Limiting molar conductivity Λ°m
Molar conductivity $\Lambda_m$ is the conductance of all the ions produced by one mole of an electrolyte held between two electrodes one centimetre apart. It is defined as $\Lambda_m = \dfrac{\kappa \times 1000}{c}$, where $\kappa$ is the conductivity and $c$ the molar concentration. As a solution is diluted, the total number of ions per mole stays the same (for strong electrolytes) or rises (for weak ones), while inter-ionic attractions fall away. Both effects push $\Lambda_m$ upward as concentration drops.
The value $\Lambda_m$ approaches as concentration tends to zero is the limiting molar conductivity, written $\Lambda^\circ_m$. It is the molar conductivity at infinite dilution — the idealised state in which ions are so far apart that they no longer feel one another, and the electrolyte (if weak) is completely dissociated. Everything in this article hangs on that single quantity, so it is worth fixing its meaning before going further.
Crucially, $\Lambda^\circ_m$ cannot be reached experimentally — you cannot prepare a solution of zero concentration. It must instead be found by extrapolation (for strong electrolytes) or by calculation (for weak electrolytes). The difference between those two routes is exactly what Kohlrausch's law resolves.
Strong vs weak electrolytes
NCERT records that for a strong electrolyte such as $\ce{KCl}$, $\Lambda_m$ rises slowly and linearly with dilution and obeys the Debye–Hückel–Onsager relation $\Lambda_m = \Lambda^\circ_m - A\,c^{1/2}$. Plotting $\Lambda_m$ against $c^{1/2}$ gives a straight line whose intercept on the conductivity axis is $\Lambda^\circ_m$ and whose slope is $-A$. For the KCl data in NCERT Example 2.6, this intercept comes out to $\Lambda^\circ_m = 150.0~\text{S cm}^2\,\text{mol}^{-1}$.
A weak electrolyte such as acetic acid behaves quite differently. Its low degree of dissociation at ordinary concentrations means that diluting the solution does not merely separate existing ions — it produces more of them. Near low concentration $\Lambda_m$ shoots up steeply, the curve never settles onto a straight line, and the intercept cannot be read. The figure below contrasts the two shapes.
The strong electrolyte (teal) extrapolates cleanly to its intercept; the weak electrolyte (coral) rises so steeply near the axis that no reliable intercept exists. This is precisely why $\Lambda^\circ_m$ of acetic acid must be obtained by Kohlrausch's law, not by plotting.
Extrapolation works for strong, not weak
A common slip is to assume $\Lambda^\circ_m$ is always read off a graph. For strong electrolytes the $\Lambda_m$ vs $c^{1/2}$ line is straight and the intercept is trustworthy. For weak electrolytes the curve is too steep near the axis, so the only reliable route is Kohlrausch's law.
Strong → extrapolate the straight line. Weak → calculate via $\Lambda^\circ_m = \nu_+\lambda^\circ_+ + \nu_-\lambda^\circ_-$.
Statement of Kohlrausch's law
Kohlrausch examined $\Lambda^\circ_m$ values for many strong electrolytes and found a striking regularity. The difference in limiting molar conductivity between any sodium salt $\ce{NaX}$ and the corresponding potassium salt $\ce{KX}$ is almost constant, independent of $\ce{X}$:
$\Lambda^\circ_m(\ce{KCl}) - \Lambda^\circ_m(\ce{NaCl}) = \Lambda^\circ_m(\ce{KBr}) - \Lambda^\circ_m(\ce{NaBr}) = \Lambda^\circ_m(\ce{KI}) - \Lambda^\circ_m(\ce{NaI}) \approx 23.4~\text{S cm}^2\,\text{mol}^{-1}$
Likewise, the difference between any bromide and the corresponding chloride was constant at about $1.8~\text{S cm}^2\,\text{mol}^{-1}$. The only way the same number can keep appearing is if each ion brings a fixed contribution that does not depend on its partner. On this basis Kohlrausch enunciated his law of independent migration of ions:
At infinite dilution, when dissociation is complete, the limiting molar conductivity of an electrolyte is the sum of the individual contributions of its constituent cation and anion, each ion migrating independently of the other.
For a 1∶1 electrolyte such as sodium chloride this is simply $\Lambda^\circ_m(\ce{NaCl}) = \lambda^\circ_{\ce{Na+}} + \lambda^\circ_{\ce{Cl-}}$. In general, if an electrolyte dissociates to give $\nu_+$ cations and $\nu_-$ anions, then
$$\Lambda^\circ_m = \nu_+\,\lambda^\circ_+ \;+\; \nu_-\,\lambda^\circ_-$$
where $\lambda^\circ_+$ and $\lambda^\circ_-$ are the limiting molar conductivities of the cation and anion. The figure below shows the physical picture: at infinite dilution the two ions drift to opposite electrodes without dragging on each other.
At infinite dilution each ion travels to its electrode with a transport rate set only by its own size, charge and hydration — never by its counter-ion. The bulk conductivity is therefore an additive sum of independent ionic shares.
Ionic contributions λ°
Because each ion's contribution is fixed, NCERT tabulates $\lambda^\circ$ values once and lets you assemble any electrolyte you need. The headline entry is $\ce{H+}$, whose value dwarfs the others because protons hop along hydrogen-bonded water (the Grotthuss mechanism) rather than physically wading through the solution; $\ce{OH-}$ is high for the same reason.
| Cation | λ° / S cm² mol⁻¹ | Anion | λ° / S cm² mol⁻¹ |
|---|---|---|---|
H⁺ | 349.6 | OH⁻ | 199.1 |
Na⁺ | 50.1 | Cl⁻ | 76.3 |
K⁺ | 73.5 | Br⁻ | 78.1 |
Ca²⁺ | 119.0 | CH₃COO⁻ | 40.9 |
Mg²⁺ | 106.0 | SO₄²⁻ | 160.0 |
Assembling a multivalent salt simply means counting ions. For calcium chloride, NCERT computes $\Lambda^\circ_m(\ce{CaCl2}) = \lambda^\circ_{\ce{Ca^2+}} + 2\lambda^\circ_{\ce{Cl-}} = 119.0 + 2(76.3) = 271.6~\text{S cm}^2\,\text{mol}^{-1}$, and for magnesium sulphate $\Lambda^\circ_m(\ce{MgSO4}) = 106.0 + 160.0 = 266~\text{S cm}^2\,\text{mol}^{-1}$. The stoichiometric coefficients $\nu_+$ and $\nu_-$ are where most arithmetic errors creep in.
Don't forget the stoichiometric multipliers
For $\ce{CaCl2}$ the chloride term is $2\lambda^\circ_{\ce{Cl-}}$, not $\lambda^\circ_{\ce{Cl-}}$. For $\ce{Al2(SO4)3}$ it would be $2\lambda^\circ_{\ce{Al^3+}} + 3\lambda^\circ_{\ce{SO4^2-}}$. Writing each ionic conductivity once, regardless of how many of that ion appear, is the single most common sign error in this topic.
Always weight each $\lambda^\circ$ by the number of that ion released on dissociation.
Λ°m of a weak electrolyte
The law's most celebrated application is reaching a number you cannot measure. Acetic acid's $\Lambda^\circ_m$ is inaccessible by extrapolation, but it can be reconstructed from three strong electrolytes whose ions, added and subtracted, leave exactly $\ce{H+}$ and $\ce{CH3COO-}$ behind. NCERT writes:
$$\Lambda^\circ_m(\ce{CH3COOH}) = \Lambda^\circ_m(\ce{HCl}) + \Lambda^\circ_m(\ce{CH3COONa}) - \Lambda^\circ_m(\ce{NaCl})$$
The bookkeeping works because $\ce{HCl}$ supplies $\ce{H+} + \ce{Cl-}$, $\ce{CH3COONa}$ supplies $\ce{CH3COO-} + \ce{Na+}$, and subtracting $\ce{NaCl}$ removes the unwanted $\ce{Na+}$ and $\ce{Cl-}$, leaving $\ce{H+} + \ce{CH3COO-}$ — which is acetic acid. With the NCERT values 425.9, 91.0 and 126.4 S cm² mol⁻¹ this gives $390.5~\text{S cm}^2\,\text{mol}^{-1}$. This exact combination, with slightly different data, was asked in NEET 2021.
Shaky on what $\Lambda_m$, $\kappa$ and cell constant actually mean? Revise conductance of electrolytic solutions before tackling these applications.
Degree of dissociation α
Once $\Lambda^\circ_m$ is in hand, the degree of dissociation of a weak electrolyte at any working concentration follows immediately. At infinite dilution the electrolyte is fully dissociated ($\alpha = 1$) and conducts at $\Lambda^\circ_m$; at concentration $c$ only a fraction $\alpha$ of the molecules have split, so the measured $\Lambda_m$ is smaller in the same proportion. Hence:
$$\alpha = \frac{\Lambda_m}{\Lambda^\circ_m}$$
The ratio is dimensionless and always lies between 0 and 1. A larger $\alpha$ means more ions, more conductance, and a $\Lambda_m$ closer to the ceiling $\Lambda^\circ_m$. This relation was tested directly in NEET 2025, where a monobasic weak acid with $\Lambda_m = 90$ and $\Lambda^\circ_m = 349.6 + 50.4 = 400~\text{S cm}^2\,\text{mol}^{-1}$ gives $\alpha = 90/400 = 0.225$.
Dissociation constant Ka
With $\alpha$ known, the dissociation constant of a weak acid such as acetic acid drops out of the equilibrium expression. For $\ce{CH3COOH <=> CH3COO^- + H+}$ at concentration $c$, the equilibrium concentrations are $c\alpha$, $c\alpha$ and $c(1-\alpha)$, so
$$K_a = \frac{c\alpha^2}{1-\alpha}$$
Because $\alpha = \Lambda_m/\Lambda^\circ_m$, the entire calculation reduces to two measured conductivities and a concentration. When $\alpha$ is small, the denominator is close to 1 and the working simplifies to $K_a \approx c\alpha^2$. This was the NEET 2021 Q.93 calculation: $\Lambda^\circ_m = 400$, $\Lambda_m = 20$, so $\alpha = 1/20$ and $K_a = 7\times10^{-3} \times (1/20)^2 = 1.75\times10^{-5}~\text{mol L}^{-1}$.
Solubility of sparingly soluble salts
A salt such as $\ce{AgCl}$ or $\ce{BaSO4}$ dissolves so little that its saturated solution is essentially at infinite dilution. Its measured molar conductivity is therefore equal, to a good approximation, to the limiting molar conductivity, which Kohlrausch's law supplies from the ionic contributions: $\Lambda^\circ_m(\ce{AgCl}) = \lambda^\circ_{\ce{Ag+}} + \lambda^\circ_{\ce{Cl-}}$. The molar solubility then follows from rearranging the molar-conductivity definition:
$$S = \frac{\kappa \times 1000}{\Lambda^\circ_m}\quad(\text{mol L}^{-1})$$
where $\kappa$ is the measured conductivity of the saturated solution (after subtracting the conductivity of the water). Once $S$ is known, the solubility product follows directly — for a 1∶1 salt, $K_{sp} = S^2$. This route turns a notoriously hard quantity to measure into two routine conductivity readings.
Worked examples
Calculate $\Lambda^\circ_m$ for $\ce{CaCl2}$ and $\ce{MgSO4}$ using the ionic data in the table above.
For $\ce{CaCl2}$: it releases one $\ce{Ca^2+}$ and two $\ce{Cl-}$, so $\Lambda^\circ_m = \lambda^\circ_{\ce{Ca^2+}} + 2\lambda^\circ_{\ce{Cl-}} = 119.0 + 2(76.3) = 271.6~\text{S cm}^2\,\text{mol}^{-1}$.
For $\ce{MgSO4}$: one $\ce{Mg^2+}$ and one $\ce{SO4^2-}$, so $\Lambda^\circ_m = 106.0 + 160.0 = 266~\text{S cm}^2\,\text{mol}^{-1}$. Note the chloride was doubled but the sulphate was not.
The molar conductivity of 0.025 mol L⁻¹ methanoic acid is 46.1 S cm² mol⁻¹. Find its degree of dissociation and dissociation constant. Given $\lambda^\circ_{\ce{H+}} = 349.6$ and $\lambda^\circ_{\ce{HCOO-}} = 54.6~\text{S cm}^2\,\text{mol}^{-1}$. (NCERT Intext 2.9)
Step 1 — Λ°m by Kohlrausch: $\Lambda^\circ_m = 349.6 + 54.6 = 404.2~\text{S cm}^2\,\text{mol}^{-1}$.
Step 2 — degree of dissociation: $\alpha = \dfrac{\Lambda_m}{\Lambda^\circ_m} = \dfrac{46.1}{404.2} = 0.114$.
Step 3 — dissociation constant: $K_a = \dfrac{c\alpha^2}{1-\alpha} = \dfrac{0.025 \times (0.114)^2}{1-0.114} \approx 3.67\times10^{-4}~\text{mol L}^{-1}$.
$\Lambda^\circ_m$ for $\ce{NaCl}$, $\ce{HCl}$ and $\ce{CH3COONa}$ are 126.4, 425.9 and 91.0 S cm² mol⁻¹. Find $\Lambda^\circ_m$ for acetic acid. (NCERT Example 2.8)
Set up the ionic bookkeeping: $\Lambda^\circ_m(\ce{CH3COOH}) = \Lambda^\circ_m(\ce{HCl}) + \Lambda^\circ_m(\ce{CH3COONa}) - \Lambda^\circ_m(\ce{NaCl})$.
Substitute: $= 425.9 + 91.0 - 126.4 = 390.5~\text{S cm}^2\,\text{mol}^{-1}$. The $\ce{Na+}$ and $\ce{Cl-}$ cancel, leaving exactly $\ce{H+} + \ce{CH3COO-}$.
Five things to carry into the exam
- The law: $\Lambda^\circ_m = \nu_+\lambda^\circ_+ + \nu_-\lambda^\circ_-$ — each ion contributes a fixed, independent share at infinite dilution.
- Strong vs weak: $\Lambda^\circ_m$ is extrapolated for strong electrolytes, but must be calculated for weak ones because their $\Lambda_m$ vs $c^{1/2}$ curve is too steep.
- Weak Λ°m: $\Lambda^\circ_m(\ce{CH3COOH}) = \Lambda^\circ_m(\ce{HCl}) + \Lambda^\circ_m(\ce{CH3COONa}) - \Lambda^\circ_m(\ce{NaCl})$.
- Dissociation: $\alpha = \Lambda_m/\Lambda^\circ_m$, then $K_a = c\alpha^2/(1-\alpha) \approx c\alpha^2$ for small $\alpha$.
- Solubility: for a sparingly soluble salt, $S = \kappa \times 1000 / \Lambda^\circ_m$, and $K_{sp} = S^2$ for a 1∶1 salt.