Ostwald's Dilution Law and Degree of Dissociation
Acetic acid does not ionise fully in water, and the fraction that does depends on how dilute the solution is. Ostwald's dilution law links the degree of dissociation (α) of a weak electrolyte with its concentration and its dissociation constant. Class 11 and Class 12 numericals use it to find α, [H+] and pH, and to explain why a weak acid becomes "stronger" on dilution.
Where the law comes from
Take a weak acid HA of initial concentration c mol/L and degree of dissociation α.
| HA | H+ | A− | |
|---|---|---|---|
| Start | c | 0 | 0 |
| Change | −cα | +cα | +cα |
| Equilibrium | c(1 − α) | cα | cα |
If α is very small (1 − α ≈ 1): Ka ≈ cα², so α ≈ √(Ka ÷ c) and [H+] = cα = √(Kac)
The same law holds for a weak base, with Kb in place of Ka and [OH−] = cα. Here c is the initial concentration in mol/L, Ka has the same units as concentration (mol/L, usually not written in school answers), and α is a pure number between 0 and 1.
Worked example 1: 0.10 M acetic acid
Question. Find α, [H+] and pH of 0.10 M acetic acid. Ka = 1.8 × 10−5.
Step 1 (approximate). α = √(Ka ÷ c) = √(1.8 × 10−5 ÷ 0.10) = √(1.8 × 10−4) = 0.01342, about 1.3%.
Step 2. [H+] = cα = 0.10 × 0.01342 = 1.342 × 10−3 M.
Step 3. pH = −log(1.342 × 10−3) = 2.87.
Check by the exact quadratic. x2 + Kax − Kac = 0 gives x = 1.333 × 10−3 M, α = 0.01333, pH = 2.875. The approximation is very close because α is small (about 1.3%).
Worked example 2: what happens on dilution
Question. The same acid is diluted to 0.0010 M. Find α and pH.
Approximate. α = √(1.8 × 10−5 ÷ 0.0010) = √(0.018) = 0.134, which is 13.4%.
Exact. Here α is no longer tiny, so check with the quadratic: x = 1.255 × 10−4 M, α = 0.1255 (12.5%), pH = 3.90.
Conclusion. Diluting from 0.10 M to 0.0010 M (100 times) raised α from about 1.3% to about 12.5%, while Ka did not change. α grows as 1/√c, so 100 times dilution should give about 10 times more α. The approximate formula gives exactly that (0.0134 → 0.134), but the exact answer is a little lower. As a rule of thumb, if α comes out above about 5%, the approximation 1 − α ≈ 1 is no longer safe and the quadratic should be used.
Worked example 3: finding Ka from α
Question. A 0.050 M solution of a weak monobasic acid is 2.0% dissociated. Find Ka.
Step 1. α = 0.020.
Step 2. Using the approximate form: Ka = cα² = 0.050 × (0.020)² = 0.050 × 4.0 × 10−4 = 2.0 × 10−5.
Step 3 (exact form). Ka = cα² ÷ (1 − α) = 2.0 × 10−5 ÷ 0.98 = 2.04 × 10−5. Both agree to two significant figures.
Why a weak acid ionises more on dilution
Ka is constant at a given temperature. If c becomes smaller, the product cα² must stay the same (when α is small), so α must increase. In Le Chatelier language: adding water shifts HA ⇌ H+ + A− towards the side with more particles. Note the difference: α increases on dilution, but [H+] goes down, because [H+] = cα and c falls faster than α rises. In the example above [H+] dropped from 1.33 × 10−3 M to 1.26 × 10−4 M.
Common mistakes
- Saying a diluted acid is stronger. Only the degree of dissociation increases. The acid strength (Ka) does not change and the pH rises.
- Using the law for strong acids. HCl and HNO3 are almost completely ionised. The idea of Ka with small α does not apply.
- Using α in percent inside the formula. Convert 2.0% to 0.020 first.
- Forgetting to take the square root. α = √(Ka/c), not Ka/c.
- Using the approximation when α is large. Check α at the end; if it is more than about 5%, solve the quadratic.
- Changing Ka with concentration. Ka changes with temperature only.
Exam summary
| You are given | You find | Use |
|---|---|---|
| Ka and c | α, [H+], pH | α = √(Ka/c); [H+] = cα |
| c and α | Ka | Ka = cα² |
| Change in c | Change in α | α ∝ 1/√c (when α is small) |
| Weak base | [OH−], pOH, pH | Same law with Kb, then pH = 14 − pOH (at 25 °C) |
Once you have [H+] from the law, the pH tool converts it to pH and pOH instantly, so you can check your logarithm.
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