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Topic: Help please  (Read 14555 times)

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Offline mass

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Re: Help please
« Reply #15 on: March 01, 2008, 07:50:22 AM »
I set up a student thing yesterday afternoon so  me and a few other students could discuss the questions.

We came up with the equation

C6H5COOH + NAOH = C6H5COONA+ + H20

Its a 1;1 ratio. However we were stuck after this stage. We attemepted the % question but we were getting ridiciolous values like 168%. Can someone take me through how to do it please.

Offline Borek

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Re: Help please
« Reply #16 on: March 01, 2008, 08:45:38 AM »
C6H5COOH + NAOH = C6H5COONA+ + H20

Close - although it is... not balanced. For the reaction to be balanced not only atoms must be balanced, but also the charge. In your case there is +1 on the right. (Not to be nitpicky about NA instead of Na).

So the correct version is

C6H5COOH + NaOH -> C6H5COONa + H2O

It gives the same ratio 1:1.

Quote
Its a 1;1 ratio. However we were stuck after this stage. We attemepted the % question but we were getting ridiciolous values like 168%. Can someone take me through how to do it please.

Have you not forgot about your blank? Your solutions contains not only benzoic acid, but also phosphoric buffer. This buffer is nothing else but a mixture of phosphoric acid and some base - but acid is in a excess. So when neutralizing your solution you have to neutralize BOTH benzoic acid and phosphoric acid from the buffer. Blank titration should tell you how much phosphoric acid was in your sample - you should substract this amount from the results of the sample titration.
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Offline mass

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Re: Help please
« Reply #17 on: March 01, 2008, 12:13:32 PM »
what question are you helping me with though ???

Offline Borek

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Re: Help please
« Reply #18 on: March 01, 2008, 03:08:13 PM »
Part of your work is to find - from titration - amount of the benzoic acid left in water solution. For now you have not yet calculated this amount properly. I am referring to this particular calculation.

Show how you got to 168%.
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Offline mass

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Re: Help please
« Reply #19 on: March 02, 2008, 05:22:14 AM »
So you are referring to the fraction of benzoic acid unionised in the aqueous phase.

As for getting the 161%, this is my calculation:

C1V1 = C2V2
0.02 V1= 0.051 *31.5

Therefore, v1= (0.051 *31.5)/ (0.02) = 80.325cm3.

Therefore the % of benzoic acid extracted is = 80.325/50*100 = 160.65%. This was question 4.

See I do have a go but I need guidance.

Offline Borek

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Re: Help please
« Reply #20 on: March 02, 2008, 05:52:48 AM »
Problem is, you are ignoring what I am telling you. 31.5 mL (shouldn't it be 31.5-0.5=30?) contains also base that was used to neutralize phosphoric acid, that's what the blank was for. I have already explained that in details.

Not to mention the fact that you are calculating amounts comparing volumes, not numbers of moles. I have no idea where to begin with the help. To help I have to build on some knowledge that you should already have :(
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Offline mass

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Re: Help please
« Reply #21 on: March 02, 2008, 06:06:13 AM »
ok yeah 30 then, my calculation is still right no?

could you not do it and show it me please, i am trying my level best but of to no use. Atleast if you do it for me, I can try and carry on next time I get a similar type of question.

Offline JGK

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Re: Help please
« Reply #22 on: March 05, 2008, 02:53:40 PM »
ok yeah 30 then, my calculation is still right no? No it's wrong, you have ignored and information provided by the blank experiment (for a start)

could you not do it and show it me please, i am trying my level best but of to no use. Atleast if you do it for me, I can try and carry on next time I get a similar type of question.

If we do your work for you, you will learn nothing. If that method of education worked, your lecturers and professors would provide full answers for all the problems you are being set, thus allowing you to learn unhindered by any intellectual challenge. :-\


As this will be my last post on this thread, au revoir
Experience is something you don't get until just after you need it.

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