Q.Λm0 for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm2 mol−1 respectively. Calculate Λ0 for HAc.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Molar Conductivity
From Resistance to Conductance: Flipping the Idea
You already know resistance (R) — it tells you how much a material opposes the flow of current. A high resistance means the wire fights the current; a low resistance means it lets current through easily.
Now flip that thought. Instead of asking "how much does it resist?", ask "how easily does it let current flow?" That's exactly what conductance measures.
Conductance (G) is the reciprocal of resistance:
G=R1
Unit: siemens (S) — named after Werner von Siemens. 1 S = 1 A/V (ampere per volt).
If a wire has R=10 Ω, its conductance is G=0.1 S. If R=0.5 Ω, G=2 S — it conducts twice as well.
Ohm's Law in Conductance Form
You know V=IR. Rearranging:
I=RV=GV
So current = conductance × voltage. A high-conductance material draws a large current for the same voltage — it's a "good conductor."
Now, Conductivity: The Material's Intrinsic Property
Resistance depends on two things: the material itself (its "resistivity" ρ) and the geometry (length L, cross-sectional area A):
R=ρAL
Conductance also depends on geometry. A thicker wire (larger A) or a shorter wire (smaller L) has higher conductance. To isolate the material's inherent ability to conduct, we define conductivity (σ):
σ=ρ1
And for a uniform wire:
G=σLA
Conductivity is the reciprocal of resistivity. It tells you how well the material itself conducts, independent of shape and size.
- Unit: siemens per metre (S/m).
- High σ → good conductor (copper: ≈5.8×107 S/m).
- Low σ → poor conductor / insulator (glass: ≈10−12 S/m).
Don't confuse conductance (property of a specific object, depends on geometry) with conductivity (property of the material, independent of geometry). A short thick copper wire has high conductance; a long thin copper wire has lower conductance — but both have the same conductivity.
The Big Picture in One Table
| Quantity | Symbol | Definition | Depends on | Unit |
|---|---|---|---|---|
| Resistance | R | V/I | Material + geometry | Ω |
| Resistivity | ρ | RA/L | Material only | Ω⋅m |
| Conductance | G | 1/R | Material + geometry | S |
| Conductivity | σ | 1/ρ | Material only | S/m |
Intuitive Analogy
Think of a water pipe:
- Resistance = how hard it is to push water through (narrow, long pipe).
- Conductance = how easily water flows (wide, short pipe). …
Why this formula?
Conductance and Conductivity: Why the Formulas Hold
Let's build this from first principles — understanding the why before the what.
1. The Core Idea: How Easily Does Current Flow?
Think of a conductor (like a copper wire). When you apply a voltage across it, electrons drift through the material. Two questions arise:
- How much current flows for a given voltage? → This is conductance (G).
- How well does the material itself allow current? → This is conductivity (σ).
The key distinction: Conductance depends on the size and shape of the object. Conductivity is an intrinsic property of the material.
2. Ohm's Law in Terms of Conductance
You know Ohm's law:
V=IR
But we can rewrite it as:
I=RV
Define conductance G as the reciprocal of resistance:
G=R1
So:
I=GV
Why this makes sense:
- A larger G means more current for the same voltage — the conductor "conducts" better.
- G has units of siemens (S) = A/V.
3. From Resistance to Conductivity: The Geometry Factor
Resistance of a uniform conductor depends on:
- Length L (longer → more resistance)
- Cross-sectional area A (thicker → less resistance)
- Material property ρ (resistivity)
The formula:
R=ρAL
Now, conductivity σ is the reciprocal of resistivity:
σ=ρ1
So:
R=σ1⋅AL
Why this form?
- If you double the length, electrons have to travel twice as far, colliding more → resistance doubles.
- If you double the area, there's twice as many "lanes" for electrons → resistance halves.
4. The Key Formula: Conductance in Terms of Conductivity
Since G=1/R, we get:
G=σLA
This is the central relationship. Let's see why it holds:
- σ tells you how well the material conducts (intrinsic).
- A/L tells you how the geometry amplifies or reduces that.
Intuition:
- A fat, short wire (A large, L small) has high conductance.
- A thin, long wire (A small, L large) has low conductance.
- A material with high σ (like copper) gives higher G than one with low σ (like iron), for the same shape.
5. Microscopic Derivation (Why σ Exists)
At the microscopic level, conductivity arises from electron motion:
σ=neμ
Where:
- n = number of free electrons per unit volume
- e = electron charge …
Concept: Molar Conductivity at infinite dilution — Kohlrausch’s law of independent migration of ions.
Step 1: Write Kohlrausch’s law for each salt:
- Λm0(NaCl)=λNa+0+λCl−0=126.4
- Λm0(HCl)=λH+0+λCl−0=425.9
- Λm0(NaAc)=λNa+0+λAc−0=91.0
Step 2: For acetic acid (HAc):
Λm0(HAc)=λH+0+λAc−0
Step 3: Combine the given values: …
Kohlrausch’s law of independent migration lets us combine the limiting molar conductivities of strong electrolytes to find that of a weak acid. Using the given values, Λm0(HAc)=Λm0(HCl)+Λm0(NaAc)−Λm0(NaCl)=425.9+91.0−126.4=390.5 S cm2mol−1.
The problem asks for the limiting molar conductivity of acetic acid (HAc), a weak electrolyte. You cannot measure it directly by extrapolation because weak acids don’t fully dissociate even at infinite dilution — the conductivity curve bends. So we need a different route.
That route is Kohlrausch’s law of independent migration of ions. At infinite dilution, each ion contributes a fixed amount to the total molar conductivity, regardless of what other ion it’s paired with. So the limiting molar conductivity of any electrolyte is simply the sum of the limiting conductivities of its constituent ions:
Λm0=λ+0+λ−0
This means we can add and subtract Λm0 values of strong electrolytes to get the Λm0 of a weak electrolyte — as long as the ionic combinations cancel out correctly.
Here’s how it works for HAc. Write the target:
Λm0(HAc)=λ0(H+)+λ0(Ac−)
We don’t know either ion’s value individually. But we do know three strong electrolytes that share these ions:
- HCl gives λ0(H+)+λ0(Cl−)=425.9
- NaAc gives λ0(Na+)+λ0(Ac−)=91.0
- NaCl gives λ0(Na+)+λ0(Cl−)=126.4
Notice: if we add the first two and subtract the third, the sodium and chloride ions cancel out, leaving exactly what we need. …
Method: Kohlrausch’s Law of Independent Migration of Ions
This law states that at infinite dilution, each ion contributes a fixed amount to the molar conductivity of an electrolyte, independent of the other ion it is paired with.
Steps
-
Write the expression for Λm0 of each given salt
At infinite dilution:
- Λm0(NaCl)=λNa+0+λCl−0=126.4
- Λm0(HCl)=λH+0+λCl−0=425.9
- Λm0(NaAc)=λNa+0+λAc−0=91.0
-
Identify the target
We need Λm0(HAc)=λH+0+λAc−0
-
Combine the given equations
Add Λm0(HCl) and Λm0(NaAc):
(λH+0+λCl−0)+(λNa+0+λAc−0)=425.9+91.0
This gives:
λH+0+λAc−0+λNa+0+λCl−0=516.9
Now subtract Λm0(NaCl): …
Here are the most common mistakes students make when solving this Kohlrausch’s law problem — and exactly how to avoid each one.
1. ✗ Forgetting the Kohlrausch’s Law Formula
Mistake: Trying to guess or recall the combination incorrectly — e.g., adding all three given values.
Why it’s wrong:
Kohlrausch’s law says:
At infinite dilution, the molar conductivity of an electrolyte is the sum of the molar conductivities of its constituent ions.
For a weak acid like HAc (H++Ac−), you cannot measure Λm0 directly — you build it from known strong electrolytes.
How to avoid:
Always write the ionic breakdown:
- Λm0(HAc)=λ0(H+)+λ0(Ac−)
- You have: Λm0(HCl)=λ0(H+)+λ0(Cl−) Λm0(NaAc)=λ0(Na+)+λ0(Ac−) Λm0(NaCl)=λ0(Na+)+λ0(Cl−)
Then combine:
Λm0(HAc)=Λm0(HCl)+Λm0(NaAc)−Λm0(NaCl)
2. ✗ Sign Errors in the Combination
Mistake: Subtracting the wrong term or adding all three.
Why it’s wrong:
You need to cancel the common ions (Na+ and Cl−). Adding all three gives a meaningless sum.
How to avoid:
Use the “add and subtract” logic:
- Start with λ0(H+)+λ0(Ac−)
- Add λ0(Cl−) and λ0(Na+) to match given data, then subtract them out.
Check:
(425.9+91.0)−126.4=390.5
If you get a negative or absurd number, you’ve swapped signs.
3. ✗ Unit Confusion
Mistake: Using values in different units (e.g., mS vs S, or cm2 vs m2).
Why it’s wrong:
All given values are in S cm2 mol−1. Mixing units gives a wrong numeric answer.
How to avoid:
- Always check units before calculation.
- If needed, convert: 1 S cm2 mol−1=10−4 S m2 mol−1.
- In this problem, all three are already consistent — just use them as given.
4. ✗ Misidentifying Strong vs Weak Electrolytes
Mistake: Thinking HAc’s Λm0 can be measured directly (like for NaCl).
Why it’s wrong: …
- AHSEC Higher Secondary (HS) Final Examination 2026Set ANNUAL2 marksQ.State the variation of conductivity and molar conductivity of acetic acid with dilution.
›Reveal solutionSolution
Diluting acetic acid lowers its conductivity but raises its molar conductivity (increasing dissociation).
Conductivity (κ) measures the conductance of ions present in unit volume of solution. On dilution the number of ions per unit volume decreases, so conductivity always decreases with dilution.
Molar conductivity (Λ_m = κ × 1000 / c) measures the conducting power of all the ions produced by one mole of electrolyte. On dilution the volume containing one mole increases, and for a weak electrolyte like acetic acid the degree of dissociation (α) rises steeply, releasing many more ions. Therefore molar conductivity increases with dilution, and the increase is very sharp near infini …
- AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL2 marksQ.Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
›Reveal solutionSolution
Conductivity falls as a solution is diluted (fewer ions per unit volume), while molar conductivity rises on dilution (each mole's ions become more independent / more of a weak electrolyte dissociates).
Conductivity (κ, specific conductance):
Conductivity is defined as the conductance of a solution of unit length (1 cm) and unit cross-sectional area (1 cm²) — i.e. it is the reciprocal of resistivity (ρ):
κ = 1/ρ
Its SI unit is S m⁻¹ (commonly expressed as S cm⁻¹). It measures how well the solution as a whole (per unit volume) conducts electricity.
Molar conductivity (Λm):
Molar conductivity is the conducting power of all the ions produced by dissolving 1 mole of an electrolyte in solution, and is related to conductivity by:
Λm = κ × 1000 / C (C = molar concentration in mol L⁻¹, κ in S cm⁻¹)
Its unit is S cm² mol⁻¹.
Variation with concentration:
- Conductivity (κ) always decreases as the solution is diluted, because the number of ions per unit volume of solution decreases with dilution, even though the degree of dissociation (for weak electrolytes) increases.
- Molar conductivity (Λm) always increases as concentration decreases (i.e. on dilution), because it is normalised per mole of electrolyte: …
- AHSEC Higher Secondary (HS) Final Examination 2023Set ANNUAL2 marksQ.The following limiting molar conductivities are given as: lambda-m-zero(H2SO4) = x S cm^2 mol^-1, lambda-m-zero(K2SO4) = y S cm^2 mol^-1, lambda-m-zero(CH3COOK) = z S cm^2 mol^-1. Calculate limiting molar conductivity of acetic acid. OR The cell potential for the following cell is 0.576V at 298K. Calculate the pH of the solution: Pt | H2(g) | H+(aq) || Cu2+(0.01M) | Cu(s). Given, E-zero(Cu2+/Cu) = 0.34V.
›Reveal solutionSolution
Option 1 combines the three given conductivities via Kohlrausch's law to cancel out K⁺ and SO₄²⁻; Option 2 uses the Nernst equation for the H₂|H⁺ vs Cu²⁺|Cu cell to back-calculate [H⁺].
Option 1 — Limiting molar conductivity of acetic acid via Kohlrausch's law:
By Kohlrausch's law of independent migration of ions, each limiting molar conductivity splits into ionic contributions:
2λ0(H+)+λ0(SO42−)=x...(i), from H2SO4
2λ0(K+)+λ0(SO42−)=y...(ii), from K2SO4
λ0(CH3COO−)+λ0(K+)=z...(iii), from CH3COOK
Subtracting (ii) from (i): 2λ0(H+)−2λ0(K+)=x−y⇒λ0(H+)=λ0(K+)+2x−y
We want λm0(CH3COOH)=λ0(CH3COO−)+λ0(H+). From (iii), λ0(CH3COO−)=z−λ0(K+), so:
λm0(CH3COOH)=[z−λ0(K+)]+[λ0(K+)+2x−y]=z+2x−y
(Equivalently, λm0(CH3COOH)=λm0(CH3COOK)+λm0(21H2SO4)−λm0(21K2SO4).)
Option 2 — pH from cell potential:
Cell: Pt∣H2(g)∣H+(aq) ∣∣ Cu2+(0.01M)∣Cu(s), with anode = H₂/H⁺ (oxidation, E0=0V) and cathode = Cu²⁺/Cu (reduction, E0=0.34V).
…
- AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL2 marksQ.Define molar conductivity of a solution. Explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte. (1+1=2)
›Reveal solutionSolution
Molar conductivity is conductivity per mole of dissolved electrolyte; it rises on dilution for both strong and weak electrolytes, but far more steeply for weak electrolytes because dilution increases their degree of dissociation.
Definition of molar conductivity
Molar conductivity (Λm) is the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution. It is related to the specific conductivity (κ) and molar concentration (C) by:
Λm = κ × 1000 / C (with κ in S cm⁻¹ and C in mol L⁻¹, giving Λm in S cm² mol⁻¹)
Variation with concentration — strong electrolytes
For strong electrolytes (fully ionised at all concentrations, e.g. NaCl, KCl), Λm increases only slowly as concentration decreases (i.e. on dilution). This is because, even though the number of ions per unit volume stays proportional to concentration (the electrolyte is always ~100% ionised), interionic attractive forces between the oppositely charged ions reduce their mobility at higher concentrations; diluting the solution weakens these interionic forces, allowing ions to move a little more freely, so Λm rises gradually. Λm for a strong electrolyte varies linearly with √C (Debye-Hückel-Onsager equation), so Λm° (the limiting molar conductivity at infinite dilution) can be found by extrapolating the Λm vs √C plot to C = 0.
Variation with concentration — weak electrolytes …
- AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL2 marksQ.(a) Following reactions occur at cathode during the electrolysis of aqueous silver chloride: Ag+(aq) + e- -> Ag(s), E° = +0.80V; H+(aq) + e- -> 1/2 H2(g), E° = 0.00V. On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?(1)(b) State Kohlrausch law of independent migration of ions. Write an expression for the molar conductivity of acetic acid at infinite dilution according to Kohlrausch law. (1/2+1/2=1)
›Reveal solutionSolution
(a) The half-reaction with the more positive standard reduction potential occurs preferentially at the cathode. (b) Kohlrausch's law lets us calculate the limiting molar conductivity of any electrolyte, including weak ones like acetic acid, by summing the independent ionic contributions.
(a) Which reaction occurs at the cathode?
Given: Ag+(aq) + e- → Ag(s), E° = +0.80 V and H+(aq) + e- → ½H2(g), E° = 0.00 V
At the cathode, reduction occurs, and between two competing reduction half-reactions, the one with the higher (more positive) standard reduction potential is thermodynamically more favourable and occurs preferentially, since it has the greater tendency to be reduced (to gain electrons). Since E°(Ag+/Ag) = +0.80 V is more positive than E°(H+/H2) = 0.00 V, silver ions are reduced in preference to hydrogen ions at the cathode:
Ag+(aq) + e- → Ag(s)
(b) Kohlrausch's law of independent migration of ions
At infinite dilution, when dissociation of an electrolyte is complete and interionic interactions vanish, each ion migrates independently of the other ion with which it is associated, and each ion makes its own definite contribution to the total molar conductivity of the electrolyte, regardless of the nature of the other ion present. Mathematically, for an electrolyte that dissociates into ν+ cations and ν- anions:
Λm° = ν+ λ°+ + ν- λ°-
where λ°+ and λ°- are the limiting (infinite-dilution) molar conductivities of the cation and anion respectively.
Molar conductivity of acetic acid at infinite dilution …
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