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7: pKa values and trends

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Léa Kowalski

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Léa Kowalski

1,554 pts

5 days ago

Choose your name

Léa Kowalski

Your opponent is

Léa Kowalski

1,554 pts
5 days ago
The quiz will be on the following text — learn it for the best chance to win.
Section 1: pKapK_a Values and Trends

Definition and Significance
pKapK_a quantifies acid strength, defined as pKa=log(Ka)pK_a = -\log(K_a), where KaK_a is the acid dissociation constant (Ka=[H+][A][HA]K_a = \frac{[H^+][A^-]}{[HA]}). A lower pKapK_a indicates a stronger acid (greater dissociation), while a higher pKapK_a denotes a weaker acid. For example, HCl (pKa7pK_a \approx -7) is stronger than acetic acid (pKa=4.8pK_a = 4.8).

Key Trends Influencing pKapK_a

  1. Element Effects

    • Electronegativity: Across a row, acidity increases with higher electronegativity of the atom bonded to H.
      Example: CH₄ (pKa=50pK_a = 50) < NH₃ (pKa=38pK_a = 38) < H₂O (pKa=15.7pK_a = 15.7) < HF (pKa=3.2pK_a = 3.2).
    • Atomic Size: Down a group, larger atoms stabilize conjugate base charge better, increasing acidity.
      Example: HF (pKa=3.2pK_a = 3.2) < HCl (pKa=7pK_a = -7) < HBr (pKa=9pK_a = -9) < HI (pKa=10pK_a = -10).
  2. Resonance Stabilization
    Delocalization of negative charge in the conjugate base dramatically lowers pKapK_a.
    Example: Acetic acid (CH₃COOH, pKa=4.8pK_a = 4.8) is stronger than ethanol (CH₃CH₂OH, pKa=15.9pK_a = 15.9) because acetate ion’s charge is resonance-stabilized over two oxygen atoms.

  3. Inductive Effects
    Electron-withdrawing groups (EWGs, e.g., \ceNO2–\ce{NO_2}, \ceCl–\ce{Cl}) stabilize the conjugate base via polarization, reducing pKapK_a. Proximity amplifies this effect.
    Example: Chloroacetic acid (ClCH₂COOH, pKa=2.9pK_a = 2.9) is stronger than acetic acid (pKa=4.8pK_a = 4.8) due to the Cl group’s I–I effect.

  4. Hybridization
    Higher s-character in the orbital holding the negative charge increases acidity.
    Example: Terminal alkynes (H–C≡C–R, pKa=25pK_a = 25) are more acidic than alkenes (H₂C=CH₂, pKa=44pK_a = 44) or alkanes (CH₄, pKa=50pK_a = 50) due to sp-hybridization (50% s-character) vs. sp² (33%) or sp³ (25%).

Practical Applications

  • Predicting Reaction Outcomes: Acid-base equilibria favor the weaker acid (higher pKapK_a). For instance, ethoxide (CH₃CH₂O⁻, conjugate base pKa=15.9pK_a = 15.9) deprotonates terminal alkynes (pKa=25pK_a = 25) because the alkyne conjugate base (pKa=25pK_a = 25) is weaker than ethanol (pKa=15.9pK_a = 15.9).
  • Structure-Acidity Relationships: Identify dominant stabilizing factors (resonance > induction > hybridization) when comparing molecules.