So far we have studied acids that had only one ionizable hydrogen (monoprotic acids). However, there are acids that have several ionizable hydrogens, called polyprotic acids. Some examples are acids: $H_3PO_4 \;H_2SO_4$

The $H_3PO_4$ molecule is a triprotic acid, with three ionizable hydrogens. The acidity of the hydrogens decreases rapidly, the first being much more acid than the second, and the latter than the third. The reason is that as we remove hydrogens, the molecule acquires a negative charge and the conjugate base is increasingly unstable (strong).

Let's see the three equilibrium in the ionization of phosphoric acid:

$H_3PO_4 + H_2O \rightleftharpoons H_2PO_4^- +H_3O^+$ $K_{a1}=7.1x10^{-3}$

$H_2PO_4^- +H_2O\rightleftharpoons HPO_4^{2-} +H_3O^+$ $K_{a2}=6.3x10^{-8}$

$HPO_4^{2-}+H_2O\rightleftharpoons PO_4^{3-} +H_3O^+$ $K_{a3}=4.2x10^{-13}$

As can be seen, the first ionization is much more displaced to the right than the rest. Being able to neglect, with a very good approximation, the protons from the second and third ionization, calculating the pH from the protons generated in the first.

Let's see an example: Calculate the concentrations of the different ions in a 3 M solution of phosphoric acid.

$H_3PO_4 + H_2O \rightleftharpoons H_2PO_4^- +H_3O^+$

3 - x x x

$H_2PO_4^- +H_2O\rightleftharpoons HPO_4^{2-} +H_3O^+$

x - y and x+y

$HPO_4^{2-}+H_2O\rightleftharpoons PO_4^{3-} +H_3O^+$

y - z z x + y + z

Due to the large difference between the ionization constants x>>y>>z, we can neglect y in the sum x+y or difference xy . We can also neglect z in the differences yz or in the sum z+y+z.

We set the equilibrium constants for the three reactions:

\begin{equation}K_{a1}=\frac{[H_2PO_4^-][H_3O^+]}{H_3PO_4]}=\frac{x^2}{3-x}=7.1x10^{-3}\ end{equation}

Clearing: $x=[H_3O^+]=[H_2PO_4^-]=0.14M$

We set the equilibrium constant of the second ionization:

\begin{equation}K_{a2}=\frac{[HPO_4^{2-}][H_3O^+]}{H_2PO_4^-]}=\frac{y(x+y)}{xy}=6.3x10 ^{-8}\end{equation}

Neglecting y in front of x, we are left with:

\begin{equation}6.3x10^{-8}=\frac{y\cdot x}{x}=y=[HPO_4^{2-}]\end{equation}

To obtain z we proceed analogously:

\begin{equation}K_{a3}=\frac{[PO_4^{3-}][H_3O^+]}{[HPO_4^{2-}]}=\frac{z(x+y+z)} {yz}=4.2x10^{-13}\end{equation}

Considering that z <

\begin{equation}4.2x10^{-13}=\frac{z\cdot x}{y}\end{equation}

Replacing z and y by the values obtained above gives z.

$z=[PO_4^{3-}]=1.9x10^{-19}$