Strong vs Weak Acids — Degree of Dissociation Explained
"Strong acid" does not mean "concentrated acid". A 0.001 M solution of hydrochloric acid is a strong acid at low concentration; a 5 M solution of acetic acid is a weak acid at high concentration. The word strong describes only one thing — how completely the acid splits into ions in water. That single idea, measured by the degree of dissociation, is what this article makes precise.
Degree of dissociation, α
Put a weak acid HA in water and an equilibrium sets up:
The degree of dissociation α is the fraction of the acid molecules that have actually broken apart:
Here C is the initial (analytical) concentration in mol L⁻¹ and α is a pure number between 0 and 1 (often quoted as a percentage). A strong acid has α ≈ 1, so writing HCl → H⁺ + Cl⁻ with a single arrow is a fair approximation. A weak acid has α well below 1 — acetic acid at ordinary laboratory concentrations is only about 1% dissociated.
Ka — the acid dissociation constant
Strength is quantified by the equilibrium constant for that dissociation:
Water does not appear because it is the solvent and its activity is taken as 1. Ka values span many powers of ten, so we usually compress them with a logarithm:
A larger Ka means a stronger acid; a larger pKa means a weaker acid. The two run in opposite directions, and that is the single most common source of confusion in this whole topic.
| Acid | Ka (25 °C, approx.) | pKa | Comment |
|---|---|---|---|
| HCl, HBr, HI, HNO₃, HClO₄ | very large | negative | Treated as fully dissociated in water |
| HSO₄⁻ (second step of H₂SO₄) | 1.2 × 10⁻² | 1.92 | Moderately strong |
| HF | 6.8 × 10⁻⁴ | 3.17 | Weak, despite fluorine's electronegativity |
| HCOOH (formic) | 1.8 × 10⁻⁴ | 3.74 | Stronger than acetic acid |
| CH₃COOH (acetic) | 1.8 × 10⁻⁵ | 4.74 | The standard textbook weak acid |
| H₂CO₃ (first step) | 4.3 × 10⁻⁷ | 6.37 | Carbonic acid in blood and rainwater |
| NH₄⁺ | 5.6 × 10⁻¹⁰ | 9.25 | Conjugate acid of ammonia |
| HCN | 4.9 × 10⁻¹⁰ | 9.31 | Very weak acid |
Different textbooks quote slightly different numbers for the same acid (acetic acid appears as pKa 4.74 or 4.76; HF as 3.17 or 3.45 depending on how activity corrections are handled). Use the value your own syllabus gives, and state the value you used in your answer.
Ostwald's dilution law — linking α and Ka
Start with C mol L⁻¹ of HA and let a fraction α dissociate. At equilibrium [HA] = C(1 − α), [H⁺] = Cα and [A⁻] = Cα. Substituting into the Ka expression:
This is Ostwald's dilution law. It is exact for a monoprotic weak acid, provided the water's own ionisation can be ignored. If α is small, the term (1 − α) is close to 1 and drops out, giving the shortcut every student uses:
Worked example 1 — 0.10 M acetic acid
Ka = 1.8 × 10⁻⁵, C = 0.10 M.
[H⁺] ≈ √(1.8 × 10⁻⁵ × 0.10) = √(1.8 × 10⁻⁶) = 1.342 × 10⁻³ M
pH = −log(1.342 × 10⁻³) = 2.87
α = 1.342 × 10⁻³ ÷ 0.10 = 0.0134 = 1.34%
Cross-check by the exact route. Solving x² + Ka·x − Ka·C = 0 without the approximation gives x = 1.333 × 10⁻³ M, pH 2.88 and α = 1.33%. The shortcut was in error by about 0.7% in [H⁺] and 0.003 in pH — completely acceptable.
Worked example 2 — where the shortcut breaks
Chloroacetic acid, Ka = 1.4 × 10⁻³, at C = 0.010 M.
Shortcut: [H⁺] ≈ √(1.4 × 10⁻³ × 0.010) = √(1.4 × 10⁻⁵) =
3.742 × 10⁻³ M → pH 2.43, α = 37.4%.
That α is nowhere near "small", so the assumption (1 − α) ≈ 1 has collapsed.
Exact: x² + (1.4 × 10⁻³)x − (1.4 × 10⁻⁵) = 0 gives x = 3.107 × 10⁻³ M, pH = 2.51, α = 31.1%.
The shortcut overestimated [H⁺] by about 20% and the pH was wrong by 0.08 units. In a numerical worth 3 marks, that is the difference between full marks and none.
The two conditions the shortcut needs
- The 5% rule. Use √(Ka·C) only if the α it predicts comes out below about 0.05 (5%). Calculate α first, then decide. If α > 5%, solve the quadratic x² + Ka·x − Ka·C = 0 instead. Roughly, the shortcut is safe when C is at least about 400 × Ka.
- The acid must out-compete water. The shortcut ignores the H⁺ that water itself supplies. That is fine while Ka·C is far larger than Kw (1.0 × 10⁻¹⁴), but it fails for very dilute or extremely weak acids. This is also why 10⁻⁸ M HCl does not have pH 8 — an acid can never make a solution basic. Its true pH is slightly below 7, because water's own H⁺ dominates.
What happens on dilution — the point students get backwards
Dilute the 0.10 M acetic acid ten-fold to 0.010 M:
α ≈ √(1.8 × 10⁻⁵ ÷ 0.010) = √(1.8 × 10⁻³) = 0.0424 = 4.24%
[H⁺] = 0.0424 × 0.010 = 4.24 × 10⁻⁴ M (pH 3.37)
So α rose from 1.34% to 4.24% — dilution pushes the equilibrium towards more ions, exactly as Le Chatelier's principle predicts. But [H⁺] fell, from 1.34 × 10⁻³ to 4.24 × 10⁻⁴ M, so the solution became less acidic. Both statements are true at once.
Common mistakes that cost marks
- Mixing up strong and concentrated. Strength is about α; concentration is about how much acid you weighed out. They are independent.
- Using √(Ka·C) for a strong acid. For a strong monoprotic acid, [H⁺] = C directly. 0.10 M HCl has pH exactly 1.00 — no Ka needed.
- Never checking the 5% rule. Always compute α, then justify the approximation in one line. Examiners award that line.
- Thinking larger pKa means stronger. It is the reverse. HCN (pKa 9.31) is far weaker than acetic acid (pKa 4.74).
- Assuming dilution raises acidity because α rises. α rises, [H⁺] falls, pH rises.
- Applying Ostwald's law to a strong acid. With α ≈ 1 the denominator (1 − α) goes to zero and the expression is meaningless.
Where this appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 11–12 | Calculate pH and α of a given weak acid; state Ostwald's dilution law |
| JEE/NEET | pH of weak acid and weak base mixtures; comparing Ka values |
| IIT-JAM / CUET-PG | Exact quadratic treatment; polyprotic acids; degree of dissociation from conductivity |
| GATE / CSIR-NET | Activity corrections, buffer design from pKa, titration-curve reasoning |
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