buffer solutions

can anyone tel me in simple english how to calculate the ph of a buffer?

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

Reply 2

The Albatross

Ka = [H+] [Salt]
..........[Acid]

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

thats all there is to it!

Reply 3

Flicker

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

Reply 4

kriztinae

hmm thanx, i kinda get what you mean

Reply 5

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

Reply 6

kriztinae

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.

Reply 7

The Albatross

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!

Reply 8

kriztinae

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???