Polyprotic acids Knowledge-Based Study

Study notes

1. What Are Polyprotic Acids?

   Acids are classified by the number of ionizable protons (hydrogen ions, $\text{H}^+$) they can donate per molecule in aqueous solution:

   • Monoprotic acids: Can donate only one proton per molecule (e.g., $\text{HCl}$, $\text{HNO}_3$, $\text{CH}_3\text{COOH}$).
   • Diprotic acids: Can donate up to two protons per molecule through successive steps (e.g., $\text{H}_2\text{SO}_4$, $\text{H}_2\text{CO}_3$, $\text{H}_2\text{C}_2\text{O}_4$).
   • Triprotic acids: Can donate up to three protons per molecule (e.g., $\text{H}_3\text{PO}_4$, $\text{H}_3\text{AsO}_4$, citric acid).

   Acids capable of donating more than one proton per molecule are collectively referred to as polyprotic acids. The loss of each proton occurs in a discrete, stepwise equilibrium rather than all at once.

   Successive Dissociation Equilibria

   Consider a general diprotic acid, $\text{H}_2\text{A}$. Its ionization in water takes place in two sequential steps, each with its own acid dissociation constant:

   First ionization step:

   $$\text{H}_2\text{A}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{HA}^-(aq) \quad K_{\text{a1}} = \frac{[\text{H}_3\text{O}^+][\text{HA}^-]}{[\text{H}_2\text{A}]}$$

   Second ionization step:

   $$\text{HA}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{A}^{2-}(aq) \quad K_{\text{a2}} = \frac{[\text{H}_3\text{O}^+][\text{A}^{2-}]}{[\text{HA}^-]}$$

   Why is $K_{\text{a1}}$ Always Greater Than $K_{\text{a2}}$?

   For any polyprotic acid, the successive ionization constants decrease drastically:

   $$K_{\text{a1}} \gg K_{\text{a2}} \gg K_{\text{a3}}$$

   This occurs because the first proton is removed from a neutral molecule ($\text{H}_2\text{A}$), whereas the second proton must be removed from a negatively charged anion ($\text{HA}^-$). Electrostatic attraction makes it significantly more difficult to separate a positively charged proton ($\text{H}^+$) from a negatively charged species, resulting in a substantially smaller equilibrium constant for subsequent steps (often by a factor of $10^4$ to $10^6$).

   Equilibrium Calculations for Weak Polyprotic Acids

   Because $K_{\text{a1}}$ is typically several orders of magnitude greater than $K_{\text{a2}}$, the hydronium ion concentration in a solution of a typical weak diprotic acid is governed almost entirely by the first ionization step.

   Key Approximations

   • Hydronium concentration and pH: We can calculate $[\text{H}_3\text{O}^+]$ by treating the solution as if it contains only the monoprotic weak acid $\text{H}_2\text{A}$ with constant $K_{\text{a1}}$. The contribution of $[\text{H}_3\text{O}^+]$ from the second dissociation step is negligible in comparison.
   • Intermediate species concentration: From the stoichiometry of the first step, $[\text{HA}^-] \approx [\text{H}_3\text{O}^+]$.
   • Divalent conjugate base concentration: By substituting $[\text{HA}^-] \approx [\text{H}_3\text{O}^+]$ into the expression for $K_{\text{a2}}$:

   $$K_{\text{a2}} = \frac{[\text{H}_3\text{O}^+][\text{A}^{2-}]}{[\text{HA}^-]} \approx \frac{[\text{H}_3\text{O}^+][\text{A}^{2-}]}{[\text{H}_3\text{O}^+]} = [\text{A}^{2-}]$$

   Therefore, for an unbuffered aqueous solution of a weak diprotic acid where $K_{\text{a1}} \gg K_{\text{a2}}$, the concentration of the fully deprotonated anion $[\text{A}^{2-}]$ is approximately equal to $K_{\text{a2}}$.

   Sulfuric Acid ($\text{H}_2\text{SO}_4$)

   Sulfuric acid is unique because it is a strong acid in its first dissociation and a weak acid in its second dissociation:

   • $\text{H}_2\text{SO}_4(aq) + \text{H}_2\text{O}(l) \rightarrow \text{H}_3\text{O}^+(aq) + \text{HSO}_4^-(aq)$ (complete ionization, $K_{\text{a1}} \to \infty$)
   • $\text{HSO}_4^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{SO}_4^{2-}(aq)$ ($K_{\text{a2}} = 1.2 \times 10^{-2}$)

   For moderately concentrated $\text{H}_2\text{SO}_4$ solutions (greater than $0.1\text{ M}$), $[\text{H}_3\text{O}^+]$ is predominantly determined by the first step. However, at lower concentrations, the second dissociation contributes significantly to total $[\text{H}_3\text{O}^+]$ and must be solved using an equilibrium expression.

   Amphiprotic Salts

   Intermediate ions such as hydrogen carbonate ($\text{HCO}_3^-$) or dihydrogen phosphate ($\text{H}_2\text{PO}_4^-$) are amphiprotic—they can act either as an acid (donating a proton) or as a base (accepting a proton). For an amphiprotic intermediate $\text{HA}^-$, the approximate pH of its salt solution is independent of initial concentration (provided the solution is reasonably concentrated) and is given by:

   $$\text{pH} \approx \frac{\text{p}K_{\text{a1}} + \text{p}K_{\text{a2}}}{2}$$