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Offline Rutherford

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Acetic acid solution
« on: March 08, 2012, 11:56:39 AM »
What volumen of pure CH3COOH (density is 1.049g/cm3) you need to add in a 500cm3 solution of CH3COOH (c1=0.1mol/dm3) so that the level of dissociation (α) in the obtained solution is 10% smaller than in the beginning solution? Density of the begginig and obtained solutions are 1g/cm3 and Ka is 1.8*10-5.
I calculated α1-->Ka=c111 so α1=(Ka/c1)1/2
Then I calculated α2-->α1*0,9
Then I calculated c2-->c2=Ka/(α22) and here I stopped. What to do now?

Offline Borek

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Re: Acetic acid solution
« Reply #1 on: March 08, 2012, 12:06:33 PM »
I calculated α1-->Ka=c111 so α1=(Ka/c1)1/2

You are using an approximated formula - do you know when it can be used? Have you checked if it can be used in this case?

If you add x mL of the pure acetic acid to teh 500 mL of the 0.1M solution, what will be the volume of this solution? How many moles of acetic acid will it contain? What will be concentration of the acetic acid in this solution?
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Offline Rutherford

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Re: Acetic acid solution
« Reply #2 on: March 08, 2012, 12:35:26 PM »
So, I can't use that approximation? How to calculate α then: Ka=(c*α2)/(1-α)? I think that the approximation has to be used.
The volume would be 500ml+x, c of the obtained solution I calculate from the α and Ka, but the number of moles is harder to calculate. I need to use the number of moles of the added CH3COOH. Can I calculate it this way: n=0.5*0.1+1.049x/60? x is the volume of the added acid.

Offline Borek

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Re: Acetic acid solution
« Reply #3 on: March 08, 2012, 01:49:49 PM »
I didn't say you can't use it - but you should check if it can be used.

See discussion here: http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

n=0.5*0.1+1.049x/60? x is the volume of the added acid.

Looks OK. Now to volume and concentration.
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Offline Rutherford

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Re: Acetic acid solution
« Reply #4 on: March 08, 2012, 02:06:57 PM »
c2=(0.5*0.1+1.049x/60)/(0.5+x) I got that c2 is 0,1235mol/dm3, but then x is a negative number  :-\ .

Offline Borek

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Re: Acetic acid solution
« Reply #5 on: March 08, 2012, 02:32:29 PM »
Watch your units, you are adding mL to L. I missed it before.
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Offline Rutherford

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Re: Acetic acid solution
« Reply #6 on: March 09, 2012, 08:23:36 AM »
Thanks for the correction, the x that I used here: 1.049x/60 is in ml and the x that I used for volume: 0.5+x is in l, so the number of moles should be 0.5*0.1+1049x/60 and I get the answer 0.6cm3 which is the right answer. Thanks so much!

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