[Rutherford]

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 K_a is 1.8*10^-5.
I calculated α₁-->K_a=c₁*α₁*α₁ so α₁=(K_a/c₁)^1/2
Then I calculated α₂-->α₁*0,9
Then I calculated c₂-->c₂=K_a/(α₂*α₂) and here I stopped. What to do now?

[Borek]

I calculated α₁-->K_a=c₁*α₁*α₁ so α₁=(K_a/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?

[Rutherford]

So, I can't use that approximation? How to calculate α then: K_a=(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 K_a, 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.

[Borek]

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.

[Rutherford]

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   .

[Borek]

Watch your units, you are adding mL to L. I missed it before.

[Rutherford]

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!