When doing a titration with a burette, the error of the burette is usually $\pm 0.05\ \mathrm {ml}$, as we have to read two times on the burette (initial/final vol), the error will be $\pm 0.07\ \mathrm{ml}$ ($\sqrt {(0.05)^2 + (0.05)^2}$)

When performing an titration, we usually do three determinations. The error of the titration is half the range of the three determinations. However, What if I get values such as $14.2\ \mathrm{ml}$, $14.3\ \mathrm{ml}$, $14.2\ \mathrm{ml}$ or even three times the same volume. The error of the measure would be less than the error in the instrument. Which rule do I have to apply here?


1 Answer 1


the error of the burette is usually ±0.05 ml

Well that depends on the quality of the burette and your technique. The "problem" with a burette is that the marking will be linear whereas the burette's diameter might not be constant.

I think that the burettes I last used were marked at the 0.05 ml mark, but you estimated between the marks. So you recorded to the 0.01 ml amount. It was easy to get at least ±0.02 ml on average for a reading.

I think the ±0.05 ml notion is from 20 drops to the ml. But you can dispense part of a drop by touching the inside of the flask to the tip of the burette.

if I get values such as 14.2 ml, 14.3 ml, 14.2 ml

Then you average and use the values calculated for the mean and the standard deviation of the measurements themselves. It is a "sanity check" to consider the precision of the titration procedure itself.

I suspect that your "problem" is that you're not carrying enough significant figures through your calculations. You only round on the the final results.

So start at 6.72 ml end at 20.92 yields 14.20 ml, not 14.2 ml.

If you get 14.20 ml, 14.20 ml, and 14.20 ml then I'd strongly suspect that your assertion that the precision of the titration is 0.07% is wrong.

The first titration I every did was with potassium permanganate to determine the equivalent weight of an unknown oxalate compound. My titration technique was horrible and I didn't have enough sample left to do more titrations. But I knew few oxalates were soluble, and the sample I had was obviously crystalline. So I did some quick cation tests and found that the cation was NH4+ which has two waters. The prof wanted how I got the right answer to four significant figures with such a poor technique.

  • $\begingroup$ @user164612 -Well I added something about a triple reading that I don't will probable be useful to you. To get more specific I need to know how your burettes are marked and how many significant figures that you are recording. $\endgroup$
    – MaxW
    Commented Oct 31, 2015 at 0:27
  • $\begingroup$ I see you did add something but you are correct in asserting that it is the least helpful thing possible. It is entirely feasable to get the same result in titration three times (I had it happen to me back in the general chem lab course in the days) so just suspecting that some error is wrong does not help. $\endgroup$
    – Jan
    Commented Nov 1, 2015 at 0:05
  • $\begingroup$ Both statements are unlikely. (1) I got three readings of 14.20, 14.20 and 14,20. (2) The error of the titration is +/- 0.07 ml. Remember that 14.20 is not equal to 14.2 in analytical chemistry. $\endgroup$
    – MaxW
    Commented Nov 1, 2015 at 0:16

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