Equivalent Conductivity

Dip two electrodes into a salt solution and measure how well it carries current. The result depends on concentration, so chemists normalize it into a conductivity per amount of substance. Equivalent conductivity is one such normalization, built around the equivalent rather than the mole. It matters because 19th- and 20th-century electrochemistry, including Kohlrausch's foundational measurements, was built on equivalents. Many textbooks, handbooks, and older data tables still report equivalent conductivities. Reading them correctly requires knowing exactly how this quantity relates to molar conductivity. This article defines equivalent conductivity carefully. It connects the quantity to molar conductivity through the charge number. It walks through the derivation, two worked examples, and the traps that catch students converting between the two units.

Key takeaways

Equivalent conductivity $\Lambda_{eq}$ measures conductivity per mole of charge equivalents, not per mole of formula units.
It relates to molar conductivity by $\Lambda_{eq} = \Lambda_m / z$ , where $z$ is the number of charges transferred per formula unit.
For 1:1 electrolytes like NaCl, $\Lambda_{eq} = \Lambda_m$ ; for 2:2 or higher charge salts they differ.
Both quantities decrease with concentration for weak electrolytes and decrease modestly for strong electrolytes, following Kohlrausch's square-root law.
Equivalent conductivity was the historically preferred quantity because it makes ionic contributions additive per equivalent.
Key idea: dividing conductivity by equivalents instead of moles absorbs the ion's charge number, so equivalent conductivity is molar conductivity rescaled by $1/z$ .

The core insight

Equivalent conductivity is defined so that one equivalent of any electrolyte, positive or negative ion, carries exactly one mole of elementary charge across the cell; dividing the measured conductivity by concentration expressed in equivalents (rather than moles) automatically divides molar conductivity by the charge number $z$ , giving $\Lambda_{eq} = \Lambda_m/z$ .

The physical picture

Consider a conductivity cell filled with an electrolyte solution and connected to an AC bridge. The measured conductance $G$ (in siemens) depends on cell geometry and on how many charge carriers are dissolved per unit volume. To strip away the geometry, chemists define the conductivity $\kappa = G \cdot (\ell/A)$ , where $\ell$ is the electrode separation and $A$ the electrode area. This is an intrinsic property of the solution. Conductivity still depends on concentration: double the salt and (roughly) double the conductivity. Molar conductivity, $\Lambda_m = \kappa/c$ , removes this concentration dependence by dividing by molar concentration. But an ion like $\text{Ca}^{2+}$ carries twice the charge of $\text{Na}^+$ . If you want a quantity where "one unit of substance" always corresponds to "one mole of charge moved," you must divide by concentration expressed in equivalents, not moles. That quantity is equivalent conductivity. An equivalent of an electrolyte is defined as the amount that supplies or accepts one mole of elementary charge in the relevant reaction. For $\text{CaCl}_2$ , one mole of the salt releases two moles of positive charge from $\text{Ca}^{2+}$ (and two moles of negative charge from the chlorides), so one mole equals two equivalents. For $\text{NaCl}$ , one mole equals one equivalent because $z=1$ .

Figure

Cell geometry and the definitions of κ, Λm, and Λeq

The second useful picture is how these quantities change with concentration. Strong electrolytes show a mild decrease as concentration rises (ion-ion interactions slow migration), following Kohlrausch's empirical square-root law. Weak electrolytes show a much steeper drop, dominated by incomplete dissociation. Because $\Lambda_{eq} = \Lambda_m/z$ is just a rescaling by a constant, the two quantities trace the same shape of curve, only offset vertically for polyvalent electrolytes.

Figure

Concentration dependence: molar vs equivalent conductivity for a 2:2 salt

The core relations

\Lambda_{eq} = \frac{\kappa}{c_{eq}}

where: $\kappa$ is conductivity (S m⁻¹), $c_{eq}$ is concentration expressed in equivalents per unit volume.

c_{eq} = z \, c

where: $z$ is the number of charge units exchanged per formula unit of electrolyte (e.g., $z=2$ for $\text{CaCl}_2$ , $\text{MgSO}_4$ ).

\Lambda_{eq} = \frac{\Lambda_m}{z}

where: $\Lambda_m = \kappa/c$ is the ordinary molar conductivity defined on the formula unit.

\Lambda_{eq}^{\circ} = \frac{\lambda_+^{\circ}}{z_+ \nu_+} \cdot \nu_+ + \frac{\lambda_-^{\circ}}{z_- \nu_-} \cdot \nu_- \;\; \Longrightarrow \;\; \Lambda_{eq}^{\circ} = \lambda_+^{eq,\circ} + \lambda_-^{eq,\circ}

where: at infinite dilution, equivalent conductivity splits additively into equivalent ionic conductivities, foreshadowing Kohlrausch's law of independent migration.

Derivation AFrom molar conductivity to equivalent conductivity

Goal. Show rigorously that equivalent conductivity is molar conductivity divided by the charge number $z$ , starting only from the definitions of conductivity and equivalent concentration.

Step 1. Start from measured conductance. In a cell of electrode separation $\ell$ and cross-sectional area $A$ , the measured conductance $G$ (siemens) relates to the intrinsic conductivity by

\kappa = G\,\frac{\ell}{A}.

Here $\kappa$ has units S m⁻¹ and depends only on the solution, not on cell geometry.

Step 2. Define molar concentration and molar conductivity. Let $c$ be the molar concentration (mol m⁻³) of the electrolyte formula unit, e.g. moles of $\text{CaCl}_2$ per cubic meter. Molar conductivity is

\Lambda_m = \frac{\kappa}{c}.

This was defined and used in the treatment of molar conductivity as the basic normalized transport quantity.

Step 3. Define the equivalent concentration. An equivalent is the amount of electrolyte that furnishes one mole of elementary positive (or negative) charge upon complete dissociation. If each formula unit releases $z$ moles of charge per mole of formula unit — where $z$ equals the magnitude of the cation charge times the number of cations per formula unit, equivalently the total positive charge released — then the equivalent concentration is

c_{eq} = z\,c.

For $\text{NaCl} \to \text{Na}^+ + \text{Cl}^-$ , one mole gives one mole of positive charge, so $z=1$ . For $\text{CaCl}_2 \to \text{Ca}^{2+} + 2\text{Cl}^-$ , one mole gives two moles of positive charge (and two moles of negative charge), so $z=2$ . For $\text{Al}_2(\text{SO}_4)_3$ , one mole gives six moles of positive charge from $2\,\text{Al}^{3+}$ , so $z=6$ .

Step 4. Substitute into the definition of equivalent conductivity. By construction,

\Lambda_{eq} = \frac{\kappa}{c_{eq}} = \frac{\kappa}{zc}.

Recognize that $\kappa/c$ is exactly $\Lambda_m$ from Step 2. Substituting gives

\Lambda_{eq} = \frac{1}{z}\left(\frac{\kappa}{c}\right) = \frac{\Lambda_m}{z}.

Step 5. Confirm the additive-ion picture survives rescaling. At infinite dilution, molar conductivity splits as $\Lambda_m^{\circ} = \nu_+\lambda_+^{\circ} + \nu_-\lambda_-^{\circ}$ , with $\nu_\pm$ the stoichiometric numbers of ions and $\lambda_\pm^{\circ}$ the molar ionic conductivities. Dividing through by $z$ and grouping each term with its own charge number defines equivalent ionic conductivities $\lambda_\pm^{eq,\circ} = \lambda_\pm^{\circ}/z_\pm$ , which for any electrolyte satisfy

\Lambda_{eq}^{\circ} = \lambda_+^{eq,\circ} + \lambda_-^{eq,\circ}.

This additive splitting per equivalent, rather than per mole, is the quantity Kohlrausch tabulated and the reason equivalent conductivity became the historical standard for tables of ionic mobilities.

Boxed result.

\boxed{\Lambda_{eq} = \frac{\Lambda_m}{z}, \qquad c_{eq} = zc}

Worked examples

Example 1 — Foundational: 0.100 M KCl. A conductivity cell filled with 0.100 mol L⁻¹ KCl solution at 25 °C is measured to have conductivity $\kappa = 1.288\ \text{S m}^{-1}$ (a standard reference value for this concentration). Step 1: convert concentration to SI. $c = 0.100\ \text{mol L}^{-1} = 100\ \text{mol m}^{-3}$ . Step 2: compute molar conductivity.

\Lambda_m = \frac{\kappa}{c} = \frac{1.288\ \text{S m}^{-1}}{100\ \text{mol m}^{-3}} = 0.01288\ \text{S m}^2\ \text{mol}^{-1} = 128.8\ \text{S cm}^2\ \text{mol}^{-1}.

Step 3: identify $z$ . KCl dissociates as $\text{K}^+ + \text{Cl}^-$ , releasing one mole of charge per mole of salt, so $z=1$ . Step 4: equivalent conductivity.

\Lambda_{eq} = \frac{\Lambda_m}{z} = \frac{128.8\ \text{S cm}^2\ \text{mol}^{-1}}{1} = 128.8\ \text{S cm}^2\ \text{eq}^{-1}.

Sanity check: for a 1:1 electrolyte, equivalent and molar conductivity must be numerically identical, and they are. This matches tabulated literature values (Λm for 0.1 M KCl is close to 128.9 S cm² mol⁻¹ at 25 °C), confirming internal consistency.

Example 2 — Applied: comparing CaCl₂ and Al₂(SO₄)₃ at the same molar concentration. Two solutions, both at $c = 0.0500\ \text{mol L}^{-1} = 50.0\ \text{mol m}^{-3}$ , are measured at 25 °C:

$\text{CaCl}_2$ : $\kappa_1 = 1.020\ \text{S m}^{-1}$
$\text{Al}_2(\text{SO}_4)_3$ : $\kappa_2 = 1.470\ \text{S m}^{-1}$

We want both $\Lambda_m$ and $\Lambda_{eq}$ for each, and to see how equivalent conductivity brings them onto a comparable footing. Step 1: molar conductivities.

\Lambda_{m,1} = \frac{1.020\ \text{S m}^{-1}}{50.0\ \text{mol m}^{-3}} = 0.02040\ \text{S m}^2\ \text{mol}^{-1} = 204.0\ \text{S cm}^2\ \text{mol}^{-1}.

\Lambda_{m,2} = \frac{1.470\ \text{S m}^{-1}}{50.0\ \text{mol m}^{-3}} = 0.02940\ \text{S m}^2\ \text{mol}^{-1} = 294.0\ \text{S cm}^2\ \text{mol}^{-1}.

Step 2: determine $z$ for each salt. $\text{CaCl}_2 \to \text{Ca}^{2+} + 2\text{Cl}^-$ : one mole releases 2 mol of positive charge, so $z_1 = 2$ . $\text{Al}_2(\text{SO}_4)_3 \to 2\text{Al}^{3+} + 3\text{SO}_4^{2-}$ : one mole releases $2\times3 = 6$ mol of positive charge, so $z_2 = 6$ . Step 3: equivalent conductivities.

\Lambda_{eq,1} = \frac{204.0}{2} = 102.0\ \text{S cm}^2\ \text{eq}^{-1}.

\Lambda_{eq,2} = \frac{294.0}{6} = 49.0\ \text{S cm}^2\ \text{eq}^{-1}.

Step 4: interpretation. Even though $\text{Al}_2(\text{SO}_4)_3$ has the larger molar conductivity (294.0 vs 204.0 S cm² mol⁻¹), its equivalent conductivity is much smaller (49.0 vs 102.0 S cm² eq⁻¹) once we divide by its much larger $z=6$ . This reflects that each mole of $\text{Al}_2(\text{SO}_4)_3$ moves far more charge through the cell per formula unit, so normalizing per unit of charge transported reveals a smaller per-equivalent transport efficiency, dominated by strong ion-pairing and the higher effective friction of highly charged, heavily hydrated $\text{Al}^{3+}$ ions. Sanity check: units are S cm² eq⁻¹ throughout; both equivalent conductivities are smaller than their parent molar conductivities, as required since $z>1$ in both cases, and $\Lambda_{eq,2} < \Lambda_{eq,1}$ is physically reasonable given the stronger electrostatic drag on trivalent aluminum ions.

Limiting cases & common traps

Situation	What happens	Common mistake
1:1 electrolyte (NaCl, KCl, HCl)	$z=1$ , so $\Lambda_{eq} = \Lambda_m$ exactly	Forgetting they are only equal for $z=1$ , not in general
2:1 or 1:2 electrolyte (CaCl₂, Na₂SO₄)	$z=2$ ; $\Lambda_{eq} = \Lambda_m/2$	Using the salt's total ion count instead of charge released
3:2 electrolyte (Al₂(SO₄)₃)	$z=6$ ; large gap between $\Lambda_m$ and $\Lambda_{eq}$	Using $z=3$ (cation charge) instead of total charge per formula unit
Infinite dilution limit	Both $\Lambda_m^{\circ}$ and $\Lambda_{eq}^{\circ}$ approach finite limits; ratio stays fixed at $z$	Assuming $\Lambda_{eq}^{\circ}$ depends on concentration like the finite- $c$ values do
Mixed electrolytes	Equivalents add linearly (charge is conserved); moles do not always add simply	Adding molar conductivities of a mixture directly instead of converting to equivalents first

The chemist behind it

Friedrich Kohlrausch spent decades in the late 19th century making exquisitely precise conductivity measurements, extrapolating molar and equivalent conductivities to infinite dilution and discovering that they split additively into independent ionic contributions. His tables of equivalent ionic conductivities, painstakingly measured with self-built AC bridges to avoid electrode polarization, remain the backbone of quantitative electrochemistry teaching today. The equivalent, rather than the mole, was his natural unit because it let him compare $\text{H}^+$ , $\text{Na}^+$ , and $\text{Ca}^{2+}$ mobilities on the same additive footing.

Origins · 1 minFriedrich Kohlrausch and how this was discovered

Friedrich Kohlrausch

1840-1910 · German physicist who established precision conductometry and the law of independent ionic migration

Ask the tutor: How does equivalent conductivity connect to Kohlrausch's law of independent migration for predicting the limiting conductivity of a weak acid?

Is equivalent conductivity still used today?

Less so in modern SI-based teaching, where molar conductivity is the default, but equivalent conductivity remains common in engineering handbooks, environmental chemistry (e.g., reporting water hardness or ionic strength), and historical data tables. Because the definitions differ only by the fixed factor $z$ , converting old equivalent-conductivity data to molar conductivity is straightforward once you identify the charge number for the electrolyte in question.

Does the charge number z depend on which ion you focus on, cation or anion?

No — $z$ is defined as the total charge (in units of elementary charge) transferred per formula unit upon complete dissociation, and charge neutrality guarantees the positive and negative charges released are equal in magnitude. For CaCl₂, the cation releases 2 units of positive charge and the two chlorides release 2 units of negative charge together; either count gives the same $z=2$ .

Why does equivalent conductivity make Kohlrausch's law of independent migration look cleaner?

Because equal equivalents of different ions carry identical charge, their conductivities can be added and subtracted directly without extra charge-weighting factors. Molar ionic conductivities still work but require tracking $z_+$ and $z_-$ separately when comparing ions of different valence; equivalent conductivities absorb that factor automatically, so tables of equivalent ionic conductivities compare directly across the periodic table.

How do I find z quickly for an unfamiliar salt?

Write the full dissociation equation and count the total positive charge released per formula unit — equivalently, multiply the cation's charge number by the number of cations in the formula. For $\text{Fe}_2(\text{SO}_4)_3 \to 2\text{Fe}^{3+} + 3\text{SO}_4^{2-}$ , $z = 2\times3 = 6$ , matching the total negative charge $3\times2=6$ as a built-in check.

Does Λeq still follow Kohlrausch's square-root law with concentration?

Yes. Since $\Lambda_{eq} = \Lambda_m/z$ is a fixed rescaling of $\Lambda_m$ , and $\Lambda_m$ for strong electrolytes follows $\Lambda_m = \Lambda_m^{\circ} - K\sqrt{c}$ , equivalent conductivity obeys the same square-root form with a rescaled coefficient, $\Lambda_{eq} = \Lambda_{eq}^{\circ} - (K/z)\sqrt{c}$ . The concentration dependence is preserved; only the vertical scale changes.