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Any one have any ideas?

Struggling with how to do the following

Calculate the pH of a 0.10 mol dm-3 solution ethanoic acid which has dissociation constant Ka of 1.8 x 10-5 mol dm-3.

I think I know the second part just not what to do with 0.10 mol dm-3. any advice welcome
Do you know the Ka formula for calculating [H+] concentration? It might have given it to you.

Ka = [H+]^2/[CH3COOH] <----- this is the concentration of the acid that you're unsure about.

Rearrange to find [H+] then do -log of that.
Original post by Alexi55
Any one have any ideas?

Struggling with how to do the following

Calculate the pH of a 0.10 mol dm-3 solution ethanoic acid which has dissociation constant Ka of 1.8 x 10-5 mol dm-3.

I think I know the second part just not what to do with 0.10 mol dm-3. any advice welcome


[H+]=square root of (ka x [acid])
then -log[H+]=pH
so the answer should be 2.9

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