Calculating Equilibrium Concentrations of Polyprotic Acids

Polyprotic acids can lose more than one proton. The first proton's dissociation may be denoted as K_a1 and the constants for successive protons' dissociations as K_a2, etc. Common polyprotic acids include sulfuric acid (H₂SO₄), and phosphoric acid (H₃PO₄).

When determining equilibrium concentrations for different ions produced by polyprotic acids, equations can become complex to account for the various components. For a diprotic acid for instance, we can calculate the fractional dissociation (alpha) of the species HA^- using the following complex equation:

Equation for finding the fractional dissociation of HA-

The above concentration can be used if pH is known, as well as the two acid dissociation constants for each dissociation step; oftentimes, calculations can be simplified for polyprotic acids, however.

We can simplify the problem, depending on the polyprotic acid. The following examples indicate the mathematics and simplifications for a few polyprotic acids under specific conditions.

Diprotic Acids With a Strong First Dissociation Step

As we are already aware, sulfuric acid's first proton is strongly acidic and dissociates completely in solution:

H₂SO₄ → H⁺ + HSO₄^-

However, the second dissociation step is only weakly acidic:

$HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}$  K_a2 = 1.20x10^-2   pKa2 = 1.92

Because the first dissociation is so strong, we can assume that there is no measurable H₂SO₄ in the solution, and the only equilibrium calculations that need be performed deal with the second dissociation step only.

Determining Predominant Species From pH and pK_a

Phosphoric acid, H₃PO₄, has three dissociation steps:

$H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^-$  pK_a1 = 2.12

$H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}$   pK_a2 = 7.21

$HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-}$  pK_a3 = 12.67

Thus, in an aqueous solution of phosphoric acid, there will theoretically be seven ionic and molecular species present: H₃PO₄, H₂PO₄^-, HPO₄^2-, PO₄^3-, H₂O, H⁺, and OH^-.

At a pH equal to the pK_a for a particular dissociation, the two forms of the dissociating species are present in equal concentrations, due to the following mathematical observation. Take for instance the second dissociation step of phosphoric acid, which has a pK_a2 of 7.21:

$pK_{a2}=-log\left(\frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]}\right)=7.21$

$pH=-log[H^+]=7.21$

By the property of logarithms, we get the following:

$pH-pK_{a2}=-log\left(\frac{[HPO_4^{2-}]}{[H_2PO_4^-]}\right)=0$

$\frac{[HPO_4^{2-}]}{[H_2PO_4^-]}=1$

Thus, when pH = pKa2, we have the ratio [HPO₄^2-]/[H₂PO₄^-] = 1.00; in a near-neutral solution, H₂PO₄^- and HPO₄^2- are present in equal concentrations. Very little undissociated H₃PO₄ or dissociated PO₄^3- will be found, as is determined through similar equations with their given K_a's.

The only phosphate species that we need to consider near pH = 7 are H₂PO₄^- and HPO₄^2-. Similarly, in strongly acidic solutions near pH = 3, the only species we need to consider are H₃PO₄ and H₂PO₄^-. As long as the pK_a values of successive dissociations are separated by three or four units (as they almost always are), matters are simplified. We need only consider the equilibrium between the two predominant acid/base species, as determined by the pH of the solution.

Phosphoric acid

The chemical structure of phosphoric acid indicates it has three acidic protons.

Diprotic Acids With a Very Weak Second Dissociation Step

When a weak diprotic acid such as carbonic acid, H₂CO₃, dissociates, most of the protons present come from the first dissociation step:

$H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$  pK_a1 = 6.37

Since the second dissociation constant is smaller by four orders of magnitude (pK_a2 = 10.25 is larger by four units), the contribution of hydrogen ions from the second dissociation will be only one ten-thousandth as large. Consequently, the second dissociation has a negligible effect on the total concentration of H⁺ in solution, and can be ignored.