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# Weak acid calculation? watch

1. I'm doing some questions from my text book and I actually can't find any answers. But my Ph calculation on part C looks very wrong.What have I done wrong?
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2. You seem to have found Ka from the pKa, then used that Ka to find pKa again - you've then used the original pKa instead of Ka
3. You use the pKa value of 4.32 rather than the Ka value.

Ka = 10-pKa

Also, when working out the new concs of HA and A-, you should have just realised that HA is a 1:4 dilution and a A- is 3:4 dilution. Notice that the conc of HA is 1/4 what it was before adding the A-
4. (Original post by Pigster)
You use the pKa value of 4.32 rather than the Ka value.

Ka = 10-pKa

Also, when working out the new concs of HA and A-, you should have just realised that HA is a 1:4 dilution and a A- is 3:4 dilution. Notice that the conc of HA is 1/4 what it was before adding the A-
(Original post by h3rmit)
You seem to have found Ka from the pKa, then used that Ka to find pKa again - you've then used the original pKa instead of Ka
Ah okay thank you
5. Actually, question a.) the Ka= [H+] because a [HA] and [A-] cancel out as both their conc is 0.1moldm-3

Since Ka= [H+] and and Ka= 4.786x10-5

Then [H+] also = 4.786x10-5
So ph = -log[H+]
-log[4.786x10-5]
Ph= 4.32

Do get my reasoning? Why would this be wrong?
6. (Original post by Onthedancefloor)
Actually, question a.) the Ka= [H+] because a [HA] and [A-] cancel out as both their conc is 0.1moldm-3

Since Ka= [H+] and and Ka= 4.786x10-5

Then [H+] also = 4.786x10-5
So ph = -log[H+]
-log[4.786x10-5]
Ph= 4.32

Do get my reasoning? Why would this be wrong?
It would me much simpler to just note that:

Ka = [H+] x [A-]/[HA] = [H+]
-log Ka = -log [H+] i.e. pKa (given in Q) = pH

This works when [HA] = [A-]. The easiest way to do this is to get a weak acid and add half the number of mol of OH- needed to fully neutralise it.

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