[Scoobiedoobiedoo!]

  I hate buffers and pH

I have made an attempt at this question, please can someone validate my answer.

Calculate the pH of a 0.100 mol dm^-3 solution of propanoic acid, for which K_a=1.3*10^-5 mol dm^-3

I rearranged Ka as follows- [H+]=Ka*[CH3CH2COOH] / [CH3CH2COO-]

I then used the assumptions to fill in the values- [H+]=1.3*10^-5 *0.100 / 0.100

Then with the value for [H+]- -log[H+] to get a pH of 4.9


Hopefully someone can follow my working and tell me if it is correct

[Astrokel]

I then used the assumptions to fill in the values- [H+]=1.3*10-5 *0.100 / 0.100

No, you can't assume the concentration of propanoate ion is 0.100. If so then there will be no dissociation at all. I understand you are doing an approximation for propanoic acid(within 5%), but not with the conjugate base!

[Scoobiedoobiedoo!]

Ok so how do I find out the conc. of the propanoate ion?

[Astrokel]

ICE table!

[Scoobiedoobiedoo!]

Initial Concentration Equilibrium

[ARGOS++]

Dear ConcH2SO4+Water!!!;

How about:   http://en.wikipedia.org/wiki/ICE_table

Good Luck!
                    ARGOS⁺⁺

[Scoobiedoobiedoo!]

Thanks you have been most helpful  Have a mole snack

[ARGOS++]

Dear Scoobiedoobiedoo!;

Thank you very much!

You’re welcome!   ─   Soon again.

Good Luck!
                    ARGOS⁺⁺