OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD

Ka=([A][H])/[HA]
Which rearranges to [H]= Ka*[HA]/[A]

It's just a rearrangement mate. You'll normally have 3 values and rearrange to find the unknown. Or 1 value and a ratio.

june 2015 4ciii) how do you get 0.12M in C2H5COOH concentration.

How do you use electrode potentials to determine whether a reaction is feasible or not? In the textbook it says E cell must be 0.4+ but in an exam paper there was a question with a feasible reaction that had an E cell of 0.37. How come its feasible?

My revision guide had Ka on the numerator not to the side of it, which made it only have 2 terms.

It is diluted. They added water until volume was 100cm^3. The original volume was 25cm^3. Therefore, concentration is 4 x less (as c=n/v and v is 4x bigger). 0.480/4=0.120

anyone have the F325 June 2015 mark scheme? Or an unofficial one?

youtube f325 2015

Post a picture. You can rearrange the equation to have Ka on any side. It's never [H+]^2... You're just confusing weak acids and buffers.The only way you can have two terms is if you're given a ratio. Most Qs will work with ratios in big buffer qs. And ratios always involve the acid and conjugate base (HA and A-).

You are actually right I was on the weak acids page but what I meant was the Ka was literally on the fraction. So instead of say A=(B/C) it just had A=B/C
With A and B being on the same line.

I'm struggling to understand! If you post a picture I'll understand better.

A=(B/C) and A=B/C are the same thing? Haha.

please help!!!

Ka = 10^-pka
pka = -logKa

low/high ka = what
low/high pka = what

which combination means its a strong acid???

june 2015 4ciii) how do you get 0.12M in C2H5COOH concentration.

june 2015 4ciii) how do you get 0.12M in C2H5COOH concentration.

The conc of the 25cm3 sol of ethanoic is 0.48*0.025 but then it is diluted up to 100cm3 so divide by 0.1

(0.48*0.025)/0.1 = 0.12