been mentioned in the comments is the impact of $\ce{CO2}$ on the equilibrium. If you look at the following figure you can see the pH diagram of carbonic acid.

The equilibrium between $\ce{CO2}$ and $\ce{H2CO3}$ $$\ce{CO2 + H2O <=> H2CO3}$$ depends very much on the pH of your aqeous system. As long as it is sufficiently acidic no carbonic acid can/will react to form hydrogen carbonate or even further to carbonate. But from $\ce{pH}\approx4$ the possibility that hydrogen carbonate will be formed rises. This means for your titration, that you are not only titrating your acid anymore but you are also titrating the dissolved $\ce{CO2}$. The longer you wait for equlibration near the the equlibrium point the more dissolved $\ce{CO2}$ reacts to hydrogen carbonate and distorts your ideal $\ce{CO2}$-free titration curve.

To show you the impact more clearly I created the following figures showing both of your systems with varying carbonic acid concentrations $\left(0.00\ldots0.05~\ce{mol~L^{-1}}\right)$.

At first your system with hydrochloric acid:

And secondly the system containing acetic acid:

There are two remarking differences between both titration curves.

1. You can determine the equlibrium point of the $\ce{HCl}$-titration (though it's not around pH 7 but at pH 4-5) no matter how many hydrogen carbonate will be formed during your titration.
2. You have to pay attention during the titration of acetic acid because the inflection points gets distorted above any detectable other inflection point. However it is not that important as it seems because titrations of acetic acid are fast and there is usually not much time to dissolve enough $\ce{CO2}$.

The other points are already mentioned before. Your pH electrode could not be calibrated correctly and your concentrations might not be as precise as you think they are. The titration curve of hydrochloric acid looks like your acid was less concentrated which results in a higher starting pH and in a lower ending pH. The titration of acetic acid indeed looks like the electrode was not properly calibrated because it looks quite good if you just raise the pH values per measuring point and it suffers from dissolved $\ce{CO2}$ which is why the inflection point is shifted to a higher volume.

The mentioned increase of volume that is most times neglected during the calculation of titration curves has no big impact unless the concentration of the base is much lower than that of the acid that has to be titrated.

On the one hand the first four images show that different base concentrations around your ideal concentration don't effect the curve itself very much but only the needed volume to reach the point of inflection.

On the other hand you can see that there is a visible influence on the titration curve if the base is 10 times as diluted as the acid.

Nicolau Saker Neto's last comment is probably what's going on here. All of your titrations have too low a pH compared to theoretical, and so are less basic than expected. This suggests that your concentration of $\ce{NaOH}$ is lower than you think, and the fact that $\ce{NaOH}$ is hydroscopic is a good explanation for this (i.e. you have a more dilute solution than you think). Neto continues to be correct in that the usual procedure to correct this is to standardize your titration solution against a standard that is known to be more reliable with respect to producing accurate concentrations. $\ce{KHPh}$ is indeed one of these standards, and is used after drying in a drying oven. So the full procedure is: prepare $\ce{NaOH}$ solution as accurately as possible, do the same with $\ce{KHPh}$, titrate $\ce{NaOH}$ against $\ce{KHPh}$, and record the resulting concentration of $\ce{NaOH}$ accurately for use in future titrations. Also, $\ce{NaOH}$ loses strength over time with exposure to the air, so you have to re-standardize over time.

• That "looses strength with exposure to air" is carbonate forming from $\ce{CO_2}$ captured from the ambient air. Which means that you have a small amount of buffer in the solution. pKA $\ce{HCO_3^2-}$ is 10.3, so this may be the explanation for the bump in the second titration curve. – cbeleites supports Monica Jan 27 '15 at 11:44