12.2 Acid-Base Equilibria (Ka, Kb), Polyprotic Acids & Salt Hydrolysis

12.2 Acid-Base Equilibria (Ka, Kb), Polyprotic Acids & Salt Hydrolysis

Quick Summary: Acid-base strength is determined by molecular structure, bond dissociation energetics, central atom electronegativity, and conjugate anion resonance stabilization. Binary acid strength increases across a period with electronegativity and down a group as bond strength weakens ($\text{HF} \ll \text{HCl} < \text{HBr} < \text{HI}$). Oxoacid acidity scales with central atom electronegativity and terminal oxygen count. Polyprotic acids dissociate stepwise ($K_{a1} \gg K_{a2} \gg K_{a3}$); solution pH depends almost exclusively on $K_{a1}$, while divalent anion concentration $[\text{A}^{2-}]$ equals $K_{a2}$. Salt hydrolysis occurs when dissolved ions perturb water's autoionization, generating acidic, basic, or neutral aqueous solutions based on conjugate acid-base equilibria and hydrated cation charge density.

---

1. Equilibrium Formulations for Weak Acids ($K_a$) and Weak Bases ($K_b$)

Proton transfer equilibria in water are defined by acid and base dissociation constants: $\text{HA}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{A}^-(aq) \implies K_a = \frac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]}$ $\text{B}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{BH}^+(aq) + \text{OH}^-(aq) \implies K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}$ Higher $K_a$ or $K_b$ values signify stronger electrolytes with lower $\text{p}K_a$ or $\text{p}K_b$ values. In weak base solutions, $[\text{OH}^-] \approx \sqrt{K_b [\text{B}]_0}$, giving $\text{pOH}$ and $\text{pH}$.

---

2. Molecular Structure and Acid Strength

Acid strength reflects the thermodynamic ease of proton transfer, governed by bond polarity, bond strength, and conjugate base anion stability.

Binary Acids ($\text{H}_n\text{X}$)

• Across a Period (Electronegativity): Moving left to right ($\text{CH}_4 < \text{NH}_3 < \text{H}_2\text{O} < \text{HF}$), electronegativity of X increases. This polarizes the H-X bond ($\text{H}^{\delta+} - \text{X}^{\delta-}$) and stabilizes the conjugate base anion, increasing acidity.
• Down a Group (Bond Strength): Moving top to bottom ($\text{HF} \ll \text{HCl} < \text{HBr} < \text{HI}$), bond dissociation energy is the dominant factor. As halogen atomic radius expands, orbital overlap with hydrogen's 1s orbital becomes diffuse and weak ($\text{H-F: } 567\text{ kJ/mol}$; $\text{H-I: } 299\text{ kJ/mol}$). Weaker bonds ionize more easily, making HI the strongest hydrohalic acid and HF a weak acid.

Oxoacids ($\text{H}_n\text{XO}_m$)

Oxoacids contain acidic protons bound to oxygen ($\text{X-O-H}$):

• Variable Central Atom X: Acidity increases with higher electronegativity of central atom X ($\text{HOI} < \text{HOBr} < \text{HOCl}$) due to inductive electron withdrawal from the O-H bond.
• Variable Terminal Oxygens: Acidity increases dramatically with more terminal oxygen atoms ($\text{HClO} < \text{HClO}_2 < \text{HClO}_3 < \text{HClO}_4$). Additional terminal oxygens ($=\text{O}$) exert strong inductive withdrawal and provide resonance stabilization in the conjugate oxoanion. In perchlorate ($\text{ClO}_4^-$), negative charge is delocalized over four equivalent oxygens (formal charge $-0.25$ each), conferring high stability and making $\text{HClO}_4$ an exceptionally strong acid.

Structural Factors Summary

---

3. Polyprotic Acids & Stepwise Ionization

Polyprotic acids donate protons in sequential equilibria (e.g., $\text{H}_3\text{PO}_4$ with $K_{a1} = 7.5 \times 10^{-3}$, $K_{a2} = 6.2 \times 10^{-8}$, $K_{a3} = 4.2 \times 10^{-13}$). Successive dissociation constants decrease by factors of $10^4$ to $10^6$ ($K_{a1} \gg K_{a2} \gg K_{a3}$) because removing a positive proton ($\text{H}^+$) from an increasingly negative anion is electrostatically unfavorable.

Polyprotic Calculation Rules

1. pH Determination: Because $K_{a1} \gg K_{a2}$, $[\text{H}_3\text{O}^+]$ and pH are governed almost exclusively by the first step: $[\text{H}_3\text{O}^+] \approx \sqrt{K_{a1} [\text{H}_2\text{A}]_0}$.
2. Divalent Anion Concentration: In step 2 ($\text{HA}^- \rightleftharpoons \text{H}^+ + \text{A}^{2-}$), $K_{a2} = ([\text{H}^+][\text{A}^{2-}]) / [\text{HA}^-]$. Because $[\text{H}^+] \approx [\text{HA}^-]$, they cancel algebraically: $[\text{A}^{2-}] \approx K_{a2}$. (Exception: $\text{H}_2\text{SO}_4$ is strong in step 1 [$K_{a1} \to \infty$] and weak in step 2 [$K_{a2} = 1.2 \times 10^{-2}$], so step 2 contributes substantially to $[\text{H}^+]$ in dilute solutions).

---

4. Salt Hydrolysis & Solution pH

Salts dissociate into hydrated ions that may hydrolyze water:

• Spectator Ions (Neutral): Group 1 cations ($\text{Li}^+, \text{Na}^+, \text{K}^+$), heavy Group 2 cations ($\text{Ca}^{2+}, \text{Sr}^{2+}, \text{Ba}^{2+}$), and conjugate bases of strong acids ($\text{Cl}^-, \text{Br}^-, \text{I}^-, \text{NO}_3^-, \text{ClO}_4^-$) do not hydrolyze.
• Acidic Cations:
  1. Conjugate acids of weak bases: $\text{NH}_4^+ + \text{H}_2\text{O} \rightleftharpoons \text{NH}_3 + \text{H}_3\text{O}^+$ ($K_a = K_w / K_b$).
  2. Small, highly charged metal ions ($\text{Al}^{3+}, \text{Fe}^{3+}$): Polarize coordinated water ligands: $[\text{Al}(\text{H}_2\text{O})_6]^{3+} + \text{H}_2\text{O} \rightleftharpoons [\text{Al}(\text{H}_2\text{O})_5(\text{OH})]^{2+} + \text{H}_3\text{O}^+$
• Basic Anions: Conjugate bases of weak acids abstract protons from water: $\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-$ ($K_b = K_w / K_a$).
• Salts with Both Acidic Cation and Basic Anion: Compare $K_a$ (cation) and $K_b$ (anion):
  • $K_a > K_b \implies$ acidic (e.g., $\text{NH}_4\text{F}$, $K_a = 5.6 \times 10^{-10} > K_b = 1.5 \times 10^{-11}$).
  • $K_b > K_a \implies$ basic (e.g., $\text{NH}_4\text{CN}$, $K_b = 2.0 \times 10^{-5} > K_a = 5.6 \times 10^{-10}$).
  • $K_a \approx K_b \implies$ neutral (e.g., $\text{NH}_4\text{CH}_3\text{COO}$).

Salt Hydrolysis Prediction Matrix

---

5. Worked Example: pH Calculation of a Basic Salt

Problem: Calculate the pH of a $0.200\text{ M}$ aqueous sodium acetate ($\text{NaCH}_3\text{COO}$) solution at 25 °C ($K_a = 1.8 \times 10^{-5}$, $K_w = 1.0 \times 10^{-14}$).

Step 1: Calculate $K_b$ for acetate $\text{Na}^+$ is a spectator. Acetate hydrolyzes: $\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^- \implies K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}$

Step 2: Solve equilibrium expression $\frac{x^2}{0.200} = 5.56 \times 10^{-10} \implies x^2 = 1.112 \times 10^{-10} \implies x = [\text{OH}^-] = 1.05 \times 10^{-5}\text{ M}$

Step 3: Determine pOH and pH $\text{pOH} = -\log_{10}(1.05 \times 10^{-5}) = 4.98 \implies \text{pH} = 14.00 - 4.98 = 9.02$ The calculated pH of 9.02 validates the basic character of aqueous acetate.