Hydrolysis of salts containing amphiprotic anion

Question

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.

Answer 1

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 [CO₃^2-]), the degree of dissociation to be α and the degree of hydrolysis to be η At equilibrium: [CO₃^2-] = c (1 - α) [H₂CO₃] = c (1 - η) Since we know α = η, we can say that [CO₃^2-] = [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.