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titration - taking average results (1 Viewer)

DepressedPenguino

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25ml of an uunknown conc. soln is titrated against a known 0.22 mol/L NaOH. We are given that 0.3, 0.35 and 0.3 of NaOH are used in the 1st, 2nd and third titration respectively. Calculate the unknown acid conc.
So to find the volume of NaOH used, do i take the average of 0.3, 0.35 and 0.3 OR just 0.3 and 0.3?
 

DepressedPenguino

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Btw, 0.3, 0.35 and 0.3 are in millitres but that is not necessary to answer my question :p
 

Crisium

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When calculating the average volume used, always disregard the first titration volume as it tends to overshoot the endpoint
 

DepressedPenguino

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When calculating the average volume used, always disregard the first titration volume as it tends to overshoot the endpoint
Yeah in this case the 2nd titration seems to be an outlier do i still include it for my averaging?
 

Ekman

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When calculating the average volume used, always disregard the first titration volume as it tends to overshoot the endpoint
That's not entirely true, when given a table of volumes required per titration, if it says titration 1: 0.3ml, you include it in the average calculation, since it did not state that it was a 'trial' run. However if the question said Trial Run: 0.3 ml then Titration 1: 0.35 ml, you disregard the 0.3ml and use the 0.35 ml instead. Basically if the question does not specify that the first run is a trial run, then you should include it as you are meant to assume that the experiment had already completed a trial run.

Take the average which is around 0.32mL
In terms of calculating averages for this question, 0.35 is massive error range in comparison to the other two results. In fact you are only meant to have a +-0.05 error range, so even if one titration said 0.32 ml, you would still not include it in the averages. So you would average out 0.3 with 0.3, which is just 0.3ml.

I say this out of experience, I got deducted a mark in my previous task for chemistry for this same mistake. The results of the titration were:
Titration 1: 23.7
Titration 2: 23.6
Titration 3: 23.9
Titration 4: 23.7
Titration 5: 24.1
I used Titration 1,2,3,4 in my average when you are meant to use Titration 1,2,4, as titration 3 has too much of an error.
 
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Librah

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25ml of an uunknown conc. soln is titrated against a known 0.22 mol/L NaOH. We are given that 0.3, 0.35 and 0.3 of NaOH are used in the 1st, 2nd and third titration respectively. Calculate the unknown acid conc.
So to find the volume of NaOH used, do i take the average of 0.3, 0.35 and 0.3 OR just 0.3 and 0.3?
Just take it as 0.30.
 

DepressedPenguino

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That's not entirely true, when given a table of volumes required per titration, if it says titration 1: 0.3ml, you include it in the average calculation, since it did not state that it was a 'trial' run. However if the question said Trial Run: 0.3 ml then Titration 1: 0.35 ml, you disregard the 0.3ml and use the 0.35 ml instead. Basically if the question does not specify that the first run is a trial run, then you should include it as you are meant to assume that the experiment had already completed a trial run.



In terms of calculating averages for this question, 0.35 is massive error range in comparison to the other two results. In fact you are only meant to have a +-0.05 error range, so even if one titration said 0.32 ml, you would still not include it in the averages. So you would average out 0.3 with 0.3, which is just 0.3ml.

I say this out of experience, I got deducted a mark in my previous task for chemistry for this same mistake. The results of the titration were:
Titration 1: 23.7
Titration 2: 23.6
Titration 3: 23.9
Titration 4: 23.7
Titration 5: 24.1
I used Titration 1,2,3,4 in my average when you are meant to use Titration 1,2,4, as titration 3 has too much of an error.
This was what i was thinking when i looked at the question :)
 

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