Challenging Buffer Question?

Question

I'll give you the question first and I'll show you what I came to....sooo....

A buffer solution is prepared by dissolving NaA and HA in pure water. The Sum of the molarities of NaA and Ha is 2.25 and the pH is 7.50. The Ka for HA is 3.0 x 10^-8. Calculate the molarities of NaA and HA in the solution.
_______________________________

HA + H_2O <=> H_3O^+ + A^- (*NaH*)

pH = pka + log [ A^- / HA]
pka = log^(-Ka) => pka = log^(-3 x 10^-8)
pka= 1

7.50 = 1 + log [ A^- / HA]
6.50 = log [ A^- / HA] => 10^6.5 = [ A^- / HA]
[ A^- / HA] = 3.162 x 10^6
[A^-] + [HA] = 2.50
x + x = 2.50?

then I get lost in translation.... please help.

Answer 1

pKa = - log 3.0 x 10^-8 = 7.52
pH = pKa + log [A-] / [HA]
7.50 = 7.52 + log [A-]/ [HA]
- 0.02 = log [A-] / [HA]
10^-0.02 = [A-]/ [HA]
0.955 = [A-] / [HA]
By the Handerson-Hasselbalch 's equation we have obtained the ratio between A- and HA

We know that [HA] + [A-] = 2.25.
To get the molarities we have to solve the system :

[ A-] / [HA] = 0.955
[HA] + [A-] = 2.25

[A-] = 0.955 [HA]
[HA] = 2.25 - 0.955 [HA]

[A-] = 0.955[HA]
[HA] + 0.955 [HA] = 2.25

[A-] = 0.955 [HA]
1.955 [HA] = 2.25

[HA] = 1.15 M
[A-] = 1.099 M

Answer 2

The long answer is as follows. As your equation is written (acid dissociation):
[H+] [A-]/[HA] = Ka (at equilibrium)
We know H+ and Ka. By the concept of equilibrium, we can select an initial starting point different from the one that is actually done as long as we get to the final end point. So we select that INITIALLY HA is 2.25 M and strong base is added to it until the pH determined is reached. Then we can set up the problem as
[H+][A-]/[HA-A-]= Ka.
Substituting in the [H+], which for pH 7.5 is 3.1x10-8, for all intent purposes [HA-A-]=[A-] or HA and A- are each 1.125.

The short answer is, of course, if [H+]=Ka, [HA]= [A-]. Unless you know what you are doing with pH and pKa and logs, stay away from them.