how do i find the ph of a buffer solution?

I'very got a chemistry degree and I can't remember how to do this I think it was a pkA equation. Ps chemistry gets you nowhere study maths or engineering instead

e.g. 75 cm3 of 0.22 moldm-3 of butanoic acid is reacted with 50 cm3 of 0.185 moldm-3 of NaOH

Ka for butanoic acid: 1.5 x 10^-5 moldm-3

ive got the ka expression for butanoic acid but i dont know what to do after that

Hey, you need to use either the ka eqaution or the henderson hasslebach equation (i think the latter speaks for itself, but is neat if you kknow it!)

But youve done the first step correctly,

as it is a buffer question, you know that butanoic acid will be in xs, so naoh moles will be limiting.

If NAOH moles is limiting, this means that the moles of NAOH used = the moles of salt formed.

You know the xs moles of acid that remains now, so work out the concentration of the resultant solution by xs moles/0.125 dm3.---this gives you the new conc of butanoic acid whc=ich youll be subbing into ka eq.

Also, as you know the ka=([Conc Butanoate Salt]x[H+]) all divided by ([Conc Butanoic Acid]) -----you need to work out the conc of both as preiously mentioned---work out conc of salt in the same way. But this would involve moles of each however, this time each will be divided by the new volume of solution i.e. 0.125dm3. (this is the volume of the buffer sol btw)--i've done the acid one for you ^..

Subbing your new conc values in and rearranging (arithmetic) leaves you with [H+], you know what to do from there...
You should get pH of the buffer easy enough...this is using the ka eq btw.

If my solution is a bit wordy, lemme know what you dont understand.

firstly, THANKU SO MUCH

ive followed all of it but im stuck on how to get the conc of the butanoate salt? ive found the new conc of butanoic acid and so ive now got h+ conc and conc of the salt as my unknowns

try chemrevise, the explanation is very clear and concise and should help you

Yea, so youve done that part correctly as well,

To get the butanoate salt is v easy too.

See the way i said earlier that NaOH would all get used up (because in buffer calculations your weak acid is in xs), the moles of NaOH used up would equal the moles of salt formed (provided they are in the same molar ratio as is the case here 1:1 - correct me if im wrong.)

now as you know moles of NaOH used = moles of Salt formed, you can work out conc of Salt using moles divided by total volume ie. (0.125dm3) as your salt is just soluble in this aqueous solution. Then sub these values into ka equation.

The only unknown you will have is [H+]--dw, we are gonna find this soon.

Basically rearrange the ka equation with your values all plugged in so that the subjecct of the formula is [H+] - once you find this...

ph=-log[H+]

and there you go!

If you dont understand any of this, feel free to vocalise yourself

I get it!! Thankyou so much for your help