# 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:

$\ce{HCO3^- + H2O <=> CO3^2- + H3O+};\quad K_\mathrm{a_2}$

Hydrolysis:

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

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.

• Can you provide an answer using the Hydrolysis Equilibrium Constant ($K_h$)? The derivation given in my book uses that too, so if I could understand the use of Equilibrium constant in this equation, I'd be able to apply it better... – AbhigyanC Sep 3 '17 at 6:10