I have five values for the volume of sodium hydroxide needed to neutralise a fixed quantity of hydrochloric acid, each trial with an uncertainty of 0.05 mL. If I take the average of these five values, in what way would the uncertainty propagate? I can think of three of the following ways, but don't understand which would be correct:
Because we can see the calculation of an average as the sum of the values for the five trials (with their uncertainties), and then a division by 5, we sum up the uncertainty to get 0.25 mL and then divide it by 5 to get the uncertainty of the average to be 0.05 mL.
Because we can see the calculation of an average as the sum of the values for the five trials (with their uncertainties), and then a division by 5 (which has no uncertainty), we sum up the uncertainty to get 0.25 mL and then divide it by 5 to get the uncertainty of the average to be 0.25 mL.
Range divided by two is often another way to calculate the uncertainty of an average value, which for my case results in a value different to the above two for the uncertainty of the average.
Which one, and why, would be the correct one to state next to the average value (with a plus-minus sign), as an uncertainty of the average value?
The first one is correct from the perspective of interval arithmetic. Meaning that if you are absolutely certain that each volume is within $0.05$ mL of the true value, then you are absolutely certain that the average volume is within $0.05$ mL of the true value as well. This works out for the reason you described.
The second one is never correct. Dividing by 5 makes the overall quantity smaller so it makes the uncertainties smaller regardless of how you conceive of the uncertainties.
I don't really understand what you mean in the third case, because I don't see how you would define the "range" of the average when there is only one average.
The more common way of propagating these uncertainties is to treat them as standard deviations of independent random variables. When you do this, the variance of the sum is the sum of the variances. Thus the standard deviation of the sum is the square root of the sum of the variances. This makes your uncertainty for the sum be $\sqrt{5} \cdot 0.05$, which means the uncertainty for the average is $\frac{\sqrt{5} \cdot 0.05}{5} = \frac{0.05}{\sqrt{5}}$.
Note that this is less than the uncertainty that you started with, whereas when we propagate uncertainties with interval arithmetic, we do not get a smaller uncertainty. This is a good feature of an uncertainty model for scientific purposes. That's because when we take a bunch of distinct measurements, we usually see some positive errors and some negative errors, which in the end will mostly cancel each other out. This is a pretty typical situation in a chemistry laboratory, and is one of the major reasons why we perform multiple trials.