mm something like this :
You work out the [H+(aq)] by doing this [H+(aq)] = Ka * acid/salt
Then you can work out the pH normally i.e pH= -log10[H+(aq)]

Ka is usually given to you - Ka represent dissociation - large value indicates strong acid , low value weak acid . Acid/salt are the concs. of the solutions
Ka = [H+] [Salt]
..........[Acid]

then find the pH using -ve log of the [H+]

thats all there is to it!
kriztinae
can anyone tel me in simple english how to calculate the ph of a buffer?

ph buffer = -log10 Ka (which they will tell you) x [HA]/[Conjugate base]

so for ethanoic acid and sodium ethanoate = Ka x ch3cooh/ch3coo-na+

i revised this today!! woo
hmm thanx, i kinda get what you mean
Hehe use my way its easiest :P - person who wrote after me told you how to work out the pH of a weak acid which is diff. from buffer solution
Kyrotec
Hehe use my way its easiest :P - person who wrote after me told you how to work out the pH of a weak acid which is diff. from buffer solution

hehe ok, its all good.
Kyrotec
Hehe use my way its easiest :P - person who wrote after me told you how to work out the pH of a weak acid which is diff. from buffer solution

um actually mine's just a rearranged version of yours!
ok well here is a question i dont understand, its an example in the book actually
calculate the ph of a buffer soln made by dissolving 18.5g of propanoic acid C2H5COOH and 12.0g of sodium propanoate, C2H5COONa in water and then making the volume up to 250cm3

they do this
[C2H5COOH] = (18.5 x 4 )/74 = 1.00 mol dm-3

[C2H5COONa] = (12.0 x 4) /96 = 0.500 mol dm-3

right i get why they divide by 96 and 74, its the RMM, but why do they multiply by 4 to get the concn???
because the question says its in 250cm3 so x4 = dm3!
misty
because the question says its in 250cm3 so x4 = dm3!

thank you so much!
i knew it was something stupid and small! argg im useless!
henderson-hasselbach equation:

pH (of a buffer) = pKa + log10 ( [salt]/[acid] )

where pKa = -log10(Ka)
visesh
henderson-hasselbach equation:

pH (of a buffer) = pKa + log10 ( [salt]/[acid] )

where pKa = -log10(Ka)

hmm ill try that thanx
kriztinae
thank you so much!
i knew it was something stupid and small! argg im useless!

heh no problem! good luck tomorrow
misty
heh no problem! good luck tomorrow

heh you too! we'r gonna make it!