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So, the other day I was carrying out a titration and got the titres $33.4cm^3$ and $33.3cm^3$. I calculated the random uncertainty to be $0.05cm^3$. Now, the burette itself had an uncertainty of and I'm wondering whether I should add these two uncertainties together to find the uncertainty for the titres. Also, if I have to use the average of the titres to calculate the percentage uncertainty or whether I need to calculate the percentage uncertainty individually for the titres.

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  • $\begingroup$ Related: Significant figures when standard deviation is 0 $\endgroup$ – Loong Nov 27 '16 at 19:56
  • $\begingroup$ Via interpolation you should be able to read a burette to 0.01 ml. It depends on the class of burette, but I'd think all would be better than 0.05 ml. $\endgroup$ – MaxW Nov 27 '16 at 19:57
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Yes - to calculate the total uncertainty when adding or subtracting measurements, it's the room-sum-squared of all the uncertainties.

$$ e_{tot} = \sqrt{(e_{meas})^2 + (e_{rand})^2} $$ The more generalized form of this will be the root-sum-squared of all sources of error and uncertainty. $$ e_{tot} = \sqrt{\sum{(e_n)^2}} $$

To get the quantity "$e_{meas}$" you apply the root-sum-squared formula of all the uncertainties of your measurement devices used to transfer liquids (transfer beaker, graduated cylinder, burette, etc...).

$$ e_{meas} = \sqrt{(e_{beaker})^2 + (e_{burette})^2+(e_{grad.cyl})^2+(e_{burette})^2...etc} $$

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To answer whether you apply the burette uncertainty to your measurement, the answer is also yes. Assuming that the burette has an uncertainty of $±0.01 cm^3$, the actual reading of your measurements will be: $33.3cm^3 ± 0.01 mL$ and $33.4cm^3 ± 0.01 mL$. The two uncertainties in the measurement will be applied to the $e_{tot}$ equation above.

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    $\begingroup$ You can't end up with a measurement that has multiple kinds of units (i.e. statement However, in the case that your measurement devices use different units). If you're adding and subtracting then you use rms of error. If you're using multiplication and division then you use the rms of the % error. $\endgroup$ – MaxW Nov 27 '16 at 22:40
  • $\begingroup$ Oh yikes, I've cleared that up to not cause confusion and specifying that these error calculations pertain only to addition and subtraction of measurements. $\endgroup$ – Patrick B. Nov 28 '16 at 0:10

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