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Acids and bases question help

At 25 °C, the acid dissociation constant Ka for ethanoic acid has the value
1.75 × 10–5 mol dm–3.
2 (c) (i) Calculate the pH of the solution formed when 10.0 cm3 of 0.154 mol dm–3 potassium
hydroxide are added to 20.0 cm3 of 0.154 mol dm–3 ethanoic acid at 25 °C.

Can someone do a walkthrough of this?
Reply 1
Avatar for Pigster
Pigster
20
If you have done ICE tables, if you have done Kc calcs, then these are similar.

You know how many mol of HA you have, for every mol of OH- added, you will create that many mol of A- (since HA + OH- -> A- + H2O) and the amount of HA will go down by that many mol.

You now have Ka, [HA] and [A-], gooooooooo calculator.
Reply 2
Avatar for MoJam
MoJam
13
Original post by Pigster
If you have done ICE tables, if you have done Kc calcs, then these are similar.

You know how many mol of HA you have, for every mol of OH- added, you will create that many mol of A- (since HA + OH- -> A- + H2O) and the amount of HA will go down by that many mol.

You now have Ka, [HA] and [A-], gooooooooo calculator.

although ur correct by markscheme, youve only worked out moles, and not [concentration]. If i divide moles by volume, i get a completely different answer, help?
Original post by MoJam
although ur correct by markscheme, youve only worked out moles, and not [concentration]. If i divide moles by volume, i get a completely different answer, help?

Well... I did say "[HA] and [A-]" which as you know meant that the OP should convert from amount to HA to conc. But... it actually doesn't matter:

Ka = [H+] x ([A-] / [HA])

But since [A-] / [HA] = (nA-/vol) / (nHA/vol) = n(A-) / n(HA), since the volume terms cancel out. So you don't need to covert the amounts of A- and HA to conc. Plug the values in and the one you're looking for, i.e. [H+], automatically is a conc. and you don't need to worry about the volume.
(edited 1 year ago)
Reply 4
Avatar for MoJam
MoJam
13
Original post by Pigster
Well... I did say "[HA] and [A-]" which as you know meant that the OP should convert from amount to HA to conc. But... it actually doesn't matter:

Ka = [H+] x ([A-] / [HA])

But since [A-] / [HA] = (nA-/vol) / (nHA/vol) = n(A-) / n(HA), since the volume terms cancel out. So you don't need to covert the amounts of A- and HA to conc. Plug the values in and the one you're looking for, i.e. [H+], automatically is a conc. and you don't need to worry about the volume.

ahhh right thanks, my mistake was that i divided [H+] by volume

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