# pH Calculation - solutions of polyprotic acids (complete)

Polyprotic acids are, as the name says, those acids which in their molecule contain more than one acidic hydrogen. Polyprotic acids ionize in more stages.

Let's take for example H2S, 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 is:

$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 overall reaction constant is just the result of the product of the two dissociation constants:$K_{tot}&space;=&space;\frac{[H_3O^+]^2[S^{2-}]}{[H_2S]}&space;=&space;K_1\cdot&space;K_2$

Generally, the first dissociation constant is way larger than the second dissociation constant, and right for this reason the majority of $[H_3O^+]{\color{Blue}&space;}$ in solution derives from the first dissociation.

There are cases in which the first dissociation is only partial (Ka1 relatively small) and other cases where the first dissociation is practically quantitative (Ka1 is great). These two different contexts require two different approaches for what concern the calculation of pH.

case 1 → first dissociation not quantitative

We're gonna start by analyzing the first case, which is right the one of hydrogen sulfide. Lets' suppose we have an aqueous solution of 0.1 M H2S .

We firstly have to calculate $[H_3O^+]{\color{Blue}&space;}$ which derives from the first dissociation:
$H_2S&space;+&space;H_2O&space;\rightleftharpoons&space;HS^-&space;+&space;H_3O^+$$K_{a1}&space;=&space;1,1\cdot10^{-7}$

In order to do this, we consider $H_2S$ as a 'weak monoprotic acid and we use the approximate formula for the calculation of $[H_3O^+]{\color{Blue}&space;}$ :

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

We have now to calculate $[H_3O^+]{\color{Blue}&space;}$ resulting from the second dissociation  equilibrium:

Substituting in the expression of the second one has dissociation constant:

$\bg_white&space;K_{a2}&space;=&space;\frac{[H_3O^+][S^{2-}]}{[HS^-]}&space;=&space;\frac{[1,4\cdot&space;10^{-5}&space;+&space;x][x]}{[1,4\cdot&space;10^{-5}&space;-&space;x]}$

It is obtained an expression of second degree of the form:

$\bg_white&space;x^2&space;+&space;(1,1\cdot&space;10^{-14}&space;+&space;1,4\cdot&space;10^{-5})x&space;-&space;(1,1\cdot&space;10^{-14}&space;\cdot&space;1,4\cdot&space;10^{-5})&space;=&space;0$

You are obtained by solving for x:

$\bg_white&space;x&space;\approx&space;10^{-14}$

Since this value is negligible with respect to $[H_3O^+]$ deriving from the first dissociation, you can be considered

$\bg_white&space;[H_3O^+]_{tot}&space;\cong&space;[H_3O^+]_{1^0diss}$

It will have that

$\bg_white&space;pH&space;=&space;-log&space;[H_3O^+]&space;=&space;-log&space;(1,4\cdot&space;10^{-5})&space;=&space;3,98$

Which it is precisely the same result that you get with the approximate calculation.