Degree of Dissociation from Conductivity — Worked Method
A weak electrolyte such as acetic acid only partly splits into ions in water. The fraction that does split is the degree of dissociation, α. You cannot see it, but you can measure it, because only the ions carry current — so the solution's conductivity is a direct report on how much of the acid has ionised. This is one of the most elegant measurements in Class 12 electrochemistry, and one of the most reliably examined.
The formula
Here Λm is the molar conductivity of the solution at the concentration you are working at, and Λ°m is the limiting molar conductivity — the value at infinite dilution, where the electrolyte would be completely dissociated. The ratio of the two is therefore the fraction actually dissociated. α has no unit and always lies between 0 and 1; multiply by 100 for a percentage.
To get Λm from a measured conductivity κ:
The 1000 is not decoration. It converts litres to cubic centimetres, because κ is per centimetre while c is per litre. Leaving it out is the single most common error in this topic. For a weak acid or base, Λ°m cannot be measured by extrapolation and must come from Kohlrausch's law; for acetic acid at 298 K the standard value is 390.5 S cm² mol⁻¹.
Worked example 1 — acetic acid at 0.0100 M
Question. The conductivity of 0.0100 M acetic acid at 298 K is 1.65 × 10⁻⁴ S cm⁻¹. Find its degree of dissociation.
Step 1 — molar conductivity.
Λm = 1000 × 1.65 × 10⁻⁴ ÷ 0.0100 = 0.165 ÷ 0.0100 =
16.5 S cm² mol⁻¹
Step 2 — degree of dissociation.
α = 16.5 ÷ 390.5 = 0.0423, that is 4.23%
So fewer than 5 molecules in every 100 have given up their proton. The other 95-plus are sitting in solution as intact CH₃COOH — which is exactly what "weak acid" means, expressed as a number.
Worked example 2 — the same acid ten times more dilute
Question. The conductivity of 0.00100 M acetic acid at 298 K is 4.95 × 10⁻⁵ S cm⁻¹. Find α.
Λm = 1000 × 4.95 × 10⁻⁵ ÷ 0.00100 = 0.0495 ÷ 0.00100 =
49.5 S cm² mol⁻¹
α = 49.5 ÷ 390.5 = 0.1268, that is 12.68%
Diluting ten-fold has roughly tripled the degree of dissociation. That is Ostwald's dilution law in action: for small α, α is proportional to 1 ÷ √c, and √10 ≈ 3.16. Notice also that κ went down (fewer ions per cm³) while Λm went up (more ions per mole). Those two moving in opposite directions confuses a lot of students, and it is a favourite one-mark question.
From α to the dissociation constant
Once you have α you have the equilibrium constant, through Ostwald's dilution law. For HA ⇌ H⁺ + A⁻ starting at concentration c:
Using example 1 (c = 0.0100, α = 0.0423):
α² = 0.0423² = 1.7853 × 10⁻³
cα² = 0.0100 × 1.7853 × 10⁻³ = 1.7853 × 10⁻⁵
1 − α = 1 − 0.0423 = 0.9577
Ka = 1.7853 × 10⁻⁵ ÷ 0.9577 = 1.86 × 10⁻⁵
Using example 2 (c = 0.00100, α = 0.1268):
α² = 0.1268² = 1.6068 × 10⁻²
cα² = 0.00100 × 1.6068 × 10⁻² = 1.6068 × 10⁻⁵
1 − α = 0.8732
Ka = 1.6068 × 10⁻⁵ ÷ 0.8732 = 1.84 × 10⁻⁵
This is the real test of the method. The two solutions differ ten-fold in concentration and give completely different values of α — but they return the same Ka, about 1.8 × 10⁻⁵, which is the accepted dissociation constant of acetic acid. α depends on dilution; Ka does not. If your two answers disagree, you have made an arithmetic error somewhere.
Why this only works for weak electrolytes
It is tempting to apply α = Λm ÷ Λ°m to sodium chloride and conclude that NaCl is "92% dissociated". That conclusion is wrong. A strong electrolyte is completely dissociated at all concentrations; its molar conductivity is lower than Λ°m not because ions are missing but because the ions get in each other's way — each ion is surrounded by an atmosphere of oppositely charged ions that slows it down. That behaviour is described by the Debye–Hückel–Onsager equation, Λm = Λ°m − A√c, and the ratio Λm ÷ Λ°m for a strong electrolyte is called the conductivity ratio, not the degree of dissociation.
| Weak electrolyte | Strong electrolyte | |
|---|---|---|
| Example | CH₃COOH, NH₄OH, HF | KCl, NaCl, HCl, NaOH |
| Why Λm < Λ°m | Only part of it is ionised | Fully ionised; inter-ionic drag |
| Λm ÷ Λ°m is | the degree of dissociation α | the conductivity ratio, not α |
| Λm against √c | Rises very steeply near c = 0 | Nearly straight; can be extrapolated |
| Λ°m obtained by | Kohlrausch's law only | Extrapolation to c = 0 |
Common mistakes that cost marks
- Dropping the factor of 1000. Without it, example 1 gives Λm = 0.0165 and α = 4.2 × 10⁻⁵ — a thousand times too small, and obviously unphysical.
- Mixing SI and practical units. κ in S m⁻¹ with c in mol L⁻¹ will not work. Either use κ in S cm⁻¹ with c in mol L⁻¹ and the 1000 factor, or work fully in SI (κ in S m⁻¹, c in mol m⁻³, Λm in S m² mol⁻¹). Remember 1 S m² mol⁻¹ = 10⁴ S cm² mol⁻¹.
- Applying α = Λm ÷ Λ°m to a strong electrolyte. See the section above — this is a conceptual error, not just an arithmetic one.
- Confusing κ with Λm. κ is conductivity, per centimetre of solution; Λm is per mole. They move in opposite directions on dilution.
- Using the approximation Ka ≈ cα² when α is not small. At α = 0.127 the approximation gives 1.61 × 10⁻⁵ instead of 1.84 × 10⁻⁵, a 13% error. Keep the (1 − α) unless α is below about 0.05.
- Forgetting that Λ°m is temperature-dependent. The value 390.5 S cm² mol⁻¹ for acetic acid is for 298 K.
Where this appears in exams
| Exam | Typical use |
|---|---|
| CBSE Class 12 | Electrochemistry: α from κ, then Ka by Ostwald's law |
| JEE / NEET | Numericals linking conductivity, α, Ka and pH |
| IIT-JAM / CUET-PG | Conductance of weak electrolytes, Kohlrausch's law applications |
| GATE / CSIR-NET | Debye–Hückel–Onsager behaviour, ion mobilities, conductometric titration |
No dedicated conductance tool exists in the suite yet, so this calculation is done by hand — but every supporting step (the ratio arithmetic, Ka and the resulting pH) can be checked among the chemistry tools. Open the suite and pick the tool you need.
Open the ABC Chemistry Calculator Suite →Electrochemistry carries a large share of the Class 12 chemistry paper and feeds directly into JAM and NET later. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre plus online classes across India — abcchemistry.in.