That's true, but they both also have specific formula which define them
Ka = [H+] [A-] / [HA]
pH = -log[H+]
are those familiar?
Yeah, I have used them before but sometimes in the question it will give a concertration (e.g. 0.100 mol dm-3). So where does that concertration go in the formula ?. Also when's it saids a time 298k that doesn't need to be in the equation?
Yeah, I have used them before but sometimes in the question it will give a concertration (e.g. 0.100 mol dm-3). So where does that concertration go in the formula ?. Also when's it saids a time 298k that doesn't need to be in the equation?
Do you know what the different terms in the two equations mean, particularly the parts in square brackets?
That's along the right lines, [H+] is the concentration of protons [A-] is concentration of conjugate base [HA] is the concentration of the undissociated acid
I'll give you an example
Ka of ethanoic acid is 1.8x10-5 pH of a solution of ethanoic acid is 2.523 find the concentration of the acid solution?
So the concentration its asking you for is [HA]
First you can use pH to find [H+] pH = -log[H+] rearranges to [H+] = 10-pH
so here [H+] = 10-2.523 = 0.003 mol dm-3
Than you can use the definition of Ka to find what you want. First you have to realise that all the protons (H+) and all the conjugate base molecules (A-) in this question come from dissociation of (HA) molecules so there must always be the same amount of H+ and A- ( ie [H+] = [A-] )
so Ka = [H+][A-] / [HA] becomes Ka = [H+]2 / [HA]
putting in the numbers you know for Ka and for [H+] 1.8x10-5= (0.003)2 / [HA]
rearranging gives
[HA] = (0.003)2 / 1.8x10-5 = 0.5 mol dm-3
So the acid solution is 0.5 mol dm-3
Hope from that example you can see that if you know two out of three of the following bits of information: [HA] pH Ka
than you can use the two equations too work out the third bit of information!
That's along the right lines, [H+] is the concentration of protons [A-] is concentration of conjugate base [HA] is the concentration of the undissociated acid
I'll give you an example
Ka of ethanoic acid is 1.8x10-5 pH of a solution of ethanoic acid is 2.523 find the concentration of the acid solution?
So the concentration its asking you for is [HA]
First you can use pH to find [H+] pH = -log[H+] rearranges to [H+] = 10-pH
so here [H+] = 10-2.523 = 0.003 mol dm-3
Than you can use the definition of Ka to find what you want. First you have to realise that all the protons (H+) and all the conjugate base molecules (A-) in this question come from dissociation of (HA) molecules so there must always be the same amount of H+ and A- ( ie [H+] = [A-] )
so Ka = [H+][A-] / [HA] becomes Ka = [H+]2 / [HA]
putting in the numbers you know for Ka and for [H+] 1.8x10-5= (0.003)2 / [HA]
rearranging gives
[HA] = (0.003)2 / 1.8x10-5 = 0.5 mol dm-3
So the acid solution is 0.5 mol dm-3
Hope from that example you can see that if you know two out of three of the following bits of information: [HA] pH Ka
than you can use the two equations too work out the third bit of information!
Sorry for the long post, hope it helped!
Thank you it helped a lot I just get a bit confused on the steps