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


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


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

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:

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.


1 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 [CO32-]), the degree of dissociation to be α and the degree of hydrolysis to be η At equilibrium: [CO32-] = c (1 - α) [H2CO3] = c (1 - η) Since we know α = η, 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.

  • $\begingroup$ 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... $\endgroup$
    – Abhigyan
    Sep 3, 2017 at 6:10
  • $\begingroup$ I ask you again, for an answer involving the Equilibrium Constant... I'd be really benefited by it $\endgroup$
    – Abhigyan
    Sep 3, 2017 at 16:47
  • 1
    $\begingroup$ That's not too much of a problem. K(h) = c eta^2 / 1 - eta (Hydrolysis) K(a) = c alpha^2 / 1 - alpha (Dissociation for the acid) Since we know the relation b/w alpha and eta, it should not be too much of a problem to incorporate this into the answer. $\endgroup$ Sep 4, 2017 at 7:57

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