#### Learning Objective

- Solve equilibrium problems using the appropriate approximations for weak and strong polyprotic acids.

#### Key Points

- Polyprotic acids contain multiple acidic protons that can sequentially dissociate from the compound with unique acid dissociation constants for each proton.
- Due to the variety of possible ionic species in solution for each acid, precisely calculating the concentrations of different species at equilibrium can be very complicated.
- Certain simplifications can make the calculations easier; these simplifications vary with the specific acid and the solution conditions.

#### Terms

- ionic specieschemical species with a residual charge; in acid-base equilibria, the charge resulting from the loss or addition of electrons from chemical compounds
- equilibriumthe state of a reaction in which the rates of the forward and reverse reactions are the same
- polyprotic acidsan acid with multiple acidic protons

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 (H_{2}SO_{4}), and phosphoric acid (H_{3}PO_{4}).

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:

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:

H_{2}SO_{4} → H^{+} + HSO_{4}^{–}

However, the second dissociation step is only weakly acidic:

[latex]HSO_4^- \rightleftharpoons H^+ + SO_4^{2-}[/latex] K_{a2} = 1.20×10^{-2} pKa2 = 1.92

Because the first dissociation is so strong, we can assume that there is no measurable H_{2}SO_{4} 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, H_{3}PO_{4}, has three dissociation steps:

[latex]H_3PO_4 \rightleftharpoons H^+ + H_2PO_4^-[/latex] pK_{a1} = 2.12

[latex]H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}[/latex] pK_{a2} = 7.21

[latex]HPO_4^{2-} \rightleftharpoons H^+ + PO_4^{3-}[/latex] pK_{a3} = 12.67

Thus, in an aqueous solution of phosphoric acid, there will theoretically be seven ionic and molecular species present: H_{3}PO_{4}, H_{2}PO_{4}^{–}, HPO_{4}^{2-}, PO_{4}^{3-}, H_{2}O, 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:

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

[latex]pH=-log[H^+]=7.21[/latex]

By the property of logarithms, we get the following:

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

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

Thus, when pH = pKa2, we have the ratio [HPO_{4}^{2-}]/[H_{2}PO_{4}^{–}] = 1.00; in a near-neutral solution, H_{2}PO_{4}^{–} and HPO_{4}^{2-} are present in equal concentrations. Very little undissociated H_{3}PO_{4} or dissociated PO_{4}^{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 H_{2}PO_{4}^{–} and HPO_{4}^{2-}. Similarly, in strongly acidic solutions near pH = 3, the only species we need to consider are H_{3}PO_{4} and H_{2}PO_{4}^{–}. 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.

## Diprotic Acids With a Very Weak Second Dissociation Step

When a weak diprotic acid such as carbonic acid, H_{2}CO_{3}, dissociates, most of the protons present come from the first dissociation step:

[latex]H_2CO_3 \rightleftharpoons H^+ + HCO_3^-[/latex] 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.