# pH Calculation - solutions of polybasic acids (approximate)

A polyprotic acid is an acid which contains in its molecule more than one acid hydrogen, ionizable.

Take for example H 2 S, a 'diprotic acid:

$H_2S&space;+&space;H_2O&space;\rightleftharpoons&space;HS^-&space;+&space;H_3O^+$$K_{a1}&space;=&space;\frac{[H_3O^+][HS^-]}{[H_2S]}&space;=&space;1,1\cdot10^{-7}$

$HS^-&space;+&space;H_2O&space;\rightleftharpoons&space;S^{2-}&space;+&space;H_3O^+$$K_{a2}&space;=&space;\frac{[H_3O^+][S^{2-}]}{[HS^-]}&space;=&space;1,3&space;\cdot&space;10^{-14}$

The overall expression will be:

$H_2S&space;+&space;H_2O&space;\rightleftharpoons&space;S^{2-}&space;+&space;2H_3O^+$$K_{tot}&space;=&space;\frac{[H_3O^+]^2[S^{2-}]}{[H_2S]}$

The time constant of the overall reaction is not more than the result of the product of the two dissociation constants:

$K_{tot}&space;=&space;\frac{[H_3O^+]^2[S^{2-}]}{[H_2S]}&space;=&space;K_1\cdot&space;K_2$

When it has to do with a polyprotic acid usually the first acid dissociation constant is much larger (several orders of magnitude) of the successive and for this reason contributes more to the pH value. Considering all $\bg_black&space;[H_3O^+]$ deriving from the first dissociation.

In principle, for the calculation of pH we must take into account two types of polybasic acids:

• 1 Those whose first dissociation is complete (K a1 very large)
• 2 → Those whose first dissociation is partial (K relatively small a1)

case 1

0.1 M solution of H 2 SO 4. Since the first dissociation

$\bg_white&space;H_2SO_4&space;+&space;H_2O&space;\rightleftharpoons&space;HSO_4^-&space;+&space;H_3O^+$

is definitely complete, you can say that:

$\bg_white&space;[H_3O^+]&space;=&space;C_a&space;=&space;0,1&space;M$

where C a is the initial molar concentration of acid. The pH will be:

$\bg_white&space;\bg_white&space;pH&space;=&space;-log&space;[H_3O^+]&space;=&space;1$

case 2

0.1 M solution of H 2 S. As in "Case 1" we neglect the second dissociation. As regards the first dissociation instead, H 2 S behaves as an acid weak. Therefore:

$\bg_white&space;\bg_white&space;\bg_white&space;[H_3O^+]&space;=&space;\sqrt{K_{a1}&space;\cdot&space;C_a}&space;=&space;\sqrt{1,1\cdot10^{-7}\cdot0,1}&space;=&space;1,05&space;\cdot&space;10^{-4}$

the pH will be:

$\bg_white&space;pH&space;=&space;-log[H_3O^+]&space;=&space;3,98$

In both cases, we have resorted to approximate solutions, if you want to know how to calculate systematically the pH of polybasic acids → systematic calculation of pH of polybasic acids