Hydrolysis of salts containing amphiprotic anion I am trying to understand the derivation of pH of salt containing amphiprotic anions, but I am facing some difficulty:   

Ionisation:  

$\ce{HCO3^- + H2O &lt;=&gt; CO3^2- + H3O+};\quad K_\mathrm{a_2} $ 

Hydrolysis: 

$\displaystyle\ce{HCO3- + H2O &lt;=&gt; H2CO3 +OH-};\quad \frac{K_\mathrm{w}}{K_\mathrm{a_1}}$    

Taking the assumption: $\text{Degree of ionisation} = \text{degree of hydrolysis}$, or, $[\ce{CO3^2-}]=[\ce{H2CO3}]$, an approximate calculation of pH can be done by the relation:
$$\mathrm{pH}(\ce{HCO3-})=\frac{\mathrm{p}K_\mathrm{a_1}+\mathrm{p}K_\mathrm{a_2}}{2}$$

I don't understand the assumption $[\ce{CO3^2-}]=[\ce{H2CO3}]$. We know that $[\ce{CO3^2-}]=[\ce{H3O+}]$ and $[\ce{OH-}]=[\ce{H2CO3}]$. But if $[\ce{CO3^2-}]=[\ce{H2CO3}]$ then $[\ce{H3O+}]=[\ce{OH-}]$. Hence pH should be 7. I don't know where my logic is going wrong and it would be great if someone could help me clarify the derivation.
If the degree of ionisation is equal to the degree of hydrolysis, then (assuming you take the same initial molar concentration of [HCO3-]):

Let's say the initial concentration was 'c' (with no [H2CO3] and no [CO32-]), the degree of dissociation to be &alpha; and the degree of hydrolysis to be &eta; 

At equilibrium:

[CO32-] = c (1 - &alpha;)

[H2CO3] = c (1 - &eta;)

Since we know &alpha; = &eta;, we can say that [CO32-] = [H2CO3] 

Only because you made the assumption that degree of dissociation = degree of ionisation, do you get pH = 7.

But, since we assume water to always be in excess, even in a real-world situation, difference in pH (from 7) would not be significant.