URGENT!!! Acid Dissociation Constant (Ka)!!

An aqueous solution of the weak acid HA, which has an initial pH of 2.50, is titrated against sodium hydroxide. When the pH reaches 4.30, [HA] = [A-]. Calculate the value of the acid dissociation constant of HA and hence the orginal concentration of the acid in solution [7]

I get silly answers, an extremely small concentration so could someone work through the problem with me.

As it says when the pH 4.30 [HA] = [A-], I thought that Ka = [H+]. So I converted the pH to a concentration. I then had determined the value for Ka.

I then thought that as:

Ka = [H+]^2/HA

So [HA] = [H+]^2 / Ka

We know [H+] to start with as we are given the pH value of 2.50 and we know Ka. So we should square the concentration of H+ and divide by Ka to get the orginal concentration of HA.

PS: I think I am wrong because of the silly answer I get and the fact that doesn't seem like work that is worth 7 marks. Also sorry for being a pain :biggrin:

Reply 1

john !!

14

Since Ka = [H+][A-]/[HA], at the point where [A-]=[HA], the equations simplifies to Ka = [H+]

You're told that this point is at a pH of 4.30. You can work out [H+] from the equation:
[H+] = 10^(-pH)
[H+] = 10^(-4.3) = 5.01*10^-5 mol dm^-3
Ka = 5.01*10^-5 mol dm^-3

Ka = [H+][A-]/[HA], since in the original concentration there is no alkali, [H+]=[A-], so Ka = [H+]^2/[HA] like you said. Then [HA]=[H+]^2/Ka.

[H+] = 10^(-2.5) = 3.16*10^-3 mol dm^-3, Ka = 5.01*10^-5 mol dm^-3, so
[HA] = (3.16*10^-3)^2/(5.01*10^-5) = 0.200 mol dm^-3 (3s.f.)

What you did looks about right.

Reply 2

supreme

OP

4

mik1a

Since Ka = [H+][A-]/[HA], at the point where [A-]=[HA], the equations simplifies to Ka = [H+]

You're told that this point is at a pH of 4.30. You can work out [H+] from the equation:
[H+] = 10^(-pH)
[H+] = 10^(-4.3) = 5.01*10^-5 mol dm^-3
Ka = 5.01*10^-5 mol dm^-3

Ka = [H+][A-]/[HA], since in the original concentration there is no alkali, [H+]=[A-], so Ka = [H+]^2/[HA] like you said. Then [HA]=[H+]^2/Ka.

[H+] = 10^(-2.5) = 3.16*10^-3 mol dm^-3, Ka = 5.01*10^-5 mol dm^-3, so
[HA] = (3.16*10^-3)^2/(5.01*10^-5) = 0.200 mol dm^-3 (3s.f.)

What you did looks about right.

Thanks I must have pressed a wrong button because as you said my working out seems sound although 7 marks seems to be very generous don't you think ? :biggrin:

URGENT!!! Acid Dissociation Constant (Ka)!!

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