0.1
eq/L
1.290000e-2
S·m²/eq
1.290000e+2
S·cm²/eq
100
eq/m³
0.1
eq/L
1.290000e-2
S·m²/eq
1.290000e+2
S·cm²/eq
100
eq/m³
Equivalent conductivity (Λeq) is a classical electrochemical parameter that expresses the conductivity of a solution per gram-equivalent of dissolved electrolyte. While modern chemistry increasingly favors molar conductivity, equivalent conductivity remains important in analytical chemistry, titration calculations, and understanding historical literature. The Equivalent Conductivity Calculator computes Λeq from specific conductivity, molar concentration, and the n-factor (number of equivalents per mole). The n-factor depends on the reaction context: for acids, it equals the number of replaceable H⁺ ions; for salts, the total positive charge; and for redox reactions, the electrons transferred. This approach normalizes conductivity to the charge transferred per formula unit, making it especially useful for comparing electrolytes with different stoichiometries. The calculator outputs both traditional CGS units (S·cm²/eq) and SI units (S·m²/eq) along with the solution's normality.
Equivalent conductivity is defined as:
$$\Lambda_{eq} = \frac{\kappa}{c_{eq}}$$
where ceq is the equivalent concentration (normality) in eq/m³. The normality is related to molarity by the n-factor:
$$N = c \times n$$
Therefore:
$$\Lambda_{eq} = \frac{\kappa}{c \times n \times 1000}$$
where κ is in S/m and c is in mol/L (the factor 1000 converts L to m³). In CGS units:
$$\Lambda_{eq} \text{(S·cm²/eq)} = \Lambda_{eq} \text{(S·m²/eq)} \times 10^4$$
The relationship between molar and equivalent conductivity is:
$$\Lambda_m = n \times \Lambda_{eq}$$
For a 1:1 electrolyte like NaCl (n = 1), molar and equivalent conductivities are identical. For H₂SO₄ (n = 2), Λm = 2 × Λeq.
Equivalent conductivity normalizes the conducting ability per unit of charge, making it easier to compare electrolytes of different valence types. For example, comparing NaCl (1:1) with CaCl₂ (2:1) is more meaningful on an equivalent basis. At infinite dilution, Λ°eq equals the sum of individual ionic equivalent conductivities: Λ°eq = λ°+ + λ°-. The n-factor is crucial: for H₂SO₄ → 2H⁺ + SO₄²⁻, n = 2; for NaOH, n = 1; for Al₂(SO₄)₃, n = 6 (based on 6+ total charge). If your result appears halved or doubled compared to expectations, verify the n-factor for your specific electrolyte and reaction.
Inputs
Results
For 1:1 electrolytes like KCl, equivalent conductivity equals molar conductivity. 0.1 N KCl gives Λeq = 129.0 S·cm²/eq.
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Results
H₂SO₄ has n=2 (diprotic acid). The normality is 0.1 N and Λeq = 414.0 S·cm²/eq. The molar conductivity would be 828 S·cm²/mol.
The n-factor is the number of equivalents per mole of substance. For acids, it is the number of H⁺ ions donated (HCl: n=1, H₂SO₄: n=2, H₃PO₄: n=3). For bases, it is the number of OH⁻ ions. For salts, it equals the total positive (or negative) charge per formula unit.
$$\Lambda_m = n \times \Lambda_{eq}$$. For NaCl (n=1), they are equal. For MgSO₄ (n=2), the molar conductivity is twice the equivalent conductivity.
IUPAC recommends using molar conductivity as the standard. However, equivalent conductivity remains in use for titration calculations, water treatment specifications, and interpreting older literature. Many textbooks still cover it for completeness.
Normality (N) = Molarity × n-factor. It expresses concentration in equivalents per liter. For 0.1 M H₂SO₄ with n=2, the normality is 0.2 N. Normality is reaction-specific—the same solution can have different normalities for acid-base vs redox reactions.
The n-factor depends on what the substance does in a reaction. H₃PO₄ has n=1, 2, or 3 depending on whether it loses one, two, or three protons. KMnO₄ has n=5 in acidic medium (Mn⁷⁺→Mn²⁺) but n=3 in neutral medium (Mn⁷⁺→Mn⁴⁺).
λ°(H⁺) = 349.8, λ°(OH⁻) = 198.0, λ°(K⁺) = 73.5, λ°(Na⁺) = 50.1, λ°(Cl⁻) = 76.3 S·cm²/eq at 25°C. These individual ionic values sum to give Λ°eq of electrolytes.
For weak electrolytes, Λ°eq cannot be obtained by extrapolation. Use Kohlrausch's law: Λ°eq = λ°₊ + λ°₋, where individual ionic conductivities come from strong electrolyte data.
Yes. For a weak acid: α = Λeq/Λ°eq, then Ka = Nα²/(1−α) where N is normality. This is Ostwald's dilution law applied using equivalent conductivity.
1 S·m²/eq = 10,000 S·cm²/eq. Multiply S·m²/eq by 10⁴ to get S·cm²/eq, or divide S·cm²/eq by 10⁴ for S·m²/eq.
Equivalent conductivity always increases with dilution, approaching Λ°eq at infinite dilution. For strong electrolytes, the increase is gradual (Onsager equation). For weak electrolytes, the increase is dramatic because dilution promotes dissociation.
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