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Chemistry · Ch 6 — Equilibrium

Di- and Polybasic Acids and Di- and Polyacidic Bases

6.11.6

Di- and Polybasic Acids and Di- and Polyacidic Bases

Polyprotic Acids: More Than One Proton to Give

So far we have dealt with acids that donate a single proton (H+\text{H}^+) per molecule — monoprotic acids like HCl\text{HCl} or CH3COOH\text{CH}_3\text{COOH}. But many important acids, such as oxalic acid (H2C2O4\text{H}_2\text{C}_2\text{O}_4), sulphuric acid (H2SO4\text{H}_2\text{SO}_4), and phosphoric acid (H3PO4\text{H}_3\text{PO}_4), contain more than one ionizable proton. These are called polybasic or polyprotic acids. A dibasic acid has two ionizable protons; a tribasic acid has three.

The key point is that these protons are not lost all at once. They are removed in a stepwise manner, and each step has its own equilibrium constant.

Stepwise Ionization of a Dibasic Acid H2X\text{H}_2\text{X}

Consider a general dibasic acid H2X\text{H}_2\text{X}. It ionizes in two distinct stages.

First ionization step:

H2X(aq)⇌H+(aq)+HX−(aq)\text{H}_2\text{X}(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{HX}^-(\text{aq})

The equilibrium constant for this step is called the first ionization constant, Ka1K_{a1}:

Ka1=[H+][HX−][H2X]K_{a1} = \frac{[\text{H}^+][\text{HX}^-]}{[\text{H}_2\text{X}]}

Second ionization step:

The anion HX−\text{HX}^- formed in the first step can itself lose a proton:

HX−(aq)⇌H+(aq)+X2−(aq)\text{HX}^-(\text{aq}) \rightleftharpoons \text{H}^+(\text{aq}) + \text{X}^{2-}(\text{aq})

The equilibrium constant for this step is the second ionization constant, Ka2K_{a2}:

Ka2=[H+][X2−][HX−]K_{a2} = \frac{[\text{H}^+][\text{X}^{2-}]}{[\text{HX}^-]}

For a tribasic acid like phosphoric acid (H3PO4\text{H}_3\text{PO}_4), there are three such steps and three ionization constants: Ka1K_{a1}, Ka2K_{a2}, and Ka3K_{a3}.

Watch out

A very common mistake is to think that Ka2K_{a2} is the equilibrium constant for the overall reaction H2X⇌2H++X2−\text{H}_2\text{X} \rightleftharpoons 2\text{H}^+ + \text{X}^{2-}. It is not. The overall constant for that reaction is the product Ka1×Ka2K_{a1} \times K_{a2}, but the stepwise constants are the ones that describe the actual, sequential loss of protons.

The Pattern of KaK_a Values

For any polyprotic acid, the successive ionization constants always decrease sharply:

Ka1>Ka2>Ka3 (if applicable)K_{a1} > K_{a2} > K_{a3} \text{ (if applicable)}

This makes physical sense. Removing the first proton from a neutral molecule is relatively easy. Removing a second proton from a negatively charged ion (HX−\text{HX}^-) is much harder because you are pulling a positive charge away from a species that already carries a negative charge. The electrostatic attraction makes the H+\text{H}^+ in HX−\text{HX}^- more tightly bound. The third step, if it exists, is even more difficult. …

Table 6.8The Ionization Constants of Some Common Polyprotic Acids (298 K)
AcidKa1K_{a_1}Ka2K_{a_2}Ka3K_{a_3}
Oxalic Acid5.9×10−25.9 \times 10^{-2}6.4×10−56.4 \times 10^{-5}
Ascorbic Acid7.4×10−47.4 \times 10^{-4}1.6×10−121.6 \times 10^{-12}
Sulphurous Acid1.7×10−21.7 \times 10^{-2}6.4×10−86.4 \times 10^{-8}
Sulphuric AcidVery large1.2×10−21.2 \times 10^{-2}
Carbonic Acid4.3×10−74.3 \times 10^{-7}5.6×10−115.6 \times 10^{-11}
Citric Acid7.4×10−47.4 \times 10^{-4}1.7×10−51.7 \times 10^{-5}4.0×10−74.0 \times 10^{-7}
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