pH of 0.05 M NaH2PO3 What is the pH of a $\pu{0.050 M}$ solution of $\ce{NaH2PO3}$? I am given that $\mathrm{p}K_\mathrm{a1} = 2.0$ and $\mathrm{p}K_\mathrm{a2} = 6.58$.

I am getting two different answers, $3.94$ and $4.30$, using two different methods that I have learned:

Method 1

$\ce{H2PO3-}$ can either be protonated or deprotonated. The protonation reaction has

$$K = \frac{10^{-14}}{10^{-2}} = 10^{-12},$$

while the deprotonation reaction has

$$K = 10^{-6.58} = 2.63\times 10^{-7}.$$

Since the protonation reaction is significantly less favorable than the deprotonation ($K$ ratio is less than $1000$), we can ignore the protonation and only focus on the deprotonation. This yields:

$$10^{-6.58} = \frac{x^2}{0.05-x} \to x = [\ce{H+}] = \pu{1.15e-4 M},$$

$$\mathrm{pH} = -\log(x) = 3.94.$$

Method 2

With a diprotic acid that can either be protonated or deprotonated, the $\mathrm{pH}$ is given by the following approximation:

$$pH = 1/2(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a1}),$$

$$\mathrm{pH} = 1/2 \times (2.0 + 6.58) = 4.29.$$

This is the correct answer as presented in the text. Why is the first method invalid?
The problem with method 1 is that you discard the protonation process, which effectively would increase pH as it produces $\ce{OH-}$ ions.

The criteria of $K_1 &gt; 1000K_2$ is applicable when you have a polyprotic acid in its completely (de)protonated state. The ionisation of H2A produces only a small amount of HA- and K2 is small as well, so ionisation of HA- would be negligible and we can ignore it.

However with an ampholytic species, while K2 is smaller than K1, the initial reactant is the same (HA-) and it is in high concentration, so both processes occur. The formula in method 2 is the result of this consideration.