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### Topic: Back Titration Calculations  (Read 27619 times)

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#### Duckie

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• Mole Snacks: +1/-0 ##### Back Titration Calculations
« on: February 23, 2008, 06:15:18 AM »
Hi, I'm doing my Aspirin Coursework at the moment and I'm a little confused on the calculations of back titrations. I've tried it following the method here: http://www.chemistry-react.org/go/Tutorial/Tutorial_21681.html but I'm not sure I used the correct numbers in the right place!

Using 0.5g of an aspirin sample I made, I added 4.375cm3 of 0.5moldm-3 NaOH until the phenolphthalein indicator remained pink. Then I added a further 14.375cm3 of NaOH in excess. (Therefore my total volume of NaOH added was 18.75cm3)
Then I backtitrated with 10.5cm3 of 0.5moldm-3 HCl to neutralise the excess base.

I need to find out how much ASA is present in the sample, and hence work out the purity of my sample.

Moles HCl required: 0.0105 x 0.5 = 0.00525 mol
Moles NaOH reacting with HCl: 0.00525mol
Moles NaOH added to hydrolyse aspirin: 0.014375 x 0.5 = 0.0071875 mol
Net moles of NaOH reacting with ASA alone: 0.0071875 - 0.00525 = 0.0019375 mol
Since 1 mole ASA reacts with 2 moles NaOH: 0.0019375/2 = 0.00096875 mol
Therefore, mass of ASA in sample: 0.00096875 x 180 = 0.174375g

Hence % purity: (0.17435/0.5) x 100 = 34.87%

Would this be correct?

#### sjb

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« Reply #1 on: February 23, 2008, 07:27:52 AM »
I need to find out how much ASA is present in the sample, and hence work out the purity of my sample.

Moles HCl required: 0.0105 x 0.5 = 0.00525 mol
Moles NaOH reacting with HCl: 0.00525mol

I'm with you until here, then I start getting confused

You had a total of 18.75cm3 0.5M NaOH, which is 9.375 mmol NaOH. 5.25 mmol of NaOH reacted with the HCl, so 4.125 mmol reacted with the aspirin, As you've decided the ratio of NaOH:aspirin is 2:1, this means there was 2.0865 mmol of aspirin in your sample, which means there was 371.25 mg of aspirin, so the purity is ~74%.

It's often easier to see disappearance of a colour, rather than appearance, and things are even more confused here as the aspirin is a weak acid.

S
Moles NaOH added to hydrolyse aspirin: 0.014375 x 0.5 = 0.0071875 mol
Net moles of NaOH reacting with ASA alone: 0.0071875 - 0.00525 = 0.0019375 mol
Since 1 mole ASA reacts with 2 moles NaOH: 0.0019375/2 = 0.00096875 mol
Therefore, mass of ASA in sample: 0.00096875 x 180 = 0.174375g

Hence % purity: (0.17435/0.5) x 100 = 34.87%

Would this be correct?

#### Duckie

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• Mole Snacks: +1/-0 ##### Re: Back Titration Calculations
« Reply #2 on: February 23, 2008, 09:23:44 AM »
I thought the same as you at first, in thinking that I should subtract 0.00525 from the moles of NaOH in TOTAL, but my friend said I should subtract from the moles of NaOH added in excess (14.375cm3) since the volume added before that (4.375cm3) only neutralised the acids in the aspirin sample. The amount added in excess contained the volume needed to actually HYDROLYSE the aspirin.... but now I don't know which is right (because my friend isn't sure either) :S

EDIT: I *THINK* I've worked it out now, with the help of a few friends!! Basically I DID do the calculation right... If you want to see my working and stuff just say and I'll be happy to paste it in, or pm you! (Then you can see if there are any point to argue about XD) Otherwise, I think I'm ok now, but thank you for you *delete me* « Last Edit: February 23, 2008, 08:18:00 PM by Duckie »