Q.Conductivity of 0.00241 M acetic acid is 7.896×10−5 S cm−1. Calculate its molar conductivity. If Λm0 for acetic acid is 390.5 S cm2 mol−1, what is its dissociation constant?
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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 — the conductivity of all ions produced from one mole of electrolyte in solution.
Step 1: Molar conductivity
Λm=Cκ×1000
where κ=7.896×10−5 S cm−1, C=0.00241 mol L−1.
Λm=0.002417.896×10−5×1000=32.76 S cm2 mol−1
Step 2: Degree of dissociation
α=Λm0Λm=390.532.76=0.0839
Step 3: Dissociation constant (for weak acid HA ⇌ H⁺ + A⁻) …
Molar conductivity is conductivity per mole of electrolyte; for a weak acid like acetic acid, its low value relative to the limiting molar conductivity gives the degree of dissociation, from which the dissociation constant is calculated using the Ostwald dilution law. The dissociation constant comes out to be 1.85×10−5.
The problem gives you the conductivity (κ) of a dilute acetic acid solution and asks for two things: the molar conductivity (Λm), and then the dissociation constant (Ka) using the given limiting molar conductivity (Λm0). This is a classic weak electrolyte problem — the key is that acetic acid does not fully dissociate, so its molar conductivity is much smaller than Λm0, and the ratio tells you the fraction of molecules that have actually split into ions.
Let’s go step by step.
- Molar conductivity from conductivity and concentration Molar conductivity is defined as the conductivity of the solution divided by the molar concentration:
Λm=cκ
But watch the units carefully. Conductivity κ is given in S cm−1, and concentration c is in mol L−1. For molar conductivity in S cm2 mol−1, you need to convert concentration from mol/L to mol/cm3:
1 L=1000 cm3⇒c=0.00241 mol/L=0.00241×10−3 mol/cm3=2.41×10−6 mol/cm3
Now plug in:
Λm=2.41×10−6 mol cm−37.896×10−5 S cm−1=32.76 S cm2 mol−1
So the molar conductivity of this solution is about 32.8 S cm2 mol−1.
A quick check: if you forget the 1000 factor, you’d get 0.0328 — which is clearly wrong because molar conductivities of even weak electrolytes are in the tens or hundreds. The 1000 factor is the most common slip here.
- Degree of dissociation from molar conductivities For a weak electrolyte, the degree of dissociation α is given by the ratio of its molar conductivity at a given concentration to its limiting molar conductivity (at infinite dilution):
α=Λm0Λm
This works because at infinite dilution, every molecule is dissociated, so Λm0 represents the conductivity if all molecules were ions. Here:
α=390.532.76=0.0839 …
Method: Conductometric Determination of Degree of Dissociation and Dissociation Constant
This method uses the definition of molar conductivity and the degree-of-dissociation relation for a weak electrolyte.
Step 1: Calculate molar conductivity (Λm)
Λm=Cκ×1000
Given κ=7.896×10−5 S cm−1 and C=0.00241 mol L−1:
Λm=0.002417.896×10−5×1000=32.76 S cm2 mol−1
Step 2: Calculate degree of dissociation (α)
For a weak electrolyte, the degree of dissociation is:
α=Λm0Λm
Using the given Λm0=390.5 S cm2 mol−1 (no Kohlrausch summation is needed here since the limiting molar conductivity is provided directly):
α=390.532.76=0.0839
So about 8.4% of acetic acid molecules are dissociated.
Step 3: Calculate dissociation constant (Ka) …
Common Mistakes & How to Avoid Them — Molar Conductivity & Dissociation Constant of Acetic Acid
Mistake 1: Forgetting the 1000 Factor in Λm=κ×1000/C
Conductivity κ is effectively per cm³, but concentration is given per litre (1000 cm3). Skipping the 1000 factor gives a value 1000× too small.
Fix: Always use Λm=C(molL−1)1000κ(Scm−1).
Mistake 2: Inverting the Degree-of-Dissociation Ratio
Some students divide Λm0 by Λm instead of the other way round, giving α>1, which is physically impossible (a fraction dissociated can never exceed 1).
Fix: α=Λm/Λm0 is always ≤1; if your ratio exceeds 1, you've inverted it.
Mistake 3: Dropping the (1−α) Term in Ostwald's Dilution Law
Writing Ka=cα2 alone is only a valid approximation when α≪1 (typically <0.05). Here α≈0.084, so omitting (1−α) changes the answer by about 8% (1.70×10−5 instead of the correct 1.85×10−5).
Fix: Always use the full expression Ka=1−αcα2 unless the question explicitly says α is small enough to neglect.
Mistake 4: Mixing Up S m−1 and S cm−1
If κ were given in S m−1 (as in some other problems), forgetting to convert (1 S m−1=10−2 S cm−1) throws the answer off by a factor of 100. In this problem κ is already in S cm−1, so no conversion is needed — but always check the given unit first.
Mistake 5: Rounding α Too Early …
- 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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