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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Rutherford on March 08, 2012, 11:56:39 AM

Title: Acetic acid solution
Post by: Rutherford 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=c1*α1*α1 so α1=(Ka/c1)1/2
Then I calculated α2-->α1*0,9
Then I calculated c2-->c2=Ka/(α2*α2) and here I stopped. What to do now?
Title: Re: Acetic acid solution
Post by: Borek on March 08, 2012, 12:06:33 PM
Quote from: Raderford on March 08, 2012, 11:56:39 AM
I calculated α1-->Ka=c1*α1*α1 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?
Title: Re: Acetic acid solution
Post by: Rutherford 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.
Title: Re: Acetic acid solution
Post by: Borek 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

Quote from: Raderford on March 08, 2012, 12:35:26 PM
n=0.5*0.1+1.049x/60? x is the volume of the added acid.

Looks OK. Now to volume and concentration.
Title: Re: Acetic acid solution
Post by: Rutherford 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  :-\ .
Title: Re: Acetic acid solution
Post by: Borek on March 08, 2012, 02:32:29 PM
Watch your units, you are adding mL to L. I missed it before.
Title: Re: Acetic acid solution
Post by: Rutherford 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!