IB DP Chemistry - R3.1.10 Acid and base strength and constants - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – R3.1.10 Acid and base strength and constants – Study Notes – New Syllabus
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Reactivity 3.1.10 — Acid and Base Strengths via Equilibrium Constants
Reactivity 3.1.10 — Acid and Base Strengths via Equilibrium Constants
Role of Ka and Kb:
- Ka (acid dissociation constant) quantifies the extent to which a weak acid ionizes in aqueous solution.
- Kb (base dissociation constant) describes the extent to which a weak base accepts protons (ionizes) in water.
- These constants are based on equilibrium concentrations, not initial ones.
- Larger values of Ka or Kb indicate greater ionization → stronger acid or base.
- For strong acids/bases, Ka or Kb is very large (often too large to measure) — they ionize almost completely, so they do not establish an equilibrium.
Equilibrium Reactions:
- Weak Acid: \( \text{HA} (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{H}_3\text{O}^+ (aq) + \text{A}^- (aq) \)
- Ka expression: \( K_a = \dfrac{[\text{H}_3\text{O}^+][\text{A}^-]}{[\text{HA}]} \)
- Weak Base: \( \text{B} (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{BH}^+ (aq) + \text{OH}^- (aq) \)
- Kb expression: \( K_b = \dfrac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]} \)
pKa and pKb: The Logarithmic Scale
- Since Ka and Kb values can vary widely in magnitude, their logarithmic forms are often used:
- \( \text{p}K_a = -\log_{10}(K_a) \)
- \( \text{p}K_b = -\log_{10}(K_b) \)
- Lower pKa or pKb values indicate stronger acids or bases, respectively.
- This is because a stronger acid has a larger Ka → smaller pKa.
Relationship Between Ka, Kb and Kw:
- For a conjugate acid–base pair:
- \( K_a \times K_b = K_w = 1.0 \times 10^{-14} \) at 25°C
- \( \text{p}K_a + \text{p}K_b = 14 \)
- This allows conversion between Ka and Kb, or pKa and pKb, for conjugate pairs.
Why Ka and pKa Matter:
Ka/pKa values help determine:
- Extent of dissociation
- Relative acid or base strength
- pH of the resulting solution
- Buffering ability (more in HL)
In reactions involving weak acids/bases, the equilibrium lies closer to the side with the weaker acid/base — this can be predicted using Ka or pKa values.
Examples of Ka and pKa Values:
- Ethanoic acid (CH₃COOH): Ka = \(1.8 \times 10^{-5}\), pKa ≈ 4.74 → Weak acid
- Hydrofluoric acid (HF): Ka = \(6.6 \times 10^{-4}\), pKa ≈ 3.18 → Stronger than CH₃COOH
- Ammonia (NH₃): Kb = \(1.8 \times 10^{-5}\), pKb ≈ 4.74 → Weak base
Interpreting Relative Strengths of Acids and Bases
Using Ka and pKa to Compare Acids:
- Ka (acid dissociation constant) represents how much an acid dissociates in water.
- To evaluate which acid is stronger, compare their Ka or pKa values:
- Higher Ka → more dissociation → stronger acid(greater ionization).
- Lower pKa → indicates stronger acidity (because pKa is the negative log of Ka)\( \text{pKa} = -\log_{10}(K_a) \).
- Lower Ka → weaker acid (less ionization).
- Acids with pKa < 0 are very strong acids and dissociate completely.
- If given multiple acids, arrange them in increasing or decreasing strength using Ka or pKa tables.
- Example:
- HF: pKa = 3.18
- CH₃COOH: pKa = 4.74
- HF is stronger than CH₃COOH because it has a lower pKa (higher Ka).
Using Kb and pKb to Compare Bases:
- Kb (base dissociation constant) shows how well a base accepts protons (ionizes in water).
- To evaluate base strength, compare Kb or pKb values:
- Higher Kb → stronger base (greater production of OH⁻).
- Lower pKb → stronger base (because pKb is the negative log of Kb: \( \text{pKb} = -\log_{10}(K_b) \).
- Lower Kb → weaker base.
- Use data tables to rank basicity or predict relative reaction favorability in aqueous solution.
- Example:
- NH₃: pKb = 4.74
- CH₃CH₂NH₂: pKb = 3.40
- CH₃CH₂NH₂ is the stronger base because its pKb is lower.
Link Between Acids and Conjugate Bases:
- Every acid–base pair has linked strengths — the stronger the acid, the weaker its conjugate base and vice versa.
- Strong acid → extremely weak conjugate base (e.g. Cl⁻ from HCl has almost no basicity).
- Weak acid → relatively stronger conjugate base (e.g. CH₃COO⁻ from CH₃COOH can act as a base).
- This relationship is mathematically represented as:
- \( K_a \times K_b = K_w = 1.0 \times 10^{-14} \)
- \( \text{pKa} + \text{pKb} = 14 \)
Relationship Between Ka and Kb:
- For any conjugate acid–base pair: \( K_a \times K_b = K_w = 1.0 \times 10^{-14} \) at 25°C.
- \( \text{pKa} + \text{pKb} = 14 \) → helps convert between acid and base strength.
- This relationship allows determination of the strength of a conjugate from the given species.
Predicting the Direction of Acid–Base Reactions:
- General Rule: Acid–base equilibria favor the formation of the weaker acid and base pair.
- The position of an acid–base equilibrium can be predicted by comparing the strengths of the acids and bases on both sides.
- Equilibrium favors the formation of the weaker acid–base pair.
- This is because weaker acids/bases are more stable and less reactive, lowering the system’s energy.
- Procedure:
- Identify the acids on both sides of the equilibrium.
- Compare their pKa values.
- The side with the higher pKa (weaker acid) is favored at equilibrium.
Examples:
- HF vs. CH₃COOH: HF has a pKa of 3.18; CH₃COOH has a pKa of 4.74 → HF is the stronger acid.
- NH₄⁺ vs. H₂CO₃: NH₄⁺ (pKa ≈ 9.25) is a weaker acid than H₂CO₃ (pKa ≈ 6.35) → equilibrium favors NH₄⁺.
Conclusion: The values of Ka, Kb, pKa, and pKb are essential to quantify the strength of weak acids and bases, and to compare their behavior in chemical reactions. These constants also help predict equilibrium positions and whether a species acts as a proton donor or acceptor more effectively.
Ranking Acids and Bases by Strength:
- To compare several acids or bases:
- Compare their Ka or Kb values → higher = stronger.
- Or compare their pKa or pKb values → lower = stronger.
- Use pKa/pKb values to arrange substances in order of increasing or decreasing strength.
- Data booklets often provide a pKa table — essential for exam-based questions.
- General trend:
- pKa < 0 → strong acid
- pKa 0–6 → moderately strong acid
- pKa 6–14 → weak acid
- pKa > 14 → extremely weak acid (or strong base conjugate)
Example:
Between benzoic acid (C₆H₅COOH) and HCN, which is the stronger acid? Predict the direction of the reaction:
C₆H₅COOH + CN⁻ ⇌ C₆H₅COO⁻ + HCN
▶️Answer/Explanation
pKa of benzoic acid ≈ 4.20; pKa of HCN ≈ 9.21.
Benzoic acid has a lower pKa → stronger acid than HCN.
Since equilibrium favors formation of the weaker acid, the reaction shifts to the right.
Example:
Given the reaction between phenol and acetate ion:
C₆H₅OH + CH₃COO⁻ ⇌ C₆H₅O⁻ + CH₃COOH
Determine which direction the equilibrium lies based on acidity data.
▶️Answer/Explanation
pKa of phenol ≈ 10.0; pKa of CH₃COOH ≈ 4.74.
CH₃COOH is the stronger acid (lower pKa), so its conjugate base (CH₃COO⁻) is weaker than C₆H₅O⁻.
Therefore, the equilibrium lies to the left, favoring phenol and acetate ion (weaker acid–base pair).
Example:
Consider this acid–base equilibrium:
NH₂⁻ + H₂O ⇌ NH₃ + OH⁻
Use Kb or pKb values to determine the stronger base.
▶️Answer/Explanation
pKa of NH₃ (its conjugate acid) ≈ 38 → NH₂⁻ is a very strong base.
pKa of H₂O (conjugate acid of OH⁻) = 15.7 → OH⁻ is a weaker base than NH₂⁻.
Thus, NH₂⁻ is the stronger base, and the equilibrium strongly favors the products (NH₃ + OH⁻).