AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/8/8-3)
Unit 8 · Topic 8.3
8.3 Weak Acid and Base Equilibria
A weak acid ionizes only partly, so most of it stays as un-ionized HA and its pH must be found with an equilibrium calculation using Ka. Weak bases work the same way with Kb, and for any conjugate acid–base pair, Ka × Kb = Kw.
Key terms
- weak acid
- Ka and pKa
- Kb and pKb
- percent ionization
- conjugate acid-base pair
- Ka × Kb = Kw
Weak acids and Ka
A weak acid, HA, transfers its proton to water only partly: HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq). At equilibrium, the vast majority of the acid molecules are still HA, and [H₃O⁺] is much smaller than the acid's starting concentration.
The equilibrium constant is the acid ionization constant, Ka = [H₃O⁺][A⁻] / [HA]. A larger Ka means a stronger acid. Ka values are often given as pKa = −log Ka, where a smaller pKa means a stronger acid.
Common examples: acetic acid, CH₃COOH (Ka = 1.8 × 10⁻⁵), and hydrofluoric acid, HF (Ka on the order of 10⁻⁴).
Finding the pH of a weak acid
- Write the ionization equation and the Ka expression.
- ICE table: [HA] = C − x, [H₃O⁺] = x, [A⁻] = x, where C is the starting concentration. (Water's own H₃O⁺ is negligible.)
- Because Ka is small, x is usually much smaller than C, so C − x ≈ C. Then x² / C = Ka and x = √(Ka × C).
- x is [H₃O⁺]; pH = −log x.
- Check the approximation: x should be less than about 5% of C.
Weak bases and Kb
A weak base, B, takes a proton from water only partly: B(aq) + H₂O(l) ⇌ HB⁺(aq) + OH⁻(aq). The base ionization constant is Kb = [HB⁺][OH⁻] / [B], and pKb = −log Kb.
The math mirrors weak acids, except x is [OH⁻]. Find pOH from x, then pH = 14.00 − pOH at 25 °C. Ammonia, NH₃, is the classic weak base (Kb = 1.8 × 10⁻⁵).
Percent ionization
Percent ionization = (amount ionized ÷ starting amount) × 100% = ([H₃O⁺] at equilibrium ÷ [HA] initial) × 100%. For a weak base, use [OH⁻] instead.
You can find it from Ka and the starting concentration, or directly from a measured pH. Diluting a weak acid increases its percent ionization: 0.10 M acetic acid is about 1.3% ionized, but 0.010 M acetic acid is about 4% ionized. The pH still rises on dilution, because the total [H₃O⁺] falls.
Ka × Kb = Kw for conjugate pairs
When an acid loses a proton it becomes its conjugate base, and these two are linked: Ka(acid) × Kb(conjugate base) = Kw. Equivalently, pKa + pKb = 14.00 at 25 °C.
So the stronger the acid, the weaker its conjugate base. Acetic acid has Ka = 1.8 × 10⁻⁵, so acetate, CH₃COO⁻, has Kb = (1.0 × 10⁻¹⁴) ÷ (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰. That's why a solution of sodium acetate is slightly basic.
This relationship comes from adding the acid's ionization reaction and its conjugate base's reaction with water: the sum is water's autoionization, so the K values multiply to Kw (topic 7.6).
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
pH and percent ionization of a weak acid
Find the pH and percent ionization of 0.10 M acetic acid, CH₃COOH (Ka = 1.8 × 10⁻⁵).
Show the solutionHide the solution
- Step 1: CH₃COOH + H₂O ⇌ H₃O⁺ + CH₃COO⁻. ICE: [CH₃COOH] = 0.10 − x; [H₃O⁺] = [CH₃COO⁻] = x.
- Step 2: Ka = x² / (0.10 − x) ≈ x² / 0.10 = 1.8 × 10⁻⁵.
- Step 3: x² = 1.8 × 10⁻⁶, so x = 1.3 × 10⁻³ M = [H₃O⁺].
- Step 4: Check: 1.3 × 10⁻³ ÷ 0.10 = 1.3%, under 5%, so the approximation holds.
- Step 5: pH = −log(1.34 × 10⁻³) = 2.87. Percent ionization = 1.3%.
Answer: pH = 2.87; about 1.3% ionized
- Example 2Calculator allowed
pH of a weak base
Find the pH of 0.20 M NH₃ (Kb = 1.8 × 10⁻⁵) at 25 °C.
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- Step 1: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻. ICE: [NH₃] = 0.20 − x; [NH₄⁺] = [OH⁻] = x.
- Step 2: Kb = x² / 0.20 = 1.8 × 10⁻⁵ (using 0.20 − x ≈ 0.20).
- Step 3: x = √(3.6 × 10⁻⁶) = 1.9 × 10⁻³ M = [OH⁻]. Check: under 1% of 0.20.
- Step 4: pOH = −log(1.9 × 10⁻³) = 2.72. pH = 14.00 − 2.72 = 11.28.
Answer: pH = 11.28
- Example 3Calculator allowed
Trap: finding Ka from a measured pH
A 0.050 M solution of a weak acid HA has a pH of 3.00. Calculate Ka.
Show the solutionHide the solution
- Step 1: [H₃O⁺] = 10^(−3.00) = 1.0 × 10⁻³ M. This is x, and [A⁻] = x as well.
- Step 2: [HA] at equilibrium = 0.050 − 0.0010 = 0.049 M.
- Step 3: Ka = (1.0 × 10⁻³)² ÷ 0.049 = 2.0 × 10⁻⁵.
- Step 4: The trap is plugging the pH (3.00) into Ka instead of [H₃O⁺], or using 0.050 M without subtracting x. Here the subtraction makes only a small difference, but it's the correct setup.
Answer: Ka ≈ 2.0 × 10⁻⁵
Common mistakes
- Treating a weak acid as if it ionized completely and setting [H₃O⁺] equal to its concentration.
- Using x from a weak base calculation as [H₃O⁺] instead of [OH⁻].
- Using the small-x approximation without checking it.
- Thinking a lower Ka means a stronger acid. Larger Ka (smaller pKa) means stronger.
On the exam
- Expect to calculate the pH of a weak acid or base from Ka or Kb, or to work backward from pH to Ka. Show the equilibrium expression with x substituted in.
- You may be asked to compare two acids at the same concentration: the one with the larger Ka has the lower pH and the higher percent ionization.
Connected topics
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Check yourself
4 questions on 8.3 Weak Acid and Base Equilibria. Pick an answer to see if you got it, and why.
A weak acid, HA, has Ka = 1.8 × 10⁻⁵. What is the pH of a 0.10 M solution of HA at 25 °C?
A 0.050 M solution of a weak monoprotic acid has a pH of 3.00 at 25 °C. Which of the following is closest to Ka for the acid?
Ka for HF is 6.8 × 10⁻⁴ at 25 °C. What is Kb for the fluoride ion, F⁻, at 25 °C?
Kb for NH₃ is 1.8 × 10⁻⁵ at 25 °C. Which of the following is closest to the pH of a 0.20 M NH₃ solution?
0 of 4 answered