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Does diluting vinegar make its acetic acid "more dissociated"? It sounds wrong, but the answer is yes: a larger fraction of the molecules split into ions as the solution gets more dilute. That idea is the degree of dissociation, and the formula that describes it is usually called Ostwald's dilution law. This post explains both, works through three calculations, and shows where the shortcut formula breaks.
What is the degree of dissociation?
For a weak acid HA at concentration c, only some molecules ionise. The NCERT Class 11 chapter Equilibrium lets α be "the extent of ionization", so at equilibrium:
[HA] = c(1 − α)
[H⁺] = [A⁻] = cα
Putting these into the equilibrium constant gives
Ka = (cα)² ÷ c(1 − α) = cα² ÷ (1 − α).
The same expression appears in the NCERT Class 12 chapter Electrochemistry (Eq. 2.27), where α is measured from conductivity. The relation is commonly called Ostwald's dilution law, after Wilhelm Ostwald, who won the 1909 Nobel Prize in Chemistry "in recognition of his work on catalysis and for his investigations into the fundamental principles governing chemical equilibria and rates of reaction". NCERT derives the formula without using the name, so be ready for either form in a question.
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list ionisation of electrolytes, ionisation constants and the common ion effect, and the CBSE 2026-27 curriculum includes ionisation of acids and bases. Check your exam year's syllabus.
When can you use α ≈ √(Ka/c)?
If α is small, 1 − α ≈ 1 and Ka ≈ cα², so α ≈ √(Ka/c). This shortcut is what makes α grow on dilution: divide c by 100 and α rises about √100 = 10 times. But the approximation gets worse as α gets bigger, which is exactly what dilution does.
The test: after using the shortcut, check that α is below about 5%. If it is not, solve the quadratic cα² + Kaα − Ka = 0.
Worked example 1: acetic acid at three dilutions
Use Ka = 1.74 × 10⁻⁵ for acetic acid, the value in NCERT's table of ionisation constants.
At c = 0.10 M. Shortcut: α = √(1.74 × 10⁻⁵ ÷ 0.10) = √(1.74 × 10⁻⁴) ≈ 0.0132. That is 1.3%, under 5%, so it is fine. The exact quadratic gives 0.0131. [H⁺] = 0.10 × 0.0131 ≈ 1.31 × 10⁻³ M.
At c = 0.010 M. Shortcut: α ≈ √(1.74 × 10⁻³) ≈ 0.0417. Exact: 0.0409. Still close. [H⁺] ≈ 4.1 × 10⁻⁴ M.
At c = 0.0010 M. Shortcut: α ≈ √(1.74 × 10⁻²) ≈ 0.132. That is over 5%, so solve the quadratic:
α² + 0.0174α − 0.0174 = 0 (after dividing by c = 0.0010)
α = [−0.0174 + √(0.0174² + 4 × 0.0174)] ÷ 2 = (−0.0174 + 0.2644) ÷ 2 ≈ 0.1235.
The shortcut overestimated α by about 7%. [H⁺] = 0.0010 × 0.1235 ≈ 1.2 × 10⁻⁴ M.
Read the three results together. From 0.10 M to 0.0010 M, α rose from about 1.3% to 12%, roughly nine times, while [H⁺] fell about ten times.
There is a satisfying cross-check. NCERT's Electrochemistry chapter (Example 2.9) uses measured conductivity of 0.001028 M acetic acid and finds α = 0.1233 and Ka = 1.78 × 10⁻⁵. That matches the calculation above. Our post on molar conductivity and Kohlrausch's law shows how the conductivity route works.
Worked example 2: finding Ka from α
A 0.050 M solution of a weak acid HA is 2.0% dissociated. Find Ka.
α = 0.020 (convert the percentage to a fraction first).
Ka = cα² ÷ (1 − α) = 0.050 × (0.020)² ÷ 0.98 = 0.050 × 4.0 × 10⁻⁴ ÷ 0.98 ≈ 2.0 × 10⁻⁵.
With the shortcut you would get 0.050 × 4.0 × 10⁻⁴ = 2.0 × 10⁻⁵ as well, because 2% is small.
How does a common ion change α?
NCERT's Equilibrium chapter describes the common ion effect: adding an ion that is already in the equilibrium shifts it back towards the un-ionised form.
Worked example 3. Find α for 0.10 M acetic acid that also contains 0.10 M sodium acetate.
Acetate from the salt is about 0.10 M, and the acid barely ionises, so [HA] ≈ 0.10 M.
Ka = [H⁺][A⁻] ÷ [HA] ≈ (0.10α)(0.10) ÷ 0.10 = 0.10α.
α ≈ Ka ÷ 0.10 = 1.74 × 10⁻⁴, or about 0.017%.
That is about 75 times smaller than α in the plain 0.10 M acid (0.0131). This mixture is a buffer, the subject of our post on pH, buffers and ionic equilibrium.
Which mistakes cost marks?
Using α ≈ √(Ka/c) without checking. Above about 5%, solve the quadratic.
Saying dilution makes the solution more acidic. α rises, but [H⁺] falls.
Leaving α as a percentage. Put 0.020, not 2, into the formula.
Applying the dilution law to strong electrolytes. HCl and NaCl are treated as fully ionised; the law is for weak electrolytes.
Forgetting the common ion. If the salt of the acid is present, α is set by Ka and the salt concentration, not by √(Ka/c).
If you keep picking the wrong formula, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can see why each case needs its own method.
A practice routine
Day 1: Derive Ka = cα²/(1 − α) from an ICE table, without notes.
Day 2: Repeat Worked example 1 for formic acid (Ka from NCERT's table) at 0.10 M and 0.0010 M, with the 5% check.
Day 3: Three "find Ka from α" problems, including one where α is above 10%.
Day 4: Two common ion problems, then explain in words why α falls.
Today, write the three α values for acetic acid on one line and explain to someone why [H⁺] still falls.
