How does Kohlrausch's law find limiting molar conductivity of a weak electrolyte?

For CH3COOH: combine strong-electrolyte data as Lambda-infinity of CH3COONa plus HCl minus NaCl. This works because at infinite dilution ionic contributions are independent and additive.

What are the units of specific, equivalent, and molar conductance?

Specific conductance: S/cm. Equivalent conductance: S cm2/eq. Molar conductance: S cm2/mol. In SI molar conductance may be S m2/mol, with a factor of 10-4 conversion.

Why does adding water to strong electrolyte increase molar conductance slightly?

Strong electrolytes are already fully dissociated. Dilution only reduces small inter-ionic interference so molar conductance rises marginally, linearly in root C per Debye-Huckel-Onsager.

Can conductance determine solubility of BaSO4?

Yes. Measure kappa of saturated solution. Since concentration is very low, use limiting molar conductivity to compute molarity, then multiply by molar mass for grams per litre.

Conductance Of Electrolytic Solutions

Q: What is a cell constant and why must it be measured?

Cell constant is the ratio of electrode separation to electrode area. Because measuring these directly is inaccurate, we calibrate with a KCl solution of known specific conductance and compute the cell constant from the measured resistance.

Q: Why is H+ ion mobility so much higher than other ions?

H+ doesn't migrate as a whole ion. A proton hops from one water molecule to the next along the hydrogen-bond network, an effect called the Grotthuss mechanism, which is faster than bodily ion transport.

Q: Difference between conductance and conductivity?

Conductance is reciprocal resistance and depends on cell shape. Conductivity is specific conductance, an intensive property independent of shape.

ChemistryElectrochemistryFor NEET aspirants

The conductance of an electrolytic solution measures how easily current flows through it, and depends on the number of ions present and how fast they migrate. Three main quantities describe this: specific conductance ( $κ$ ), equivalent conductance ( $Λ_{e q}$ ), and molar conductance ( $Λ_{m}$ ). Their variation with concentration, captured by Kohlrausch's law of independent migration of ions, lets us calculate degree of dissociation of weak electrolytes, solubility products of sparingly soluble salts, and limiting conductivities.

Key Formulas - Quick Reference

Resistance: $R = ρ \frac{ℓ}{A}$ where $ρ$ = resistivity ( $Ω$ cm), $ℓ$ = length, $A$ = area of cross-section
Conductance: $G = \frac{1}{R}$ (unit: siemens S $= Ω^{- 1}$ )
Specific conductance (conductivity): $κ = \frac{1}{ρ} = G \times \frac{ℓ}{A}$ (S cm $^{- 1}$ )
Cell constant: $G^{*} = \frac{ℓ}{A}$ (unit: cm $^{- 1}$ )
Equivalent conductance: $Λ_{e q} = \frac{κ \times 1000}{N}$ (S cm $^{2}$ eq $^{- 1}$ ), where $N$ = normality
Molar conductance: $Λ_{m} = \frac{κ \times 1000}{M}$ (S cm $^{2}$ mol $^{- 1}$ ), where $M$ = molarity
Kohlrausch's law: $Λ_{m}^{\infty} = ν_{+} λ_{+}^{\infty} + ν_{-} λ_{-}^{\infty}$
Degree of dissociation (weak electrolyte): $α = \frac{Λ _{m}^{c}}{Λ _{m}^{\infty}}$
Dissociation constant: $K_{a} = \frac{C α ^{2}}{1 - α}$

1. Conductance in Electrolytic Solutions

Electrolytic (ionic) conduction differs from metallic conduction in three key ways:

Feature	Metallic Conduction	Electrolytic Conduction
Charge carriers	Free electrons	Cations and anions
Effect of temperature	Conductance decreases	Conductance increases
Chemical change	None (metal is unchanged)	Electrolyte decomposes at electrodes
Obeys Ohm's law	Yes	Yes (at low potential)

1.1 Basic Definitions

Resistance (

R

): opposition to current flow.

R = ρ \frac{ℓ}{A}

where

ρ

= resistivity (specific resistance).

Conductance (

G

): reciprocal of resistance.

G = \frac{1}{R}

. SI unit: siemens (S)

= Ω^{- 1}

= mho.

Specific conductance (

κ

, kappa): reciprocal of resistivity.

κ = \frac{1}{ρ}

. Physically, the conductance of a solution held between two parallel electrodes 1 cm apart, each of 1 cm

^{2}

area. Unit: S cm

^{- 1}

or S m

^{- 1}

Cell constant (

G^{*}

G^{*} = \frac{ℓ}{A}

; a geometric constant of the conductivity cell (unit: cm

^{- 1}

). Determined by calibration with a standard solution (usually KCl) whose

κ

is known:

G^{*} = κ_{KCl} \times R_{measured}

κ = G \times G^{*} = \frac{1}{R} \times \frac{ℓ}{A}

Figure 1: A conductivity cell.

ℓ

and

A

are fixed by the hardware, so their ratio

G^{*} = ℓ / A

is a property of the cell, not of the liquid. Because both are awkward to measure directly,

G^{*}

is found once with standard KCl and then reused.

2. Equivalent and Molar Conductance

Specific conductance depends only on ion concentration in a unit volume. But different electrolytes have different valences, so a fairer comparison uses conductance normalised per mole (or equivalent) of solute:

Equivalent conductance (

Λ_{e q}

): conductance of solution containing 1 gram-equivalent of electrolyte, placed between electrodes 1 cm apart.

Λ_{e q} = \frac{κ \times 1000}{N} (S cm^{2} eq^{- 1})

Molar conductance (

Λ_{m}

): conductance per mole.

Λ_{m} = \frac{κ \times 1000}{M} (S cm^{2} mol^{- 1})

The factor 1000 converts $M$ (mol/L) to mol/cm $^{3}$ : since $κ$ is per cm $^{3}$ but concentration is per litre, $Λ_{m} = \frac{κ (S/cm)}{C (mol/cm ^{3} )} = \frac{κ \times 1000}{C (mol/L)}$ .

Relation:

Λ_{m} = n \times Λ_{e q}

, where

n

is the number of equivalents per mole (e.g., 2 for

CuSO_{4}

, 3 for

AlCl_{3}

3. Effect of Dilution on Conductance

Dilution has opposite effects on specific and molar conductance:

Specific conductance ( $κ$ ) decreases on dilution, because the number of ions per cm $^{3}$ falls. Fewer charge carriers per unit volume mean less current.
Equivalent and molar conductance ( $Λ_{e q}$ , $Λ_{m}$ ) increase on dilution. Even though ion concentration drops, the total volume of solution containing the same number of ions grows faster, and ions become more mobile as ionic interference reduces.

Figure 2: Both curves are for the same KCl solutions. Read them right to left to see what dilution does.

κ

is counted per cm

^{3}

of solution,

Λ_{m}

per mole of solute, which is the whole reason they move in opposite directions.

Figure 3: Both curves come straight from their equations,

Λ_{m} = Λ_{m}^{\infty} - b C

for KCl and

Λ_{m} = α Λ_{m}^{\infty}

with

α

from Ostwald's law for acetic acid. The KCl line is drawn from the limiting law itself, so it stays perfectly straight further out than measured data would. The weak curve is steep near the axis because

α

itself is collapsing, not because the ions slow down.

Why the difference?

Strong electrolyte: already fully dissociated at all concentrations, so $Λ_{m}$ changes only slightly (due to reduced inter-ionic interference on dilution). Extrapolation of the linear plot gives $Λ_{m}^{\infty}$ .
Weak electrolyte: dissociation is incomplete and increases sharply on dilution. Near infinite dilution, $α \to 1$ and $Λ_{m}$ suddenly shoots up towards $Λ_{m}^{\infty}$ . Extrapolation is unreliable, so we use Kohlrausch's law instead.

4. Kohlrausch's Law of Independent Migration

At infinite dilution, when the dissociation of an electrolyte is complete and inter-ionic interactions vanish, each ion contributes independently to the total conductivity. The limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of its constituent ions, weighted by their stoichiometric coefficients:

Λ_{m}^{\infty} = ν_{+} λ_{+}^{\infty} + ν_{-} λ_{-}^{\infty}

For example:

$Λ_{m}^{\infty} (NaCl) = λ^{\infty} (Na^{+}) + λ^{\infty} (Cl^{-})$
$Λ_{m}^{\infty} (Na_{2} SO_{4}) = 2 λ^{\infty} (Na^{+}) + λ^{\infty} (SO_{4}^{2 -})$
$Λ_{m}^{\infty} (AlCl_{3}) = λ^{\infty} (Al^{3 +}) + 3 λ^{\infty} (Cl^{-})$

4.1 Applications of Kohlrausch's Law

(a) Calculation of $Λ_{m}^{\infty}$ for weak electrolytes

Since direct extrapolation fails for weak electrolytes, use strong-electrolyte data to compute $Λ_{m}^{\infty}$ indirectly:

Λ_{m}^{\infty} (CH_{3} COOH) = Λ_{m}^{\infty} (CH_{3} COONa) + Λ_{m}^{\infty} (HCl) - Λ_{m}^{\infty} (NaCl)

Verify by expanding: $[Na^{+} + CH_{3} COO^{-}] + [H^{+} + Cl^{-}] - [Na^{+} + Cl^{-}] = H^{+} + CH_{3} COO^{-}$ . ✓

Figure 4: Kohlrausch's law is ion bookkeeping. Pick strong electrolytes whose spectator ions cancel, and the leftover is exactly the weak electrolyte you could not measure directly.

(b) Degree of dissociation of weak electrolyte

α = \frac{Λ _{m}^{c}}{Λ _{m}^{\infty}}

(c) Dissociation constant of weak acid/base (via Ostwald's dilution law)

K_{a} = \frac{C α ^{2}}{1 - α}

(d) Solubility of sparingly soluble salts

For salts like $AgCl$ , $BaSO_{4}$ : measure $κ$ of saturated solution, use $Λ_{m}$ (assumed = $Λ_{m}^{\infty}$ at such low concentration) to compute molarity, which equals solubility.

Solubility = \frac{κ \times 1000}{Λ _{m}^{\infty}} mol/L

5. Ionic Mobility and Transport Number

Ionic mobility (

μ

): drift velocity of an ion per unit electric field (cm

^{2}

^{- 1}

^{- 1}

). Related to limiting ionic conductivity by

λ^{\infty} = F \times μ

Ions with high mobility are $H^{+}$ ( $λ^{\infty} = 350$ ) and $OH^{-}$ ( $λ^{\infty} = 198$ ) due to their unique Grotthuss mechanism (proton hopping through the hydrogen-bond network of water, faster than bodily migration). Most other ions cluster in the 40-80 range.

Transport number

t_{+}

= fraction of total current carried by cations

= \frac{λ _{+}^{\infty}}{λ _{+}^{\infty} + λ _{-}^{\infty}}

. Similarly

t_{-}

for anions, and

t_{+} + t_{-} = 1

6. Solved Examples

Solved Example 1

The equivalent conductance of a monobasic acid at infinite dilution is 350 S cm

^{2}

^{- 1}

. At a certain concentration, the equivalent conductance is 40 S cm

^{2}

^{- 1}

. Find the degree of dissociation.

Solution:

$α = \frac{Λ _{e q}^{c}}{Λ _{e q}^{\infty}} = \frac{40}{350} = 0.114$ or $11.4%$ .

Solved Example 2

A 0.5 N solution of a salt placed between two platinum electrodes 2.0 cm apart, each of area 4.0 cm

^{2}

, has a resistance of 25

Ω

. Calculate the equivalent conductance.

Solution:

Cell constant $G^{*} = \frac{ℓ}{A} = \frac{2.0}{4.0} = 0.5$ cm $^{- 1}$ .

Specific conductance $κ = \frac{1}{R} \times G^{*} = \frac{1}{25} \times 0.5 = 0.02$ S cm $^{- 1}$ .

$Λ_{e q} = \frac{κ \times 1000}{N} = \frac{0.02 \times 1000}{0.5} = 40$ S cm $^{2}$ eq $^{- 1}$ .

Solved Example 3

The

Λ_{m}^{\infty}

values for

NaCl

and

KCl

are 126.5 and 149.9 S cm

^{2}

mol

^{- 1}

respectively. If

Λ_{m}^{\infty}

for

NaBr

is 128.5, find

Λ_{m}^{\infty}

for

KBr

Solution:

By Kohlrausch's law: $Λ_{m}^{\infty} (KBr) - Λ_{m}^{\infty} (NaBr) = λ^{\infty} (K^{+}) - λ^{\infty} (Na^{+}) = Λ_{m}^{\infty} (KCl) - Λ_{m}^{\infty} (NaCl) = 149.9 - 126.5 = 23.4$ .

Therefore $Λ_{m}^{\infty} (KBr) = 128.5 + 23.4 = 151.9$ S cm $^{2}$ mol $^{- 1}$ .

Solved Example 4

Calculate the limiting value of equivalent conductance of

AgCl

given

Λ^{\infty}

AgNO_{3} = 133.4

NaCl = 126.5

, and

NaNO_{3} = 121.5

S cm

^{2}

^{- 1}

Solution:

$Λ^{\infty} (AgCl) = Λ^{\infty} (AgNO_{3}) + Λ^{\infty} (NaCl) - Λ^{\infty} (NaNO_{3})$ .

$= 133.4 + 126.5 - 121.5 = 138.4$ S cm $^{2}$ eq $^{- 1}$ .

Verify: $[Ag^{+} + NO_{3}^{-}] + [Na^{+} + Cl^{-}] - [Na^{+} + NO_{3}^{-}] = Ag^{+} + Cl^{-}$ . ✓

Solved Example 5

Which will allow greater conductance: 0.1 M

NaCl

or 0.1 M

CH_{3} COOH

Solution:

$NaCl$ is a strong electrolyte, fully dissociated at 0.1 M. $CH_{3} COOH$ is weak, only about 1% dissociated. NaCl has vastly more free ions, so its conductance is much higher.

Solved Example 6

The molar conductivity of a 0.01 M acetic acid solution is 16.3 S cm

^{2}

mol

^{- 1}

. If

Λ_{m}^{\infty}

for acetic acid

= 390.7

S cm

^{2}

mol

^{- 1}

, calculate the dissociation constant.

Solution:

$α = \frac{16.3}{390.7} = 0.0417$ .

$K_{a} = \frac{C α ^{2}}{1 - α} = \frac{0.01 \times ( 0.0417 ) ^{2}}{1 - 0.0417} = \frac{0.01 \times 1.74 \times 1 0 ^{- 3}}{0.958} = 1.82 \times 1 0^{- 5}$ .

Solved Example 7

The specific conductance of a saturated

AgCl

solution at 25°C is

2.28 \times 1 0^{- 6}

S cm

^{- 1}

. Given

λ^{\infty} (Ag^{+}) = 61.9

and

λ^{\infty} (Cl^{-}) = 76.3

S cm

^{2}

mol

^{- 1}

, find the solubility of

AgCl

Solution:

$Λ_{m}^{\infty} (AgCl) = 61.9 + 76.3 = 138.2$ S cm $^{2}$ mol $^{- 1}$ .

Since $AgCl$ is sparingly soluble, $Λ_{m} \approx Λ_{m}^{\infty}$ .

Molarity $= \frac{κ \times 1000}{Λ _{m}^{\infty}} = \frac{2.28 \times 1 0 ^{- 6} \times 1000}{138.2} = 1.65 \times 1 0^{- 5}$ mol/L.

Solubility $= 1.65 \times 1 0^{- 5} \times 143.5 = 2.37 \times 1 0^{- 3}$ g/L.

7. Debye-Huckel-Onsager Equation

JEE Advanced For strong electrolytes, the linear decrease of

Λ_{m}

with

C

is described quantitatively by:

Λ_{m} = Λ_{m}^{\infty} - b C

where

b

is a constant depending on the solvent and temperature. This is a limiting law, valid at low concentrations. It arises from two effects:

Relaxation effect: an ion moving through solution drags an oppositely charged ionic atmosphere behind, which slows it down.
Electrophoretic effect: the ionic atmosphere itself moves in the opposite direction, further reducing net ion velocity.

Common Mistakes to Avoid

Watch out

Confusing $κ$ (specific conductance, per cm $^{3}$ ) with $Λ_{m}$ (molar conductance, per mole).
Forgetting the factor of 1000 in $Λ_{m} = κ \times 1000/ M$ : without it, units don't cancel to S cm $^{2}$ mol $^{- 1}$ .
Applying "extrapolate to $C = 0$ " to a weak electrolyte: this fails because the curve rises steeply. Use Kohlrausch's law with strong-electrolyte data instead.
Using ionic conductance ( $λ^{\infty}$ ) without matching the stoichiometry: $Λ_{m}^{\infty} (Na_{2} SO_{4}) = 2 λ (Na^{+}) + λ (SO_{4}^{2 -})$ , not just the sum.
Forgetting to convert between equivalent and molar: $Λ_{m} = n \times Λ_{e q}$ for a salt with $n$ equivalents per mole.
Assuming metals and electrolytes both increase conductance with temperature. Metals decrease; electrolytes increase (because ion mobility rises).
For sparingly soluble salts, forgetting to convert molarity to grams per litre when the question asks for "solubility in g/L".

Frequently Asked Questions

Q1. Why does specific conductance decrease but molar conductance increase on dilution?

Specific conductance is per unit volume; dilution reduces ions per cm $^{3}$ , so $κ$ falls. Molar conductance is per mole of electrolyte; even though ion density falls, the total volume containing 1 mole grows faster, and inter-ionic interference decreases, so $Λ_{m}$ rises.

Q2. What is a cell constant and why must it be measured?

The cell constant $G^{*} = ℓ / A$ is a geometric property of the conductivity cell (distance between electrodes over their area). Because it's hard to measure $ℓ$ and $A$ exactly, we calibrate the cell using a KCl solution of known $κ$ : $G^{*} = κ_{KCl} \times R_{measured}$ . Once determined, it multiplies conductance to give $κ$ for any unknown solution.

Q3. Why is $H^{+}$ ion mobility so much higher than other ions?

$H^{+}$ (actually $H_{3} O^{+}$ ) doesn't migrate as a whole ion. Instead, a proton "hops" from one water molecule to the next along the hydrogen-bond network, an effect called the Grotthuss mechanism. This is faster than bodily ion transport. Same reason $OH^{-}$ has anomalously high mobility.

Q4. How does Kohlrausch's law let us find $Λ_{m}^{\infty}$ of a weak electrolyte?

Since a weak electrolyte's $Λ_{m}^{\infty}$ can't be found by extrapolation (curve is too steep near zero), we combine limiting conductivities of related strong electrolytes. For $CH_{3} COOH$ : $Λ_{m}^{\infty} = Λ^{\infty} (CH_{3} COONa) + Λ^{\infty} (HCl) - Λ^{\infty} (NaCl)$ . This works because at infinite dilution, ionic contributions are additive and independent.

Q5. What are the units of $κ$ , $Λ_{e q}$ , and $Λ_{m}$ ?

$κ$ : S cm $^{- 1}$ or S m $^{- 1}$ . $Λ_{e q}$ : S cm $^{2}$ eq $^{- 1}$ . $Λ_{m}$ : S cm $^{2}$ mol $^{- 1}$ . In SI, $Λ_{m}$ can also be S m $^{2}$ mol $^{- 1}$ ; be careful with cm-to-m conversions (factor of $1 0^{- 4}$ ).

Q6. What is the difference between conductance and conductivity?

Conductance ( $G = 1/ R$ ) depends on the cell shape and quantity of solution. Conductivity (specific conductance $κ = 1/ ρ$ ) is an intensive property independent of shape. Always specify which one is being computed.

Q7. Why does adding water to a strong electrolyte solution increase molar conductivity only slightly?

A strong electrolyte is already fully dissociated. Dilution only reduces the small inter-ionic interference (relaxation and electrophoretic drag), so $Λ_{m}$ rises marginally from its value at moderate concentration to $Λ_{m}^{\infty}$ . The rise is linear in $C$ (Debye-Huckel-Onsager).

Q8. Can conductance measurements determine the solubility of $BaSO_{4}$ ?

Yes. Measure $κ$ of a saturated $BaSO_{4}$ solution. Since it's sparingly soluble, concentration is very low and $Λ_{m} \approx Λ_{m}^{\infty}$ . Then molarity $= κ \times 1000/ Λ_{m}^{\infty}$ , and solubility (in g/L) $=$ molarity $\times$ molar mass. Very accurate for salts with solubility below $1 0^{- 3}$ M.

Previous year questions on Conductance Of Electrolytic Solutions

2 questions from past papers, each with a step-by-step solution.

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1. Conductance in Electrolytic Solutions

1.1 Basic Definitions

2. Equivalent and Molar Conductance

3. Effect of Dilution on Conductance

4. Kohlrausch's Law of Independent Migration

4.1 Applications of Kohlrausch's Law

(a) Calculation of Λm∞​ for weak electrolytes

(b) Degree of dissociation of weak electrolyte

(c) Dissociation constant of weak acid/base (via Ostwald's dilution law)

(d) Solubility of sparingly soluble salts

5. Ionic Mobility and Transport Number

6. Solved Examples

7. Debye-Huckel-Onsager Equation

Common Mistakes to Avoid

Frequently Asked Questions

Q1. Why does specific conductance decrease but molar conductance increase on dilution?

Q2. What is a cell constant and why must it be measured?

Q3. Why is H+ ion mobility so much higher than other ions?

Q4. How does Kohlrausch's law let us find Λm∞​ of a weak electrolyte?

Q5. What are the units of κ, Λeq​, and Λm​?

Q6. What is the difference between conductance and conductivity?

Q7. Why does adding water to a strong electrolyte solution increase molar conductivity only slightly?

Q8. Can conductance measurements determine the solubility of BaSO4​?

Previous year questions on Conductance Of Electrolytic Solutions

Ready to master Electrochemistry?

(a) Calculation of $Λ_{m}^{\infty}$ for weak electrolytes

Q3. Why is $H^{+}$ ion mobility so much higher than other ions?

Q4. How does Kohlrausch's law let us find $Λ_{m}^{\infty}$ of a weak electrolyte?

Q5. What are the units of $κ$ , $Λ_{e q}$ , and $Λ_{m}$ ?

Q8. Can conductance measurements determine the solubility of $BaSO_{4}$ ?