pH = 14 - pOH = 14 + lg x = 12.977 ~ 13
This result is much worse than the previous one - and it is so obvious I don't understand how you have overlooked it. H2
is an acid - weak, but acid. Yet you have found the pH to be 13...
AWK referred rather to the page 29 of the pdf.
And here is my $.02, similar to the approach from the pdf, but different in one important aspect:
assume H2A is our acid, HA, A and H being ions (charges neglected to speed up writing):
Ka1 = [H][HA]/[H2A]
Ka2 = [H][A]/[HA]
three unknowns, two equations, it doesn't look good, but lets forget about the problem for now. Solve first equation for [HA]
[HA] = Ka1[H2A]/[H]
insert into second equation:
Ka2 = [H]^2[A]/(Ka1[H2A])
solve for [H]
[H] = sqrt(Ka1 Ka2 [H2A]/[A])
What is important here is the fact, that this is very similar to the equation from the page 29 of the pdf, but it also clearly shows that assumptions done in pdf boil down to one - [H2A] = [A]. If so
[H] = sqrt(Ka1 Ka2)
pH = (pKa1+pKa2)/2
For the phosphoric acid situation is exactly the same - third proton behaves like nailed and Ka3 doesn't disturb above equlibrium.
How far is this from the real results?
Not too far. pH = 4.7 if it is calculated as avearge of pKa1 and pKa2. If calculated using BATE for different NaH2
and so on.