[Matt17]

Hello!
The last problem with pH. Solutions of NaH₂PO₄ and Na₂HPO₄ (both are 0,1 mol/dm³) are mixed together in ratio 1 : 1. Values given:
K_a(H₃PO₄) = 6*10^-3
K_a(H₂PO₄^- = 6*10^-8
K_a(HPO₄^2-) = 5*10^-13

I tried to solve it in the following way:
Dissociation:
NaH₂PO₄  Na⁺ + H₂PO₄^-
Na₂HPO₄  2Na⁺ + HPO₄^2-
And then, according to Bronsted-Lowry's theory:
H₂PO₄^- + H₂O  HPO₄^2- + H₃O⁺

And now I think that I can calculate the concentration of H₂PO₄^- and HPO₄^2- after dissociation using ICE table, or maybe dissociation constants (though I don't know how to use them in this case), and than use it to calculate the concentraltion of H₃O⁺ from K_a (H₂PO₄^-).
I don't know if this is correct so please check me.

[Borek]

NaH₂PO₄ and Na₂HPO₄ (both are 0,1 mol/dm³) are mixed together in ratio 1 : 1

Plug and chug with a Henderson-Hasselbalch equation. First example on the buffer calculation page that I already linked to. It is just a matter of selecting the correct pKa value (the one related to the acid and its conjugate base dominating the solution).

[Matt17]

So it seems that I can assume that:
[tex]\frac{HPO_4^{2-}}{H_2PO_4^-} = 1[/tex] can I?
Which means that
[H⁺] = K_a(H₂PO₄^-)
Isn't it to easy?

Well, it seems OK because I read on the Internet that this buffer has always pH = 7,22 and from these calculations we also have pH = 7,22.

[Borek]

OK. Checked with the pH calculator built into Buffer Maker:

http://www.youtube.com/watch?v=ReKHRo7I9x0

By mistake I used 1M solutions instead of 0.1M, but it doesn't change anything. pH equals pK_a2 (value of pK_a2 is necessarily identical to the one you were given, as it was taken from the program database). At the beginning I had to change the defaults so that the program sums the volumes of solutions mixed and ignores ionic strength of the solution.

It is application of the same general rule as in the case of KF/HCl solution you asked earlier - if the concentrations of acid and conjugate base are identical, pH=pKa.

Note - it also means pKa=pH at half titration!

[Matt17]

Thank you very much.