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.


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.

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  • $\begingroup$ 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 Chattopadhyay Sep 3 '17 at 6:10
  • $\begingroup$ I ask you again, for an answer involving the Equilibrium Constant... I'd be really benefited by it $\endgroup$ – Abhigyan Chattopadhyay Sep 3 '17 at 16:47
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    $\begingroup$ 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. $\endgroup$ – SupremeFiend Sep 4 '17 at 7:57

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