Kohlrausch's Law Calculator
Calculate the limiting molar conductivity of any electrolyte at infinite dilution from its ionic conductivities and stoichiometric coefficients, using Kohlrausch's law of independent ion migration.
Predict the Electrolyte's Limiting Molar Conductivity from Its Ions
Kohlrausch's Law Formula
Where:
- \(\Lambda^\circ_m\) — limiting molar conductivity of the electrolyte at infinite dilution (S·cm²/mol)
- \(\lambda^\circ_i\) — limiting molar conductivity (ionic conductivity) of ion \(i\) at infinite dilution (S·cm²/mol)
- \(\nu_i\) — stoichiometric coefficient of ion \(i\) in the formula unit (e.g. 2 for the Cl⁻ in CaCl₂)
The formula expresses Kohlrausch's law of independent migration of ions: at infinite dilution each ion contributes to the total conductivity independently of every other ion, so the electrolyte's limiting molar conductivity is just the sum of the ionic conductivities weighted by their stoichiometric coefficients.
Kohlrausch's Law – Calculation Example
Example: an electrolyte whose formula unit contains one Na⁺ ion, one Ca²⁺ ion, one Cl⁻ ion and one SO₄²⁻ ion. All four ions have a stoichiometric coefficient of 1, and their limiting molar conductivities at 25 °C are λ°(Na⁺) = 50.11, λ°(Ca²⁺) = 119.0, λ°(Cl⁻) = 76.35 and λ°(SO₄²⁻) = 160.0 S·cm²/mol — exactly the calculator's default values.
\[ \Lambda^\circ_m = 1 \times 50.11 + 1 \times 119.0 + 1 \times 76.35 + 1 \times 160.0 = 405.46 \ \text{S·cm}^2\!/\text{mol} \]
Result: \(\Lambda^\circ_m = 405.46\) S·cm²/mol (equivalently 0.040546 S·m²/mol).
If instead you uncheck both "Add a second cation" and "Add a second anion" boxes, the same inputs reduce to NaCl: \(\Lambda^\circ_m(\text{NaCl}) = 50.11 + 76.35 = 126.46\) S·cm²/mol, the accepted tabulated value.
How to use the calculator
- Enter the Cation Name (optional, used for labeling) and its Limiting Molar Conductivity (λ°₊) in S·cm²/mol, e.g. 50.11 for Na⁺.
- Set the cation's Stoichiometric Coefficient (ν₊) — the number of such cations per formula unit, e.g. 2 for Na⁺ in Na₂SO₄.
- If the salt has two different cations (e.g. KAl(SO₄)₂), tick Add a second cation and fill in the second cation's name, λ° and coefficient; otherwise leave the box unticked.
- Repeat the same steps for the anion: Anion Name, Limiting Molar Conductivity (λ°₋), Stoichiometric Coefficient (ν₋), and optionally Add a second anion.
- Choose the Output Unit — S·cm²/mol (the unit used in most data tables) or S·m²/mol (the SI unit).
- Read the result: the Limiting Molar Conductivity (Λ°ₘ), a per-ion breakdown of contributions, and the full substitution are shown immediately.
What is Kohlrausch's law of independent ion migration?
When an electric field is applied to an electrolyte solution, both the cations and the anions migrate and carry current. Friedrich Kohlrausch discovered (around 1875) that at infinite dilution — where the ions are so far apart that they no longer interact — each ion migrates independently of its counter-ion. The molar conductivity of the whole electrolyte is therefore simply the sum of the individual ionic conductivities, each multiplied by how many of that ion the formula unit produces on dissociation.
A classic experimental proof is the difference rule: \(\Lambda^\circ_m(\text{KCl}) - \Lambda^\circ_m(\text{NaCl})\) equals \(\Lambda^\circ_m(\text{KBr}) - \Lambda^\circ_m(\text{NaBr})\), because the anion's contribution cancels out and only the difference between K⁺ and Na⁺ remains. This additivity is exactly what the calculator reproduces.
The law is most valuable for weak electrolytes such as acetic acid. Strong electrolytes obey Kohlrausch's square-root law \(\Lambda_m = \Lambda^\circ_m - K\sqrt{c}\), so their limiting conductivity can be found by extrapolating measured values to zero concentration. Weak electrolytes are almost fully associated even at low concentration, so extrapolation fails — but Kohlrausch's law gives the answer directly: \(\Lambda^\circ_m(\text{CH}_3\text{COOH}) = \lambda^\circ(\text{H}^+) + \lambda^\circ(\text{CH}_3\text{COO}^-) = 349.65 + 40.9 \approx 390.6\) S·cm²/mol.
Once you know \(\Lambda^\circ_m\), comparing it with a measured molar conductivity \(\Lambda_m\) at concentration \(c\) gives the degree of dissociation \(\alpha = \Lambda_m / \Lambda^\circ_m\) — the standard route in dissociation-constant and solubility problems. Our Molar Conductivity Calculator handles the \(\Lambda_m = \kappa / c\) step, and the Solubility Product Calculator covers the sparingly-salt applications.
Assumptions and limitations
- Infinite dilution: the law strictly holds only as \(c \to 0\). At real concentrations, ion–ion interactions (ion atmospheres) lower \(\Lambda_m\) below \(\Lambda^\circ_m\); for strong electrolytes this follows \(\Lambda_m = \Lambda^\circ_m - K\sqrt{c}\). For how activities deviate at finite concentration, see the Debye–Hückel Activity Coefficient Calculator.
- Temperature: tabulated λ° values refer to 25 °C in water. Conductivity rises roughly 2–3 % per degree, so use values measured at your temperature if high accuracy matters.
- Per mole of ions: λ° is defined per mole of the ion as written, not per unit charge. For Ca²⁺, λ° = 119.0 S·cm²/mol already accounts for the double charge — do not multiply by 2 again.
- Charge neutrality: the stoichiometric coefficients must describe a neutral formula unit (\(\sum \nu_+ z_+ = \sum \nu_- z_-\)), otherwise the entered "salt" cannot exist and the sum is meaningless.
- Complete dissociation at infinite dilution: even weak electrolytes are treated as fully dissociated in the limit \(c \to 0\); partial dissociation only matters at finite concentration.
- Unit conversion: 1 S·cm²/mol = 10⁻⁴ S·m²/mol. Data tables almost always use S·cm²/mol, which is the calculator's default input unit.
Reference: limiting molar conductivities of common ions (25 °C, S·cm²/mol)
| Cation | λ° | Anion | λ° |
|---|---|---|---|
| H⁺ | 349.65 | OH⁻ | 198.3 |
| Li⁺ | 38.7 | Cl⁻ | 76.35 |
| Na⁺ | 50.11 | Br⁻ | 78.1 |
| K⁺ | 73.5 | I⁻ | 76.8 |
| NH₄⁺ | 73.5 | NO₃⁻ | 71.5 |
| Ag⁺ | 61.9 | HCO₃⁻ | 44.5 |
| Mg²⁺ | 106.1 | CH₃COO⁻ | 40.9 |
| Ca²⁺ | 119.0 | SO₄²⁻ | 160.0 |
| Ba²⁺ | 127.2 | ½SO₄²⁻ (per equivalent) | 80.0 |
Use these values directly in the calculator's λ° fields. Note the remarkably high values for H⁺ and OH⁻ — a consequence of the Grotthuss proton-hopping mechanism in water, which also explains why strong acids and bases conduct so much better than their salts.
Frequently Asked Questions
How do I calculate the limiting molar conductivity by hand?
Write the dissociation, count the ions, then add their tabulated λ° values. For CaCl₂ → Ca²⁺ + 2 Cl⁻: \( \Lambda^\circ_m = 1 \times 119.0 + 2 \times 76.35 = 271.7 \) S·cm²/mol. For each ion, multiply its limiting ionic conductivity by how many of that ion appear per formula unit, and sum everything. The calculator just automates this weighted sum, so a hand calculation is a good way to verify your inputs.
What is a normal value for the limiting molar conductivity?
For simple 1:1 salts in water at 25 °C, expect roughly 100–150 S·cm²/mol — NaCl is 126.46, KCl about 149.9. Divalent salts run higher (CaCl₂ ≈ 271.7, CuSO₄ ≈ 267) because they release three ions per formula unit. Acids and bases are far more conductive: HCl ≈ 426 and NaOH ≈ 248 S·cm²/mol, thanks to the very mobile H⁺ and OH⁻ ions. A result far outside 50–500 S·cm²/mol usually signals an input error.
How do I convert between S·cm²/mol and S·m²/mol?
Multiply by 10⁻⁴ to go from S·cm²/mol to the SI unit S·m²/mol: 126.46 S·cm²/mol = 0.012646 S·m²/mol. To go the other way, multiply by 10⁴. Older tables may list values in Ω⁻¹·cm²/mol, which is identical to S·cm²/mol since 1 S = 1 Ω⁻¹. The calculator's output unit dropdown does this conversion for you automatically.
Can I use this calculator for weak electrolytes like acetic acid or ammonia?
Yes — this is actually the law's most important application. Weak electrolytes are only partially dissociated at any measurable concentration, so you cannot extrapolate measured conductivities to zero concentration. Kohlrausch's law bypasses the problem: for CH₃COOH, \( \Lambda^\circ_m = 349.65 + 40.9 \approx 390.6 \) S·cm²/mol, using the fully dissociated ions' values. Use coefficient 1 for both ions of a monobasic weak acid.
What if my ion is not in the reference table?
Derive its λ° from tabulated salts using the additivity itself — the classic difference method. For example, if you know Λ°ₘ(NaCl) = 126.46 and Λ°ₘ(KCl) = 149.86, then λ°(K⁺) − λ°(Na⁺) = 23.40 S·cm²/mol. Combine two or three such salt values to isolate the unknown ion. For many polyatomic ions (NO₃⁻, SO₄²⁻, CH₃COO⁻) values are tabulated directly; otherwise consult a physical chemistry handbook for rarer ions.
Why shouldn't I multiply by the ion's charge?
Because tabulated λ° values are already defined per mole of ions as written, not per unit charge. λ°(Ca²⁺) = 119.0 S·cm²/mol already reflects the doubly charged calcium ion, so multiplying by 2 would double-count. Be alert for older tables that list values per equivalent (e.g. '½SO₄²⁻ = 80.0'); those must be multiplied by 2 before entering them as SO₄²⁻, or you will get half the correct answer.
How do I find the degree of dissociation and Ka once I know Λ°ₘ?
Measure the solution's conductivity κ, convert it to the molar conductivity \( \Lambda_m = \kappa / c \), then compute \( \alpha = \Lambda_m / \Lambda^\circ_m \). For a weak monobasic acid, \( K_a = c\alpha^2 / (1-\alpha) \). Note that α from conductance is an approximation (ion mobilities change slightly with concentration), but it is the standard textbook method and accurate enough for dilute solutions.