3.4 Polyprotic acids and bases

Polyprotic Acids and Bases

Polyprotic acids and bases can donate or accept more than one proton per molecule, and they do so in a stepwise fashion. Each successive step is less favorable than the one before it because the molecule carries more charge, making it harder to lose (or gain) yet another proton.

This stepwise behavior matters for pH calculations, buffer design, and understanding biological systems like amino acid chemistry. The key idea throughout: treat each dissociation step separately, with its own equilibrium constant.

Polyprotic Acids and Bases

A polyprotic acid donates more than one $H^+$ per molecule. Common examples include sulfuric acid ($H_2SO_4$, diprotic) and phosphoric acid ($H_3PO_4$, triprotic).

A polyprotic base accepts more than one $H^+$ per molecule. Ethylenediamine ($H_2NCH_2CH_2NH_2$) is a classic example because it has two nitrogen lone pairs that can each pick up a proton.

Each successive dissociation or protonation step is less favorable than the previous one. Why? After the first proton leaves an acid, the remaining species carries a negative charge. Pulling a second proton away from a negatively charged ion takes more energy than pulling one from a neutral molecule.

Stepwise Dissociation Reactions

Polyprotic acids dissociate one proton at a time, producing a new conjugate base at each step. Here's phosphoric acid as the classic triprotic example:

1. $H_3PO_4 + H_2O \rightleftharpoons H_2PO_4^- + H_3O^+$ (phosphoric acid → dihydrogen phosphate)

2. $H_2PO_4^- + H_2O \rightleftharpoons HPO_4^{2-} + H_3O^+$ (dihydrogen phosphate → hydrogen phosphate)

3. $HPO_4^{2-} + H_2O \rightleftharpoons PO_4^{3-} + H_3O^+$ (hydrogen phosphate → phosphate)

Polyprotic bases accept protons stepwise, forming conjugate acids. For ethylenediamine:

1. $H_2NCH_2CH_2NH_2 + H^+ \rightleftharpoons {}^+H_3NCH_2CH_2NH_2$ (one nitrogen protonated)
2. ${}^+H_3NCH_2CH_2NH_2 + H^+ \rightleftharpoons {}^+H_3NCH_2CH_2NH_3^+$ (both nitrogens protonated)

Notice the pattern: each step produces a species with one more (or one fewer) proton, and each intermediate species is itself an acid and a base (amphoteric).

Ka Values for Dissociation Steps

Each dissociation step has its own acid dissociation constant: $K_{a1}$, $K_{a2}$, $K_{a3}$, and so on. These values always decrease with each step because removing a proton from an increasingly negative ion is progressively harder.

The general form of a $K_a$ expression is:

$K_a = \frac{[H^+][A^-]}{[HA]}$

For phosphoric acid, the three constants are:

• $K_{a1} = \frac{[H^+][H_2PO_4^-]}{[H_3PO_4]} = 7.1 \times 10^{-3}$
• $K_{a2} = \frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]} = 6.3 \times 10^{-8}$
• $K_{a3} = \frac{[H^+][PO_4^{3-}]}{[HPO_4^{2-}]} = 4.2 \times 10^{-13}$

The corresponding $pK_a$ values ($pK_a = -\log K_a$) are approximately 2.15, 7.20, and 12.38. Since $K_a$ decreases at each step, $pK_a$ increases: $pK_{a1} < pK_{a2} < pK_{a3}$.

A useful rule of thumb: if $K_{a1}$ is at least 1000 times larger than $K_{a2}$ (which it usually is), the first dissociation dominates and you can often ignore the later steps when calculating pH.

pH of Polyprotic Solutions

Because $K_{a1}$ is typically much larger than $K_{a2}$, the first dissociation produces nearly all the $H^+$ in solution. This lets you simplify the calculation by treating only the first step.

Procedure for calculating pH of a polyprotic acid solution:

1. Write the first dissociation equilibrium and its $K_{a1}$ expression.
2. Set up an ICE table using the initial acid concentration and solve for $[H^+]$ from $K_{a1}$ alone.
3. Calculate $pH = -\log[H^+]$.
4. If you need the concentration of later species (like $HPO_4^{2-}$), plug your $[H^+]$ from step 2 into the $K_{a2}$ expression and solve. For most polyprotic acids, $[A^{2-}] \approx K_{a2}$ numerically, which is a handy shortcut.

Example: Find the pH of 0.10 M $H_3PO_4$.

Since $K_{a1} = 7.1 \times 10^{-3}$ is not negligibly small compared to 0.10 M, you should use the quadratic or the approximation carefully. Using the simplified square-root approximation:

$[H^+] \approx \sqrt{K_{a1} \times C_0} = \sqrt{(7.1 \times 10^{-3})(0.10)} = \sqrt{7.1 \times 10^{-4}} \approx 0.027 \text{ M}$

$pH \approx -\log(0.027) \approx 1.57$

(A more exact treatment with the quadratic gives $[H^+] \approx 0.024$ M and $pH \approx 1.62$, since the 5% assumption is borderline here. Your instructor may want you to check that assumption.)

The second and third dissociations contribute negligibly to $[H^+]$ because $K_{a2}$ and $K_{a3}$ are so much smaller. You'd only need them if you wanted the concentrations of $HPO_4^{2-}$ or $PO_4^{3-}$ specifically.