Acid-base chemistry is a direct application of the chemical equilibrium concepts covered earlier, applied specifically to proton transfer reactions. This guide covers pH calculations, weak acid/base equilibria, buffers, and titrations — with detailed worked examples for the calculation types most commonly assigned in university courses.
1. Defining Acids and Bases
- Arrhenius definition: acids produce H⁺ in water; bases produce OH⁻.
- Brønsted-Lowry definition: acids are proton (H⁺) donors; bases are proton acceptors. This is the most widely used definition in university courses.
- Lewis definition: acids are electron-pair acceptors; bases are electron-pair donors — the broadest definition, important for connecting to coordination chemistry, where metal ions act as Lewis acids toward ligands.
1.1 Conjugate Acid-Base Pairs
In a Brønsted-Lowry reaction, an acid and base react to form their conjugate base and conjugate acid, differing by one H⁺.
Example: HCl + H₂O → H₃O⁺ + Cl⁻. Here, HCl is the acid, H₂O is the base, H₃O⁺ is the conjugate acid of water, and Cl⁻ is the conjugate base of HCl.
2. The pH Scale
pH = −log[H⁺] and pOH = −log[OH⁻]
At 25°C, water autoionizes: H₂O ⇌ H⁺ + OH⁻, with Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴
This gives the crucial relationship: pH + pOH = 14 (at 25°C)
- pH = 7: neutral
- pH < 7: acidic
- pH > 7: basic
Worked Example: Find the pH of a solution with [H⁺] = 3.2 × 10⁻⁴ M.
pH = −log(3.2×10⁻⁴) = 3.49
Worked Example: Find [OH⁻] for a solution with pH = 9.25.
pOH = 14 − 9.25 = 4.75 [OH⁻] = 10⁻⁴·⁷⁵ = 1.78 × 10⁻⁵ M
3. Strong Acids and Strong Bases
Strong acids (HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄) and strong bases (Group 1 hydroxides, and Group 2 hydroxides like Ca(OH)₂, Ba(OH)₂) dissociate completely in water, so [H⁺] or [OH⁻] equals the analytical concentration directly (accounting for stoichiometry).
Worked Example: Find the pH of 0.025 M HCl.
Since HCl is a strong acid, [H⁺] = 0.025 M pH = −log(0.025) = 1.60
Worked Example: Find the pH of 0.010 M Ca(OH)₂.
Each formula unit releases 2 OH⁻, so [OH⁻] = 2 × 0.010 = 0.020 M pOH = −log(0.020) = 1.70 pH = 14 − 1.70 = 12.30
4. Weak Acids and Bases: Ka and Kb
Weak acids and bases only partially dissociate, requiring equilibrium calculations exactly like those covered in chemical equilibrium.
HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA]
Worked Example: Find the pH of 0.20 M acetic acid (CH₃COOH), Ka = 1.8 × 10⁻⁵.
| CH₃COOH | H⁺ | CH₃COO⁻ | |
|---|---|---|---|
| Initial | 0.20 | 0 | 0 |
| Change | −x | +x | +x |
| Equilibrium | 0.20−x | x | x |
Ka = x²/(0.20−x) ≈ x²/0.20 (assuming x << 0.20) x² = 1.8×10⁻⁵ × 0.20 = 3.6×10⁻⁶ x = 1.897×10⁻³ M
Check approximation: 1.897×10⁻³/0.20 = 0.95% < 5% ✓ valid
pH = −log(1.897×10⁻³) = 2.72
4.1 The Relationship Between Ka and Kb for Conjugate Pairs
Ka × Kb = Kw = 1.0 × 10⁻¹⁴ (at 25°C)
Worked Example: Find Kb for the acetate ion (CH₃COO⁻), given Ka(CH₃COOH) = 1.8 × 10⁻⁵.
Kb = Kw/Ka = 1.0×10⁻¹⁴ / 1.8×10⁻⁵ = 5.6 × 10⁻¹⁰
4.2 Percent Ionization
% ionization = ([H⁺]/[HA]₀) × 100%
Worked Example: From the example above, % ionization = (1.897×10⁻³/0.20) × 100% = 0.95%. Note that percent ionization increases as the initial concentration decreases (dilution shifts weak acid equilibria toward more dissociation, consistent with Le Chatelier’s principle).
5. Polyprotic Acids
Polyprotic acids (e.g., H₂SO₄, H₃PO₄) lose protons in successive steps, each with its own Ka, where Ka1 >> Ka2 >> Ka3. For most calculations, only the first ionization needs to be considered because subsequent Ka values are dramatically smaller.
Worked Example: For H₂CO₃, Ka1 = 4.3×10⁻⁷ and Ka2 = 4.8×10⁻¹¹. For a 0.10 M solution, the pH calculation uses only Ka1, since Ka2 contributes a negligible additional [H⁺].
6. Salts and Hydrolysis
Salts formed from the reaction of an acid and base can be acidic, basic, or neutral in solution, depending on the strength of their parent acid/base:
- Strong acid + strong base salt (e.g., NaCl): neutral (neither ion hydrolyzes).
- Weak acid + strong base salt (e.g., CH₃COONa): basic (the conjugate base, CH₃COO⁻, hydrolyzes to produce OH⁻).
- Strong acid + weak base salt (e.g., NH₄Cl): acidic (the conjugate acid, NH₄⁺, hydrolyzes to produce H⁺).
- Weak acid + weak base salt: depends on relative Ka and Kb values.
Worked Example: Find the pH of 0.15 M NH₄Cl, Ka(NH₄⁺) = 5.6 × 10⁻¹⁰.
NH₄⁺ ⇌ NH₃ + H⁺ Ka = x²/(0.15−x) ≈ x²/0.15 x² = 5.6×10⁻¹⁰ × 0.15 = 8.4×10⁻¹¹ x = 9.17×10⁻⁶ M pH = −log(9.17×10⁻⁶) = 5.04
7. Buffer Solutions
A buffer resists changes in pH upon addition of small amounts of acid or base. It consists of a weak acid and its conjugate base (or a weak base and its conjugate acid) in comparable amounts.
7.1 The Henderson-Hasselbalch Equation
pH = pKa + log([A⁻]/[HA])
Worked Example: Calculate the pH of a buffer containing 0.30 M CH₃COOH and 0.20 M CH₃COONa, Ka = 1.8 × 10⁻⁵.
pKa = −log(1.8×10⁻⁵) = 4.745 pH = 4.745 + log(0.20/0.30) = 4.745 + log(0.667) = 4.745 − 0.176 = 4.57
7.2 Buffer Capacity After Adding Strong Acid/Base
Worked Example: To the buffer above (1.00 L, 0.30 mol CH₃COOH, 0.20 mol CH₃COO⁻), add 0.050 mol NaOH. Find the new pH.
NaOH reacts completely with the acid component: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O
New moles: CH₃COOH = 0.30 − 0.050 = 0.25 mol; CH₃COO⁻ = 0.20 + 0.050 = 0.25 mol
pH = pKa + log(0.25/0.25) = 4.745 + log(1) = 4.745 + 0 = 4.75
Notice how little the pH changed (4.57 → 4.75) compared to how much it would change if 0.050 mol NaOH were added to pure water — this demonstrates buffering action.
7.3 Choosing a Buffer
The most effective buffer is chosen so that pKa is close to the desired pH (ideally within ±1 pH unit), and buffer capacity is maximized when the acid and conjugate base concentrations are equal (pH = pKa exactly at that point).
8. Acid-Base Titrations
8.1 Strong Acid–Strong Base Titration
The equivalence point occurs at pH = 7, where moles of acid = moles of base.
Worked Example: 25.0 mL of 0.100 M HCl is titrated with 0.100 M NaOH. Find the pH after adding 10.0 mL of NaOH.
Moles HCl initial = 0.0250 L × 0.100 M = 2.50×10⁻³ mol Moles NaOH added = 0.0100 L × 0.100 M = 1.00×10⁻³ mol Excess HCl = 2.50×10⁻³ − 1.00×10⁻³ = 1.50×10⁻³ mol Total volume = 25.0 + 10.0 = 35.0 mL = 0.0350 L [H⁺] = 1.50×10⁻³/0.0350 = 0.0429 M pH = −log(0.0429) = 1.37
8.2 Weak Acid–Strong Base Titration
The equivalence point occurs at pH > 7, because the resulting solution contains the conjugate base of the weak acid, which hydrolyzes to produce OH⁻ (as covered in Section 6). Before the equivalence point, the solution behaves as a buffer, and the half-equivalence point is especially useful because at this point [HA] = [A⁻], so pH = pKa exactly — a frequently tested shortcut.
Worked Example: For the titration of a weak acid with Ka = 1.8×10⁻⁵, at the half-equivalence point, pH = pKa = −log(1.8×10⁻⁵) = 4.74, regardless of the exact volumes involved.
8.3 Indicators
Acid-base indicators are themselves weak acids/bases that change color over a specific pH range close to their own pKa. The indicator should be chosen so its color-change range brackets the pH at the equivalence point (e.g., phenolphthalein, range ~8.2–10, for weak acid–strong base titrations).
9. Common Assignment Pitfalls
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- Forgetting that pH + pOH = 14 only holds at 25°C — Kw changes with temperature.
- Using the strong-acid shortcut ([H⁺] = concentration) for a weak acid — always check whether the acid/base is listed as strong; if not, it must be treated as weak using Ka/Kb.
- Forgetting to check the 5% approximation rule in weak acid/base ICE tables (same as in chemical equilibrium).
- Confusing pKa (a fixed property of the acid) with pH (which depends on concentration and the amount of conjugate base present).
- For polyprotic acids, forgetting to use only Ka1 for the initial pH estimate.
10. Full Worked Problem
Question: What is the pH of a buffer prepared by mixing 0.40 mol NH₃ and 0.25 mol NH₄Cl in 1.00 L of solution? Kb(NH₃) = 1.8 × 10⁻⁵.
Solution: First find Ka of the conjugate acid, NH₄⁺: Ka = Kw/Kb = 1.0×10⁻¹⁴/1.8×10⁻⁵ = 5.56×10⁻¹⁰ pKa = −log(5.56×10⁻¹⁰) = 9.255
Using Henderson-Hasselbalch (with NH₃ as the base “A⁻” and NH₄⁺ as the acid “HA”): pH = pKa + log([NH₃]/[NH₄⁺]) = 9.255 + log(0.40/0.25) = 9.255 + log(1.6) = 9.255 + 0.204 = 9.46
Acid-base chemistry is one of the richest applications of chemical equilibrium principles, and buffer calculations in particular appear constantly in biochemistry and analytical chemistry courses. For a deeper look at electron-pair based (Lewis) acid-base behavior in metal complexes, see coordination chemistry. Practice both direction of calculation — given concentration find pH, and given pH find concentration — until both feel equally natural.
