Lambda⁰m(H_2O) is equal to ________. (Two or more than two options may be correct.) (i) Lambda⁰m(HCl) + Lambda

NCERT Exemplar · Q47

\Lambda^0_{m(H_2O)}

(i)

\Lambda^0_{m(HCl)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaCl)}

(ii)

\Lambda^0_{m(HNO_3)} + \Lambda^0_{m(NaNO_3)} - \Lambda^0_{m(NaOH)}

(iii)

\Lambda^0_{m(HNO_3)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaNO_3)}

(iv)

\Lambda^0_{m(NH_4OH)} + \Lambda^0_{m(HCl)} - \Lambda^0_{m(NH_4Cl)}

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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.

Note

Conductance ( $G$ ) is the reciprocal of resistance:

$G = \frac{1}{R}$

Unit: siemens (S) — named after Werner von Siemens. 1 S = 1 A/V (ampere per volt).

If a wire has $R = 10\ \Omega$ , its conductance is $G = 0.1\ \text{S}$ . If $R = 0.5\ \Omega$ , $G = 2\ \text{S}$ — it conducts twice as well.

Ohm's Law in Conductance Form

You know $V = IR$ . Rearranging:

I = \frac{V}{R} = G V

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" $\rho$ ) and the geometry (length $L$ , cross-sectional area $A$ ):

R = \rho \frac{L}{A}

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 ( $\sigma$ ):

\sigma = \frac{1}{\rho}

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 $\sigma$ → good conductor (copper: $\approx 5.8 \times 10^7\ \text{S/m}$ ).
Low $\sigma$ → poor conductor / insulator (glass: $\approx 10^{-12}\ \text{S/m}$ ).

Watch out

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	$\Omega$
Resistivity	$\rho$	$R A / L$	Material only	$\Omega \cdot \text{m}$
Conductance	$G$	$1/R$	Material + geometry	S
Conductivity	$\sigma$	$1/\rho$	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 ( $\sigma$ ).

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 = \frac{V}{R}

Define conductance $G$ as the reciprocal of resistance:

G = \frac{1}{R}

So:

I = G V

Why this makes sense:

A larger $G$ means more current for the same voltage — the conductor "conducts" better.
$G$ has units of siemens (S) = $\text{A/V}$ .

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 $\rho$ (resistivity)

The formula:

R = \rho \frac{L}{A}

Now, conductivity $\sigma$ is the reciprocal of resistivity:

\sigma = \frac{1}{\rho}

So:

R = \frac{1}{\sigma} \cdot \frac{L}{A}

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 = \sigma \frac{A}{L}

This is the central relationship. Let's see why it holds:

$\sigma$ 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 $\sigma$ (like copper) gives higher $G$ than one with low $\sigma$ (like iron), for the same shape.

5. Microscopic Derivation (Why $\sigma$ Exists)

At the microscopic level, conductivity arises from electron motion:

\sigma = n e \mu

Where:

$n$ = number of free electrons per unit volume
$e$ = electron charge …

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Concept: Molar Conductivity at infinite dilution — Kohlrausch’s law states that $\Lambda_m^0$ of an electrolyte is the sum of the limiting molar conductivities of its constituent ions. For water, $\Lambda_{m(H_2O)}^0 = \lambda_{H^+}^0 + \lambda_{OH^-}^0$ . Only combinations built from strong electrolytes count, since a strong electrolyte's $\Lambda_m^0$ is the only kind that can be measured directly.

We need to combine strong electrolytes so that the net ionic sum equals $\lambda_{H^+}^0 + \lambda_{OH^-}^0$ .

Step 1: Write the ionic contributions for each option.

For (i):

$\Lambda_{m(HCl)}^0 = \lambda_{H^+}^0 + \lambda_{Cl^-}^0$

$\Lambda_{m(NaOH)}^0 = \lambda_{Na^+}^0 + \lambda_{OH^-}^0$

$\Lambda_{m(NaCl)}^0 = \lambda_{Na^+}^0 + \lambda_{Cl^-}^0$

Sum: $(\lambda_{H^+}^0 + \lambda_{Cl^-}^0) + (\lambda_{Na^+}^0 + \lambda_{OH^-}^0) - (\lambda_{Na^+}^0 + \lambda_{Cl^-}^0) = \lambda_{H^+}^0 + \lambda_{OH^-}^0$ ✓

Step 2: Check (ii):

$\Lambda_{m(HNO_3)}^0 = \lambda_{H^+}^0 + \lambda_{NO_3^-}^0$

$\Lambda_{m(NaNO_3)}^0 = \lambda_{Na^+}^0 + \lambda_{NO_3^-}^0$

$\Lambda_{m(NaOH)}^0 = \lambda_{Na^+}^0 + \lambda_{OH^-}^0$

Sum: $(\lambda_{H^+}^0 + \lambda_{NO_3^-}^0) + (\lambda_{Na^+}^0 + \lambda_{NO_3^-}^0) - (\lambda_{Na^+}^0 + \lambda_{OH^-}^0) = \lambda_{H^+}^0 + 2\lambda_{NO_3^-}^0 - \lambda_{OH^-}^0$ ✗

Step 3: Check (iii):

$\Lambda_{m(HNO_3)}^0 = \lambda_{H^+}^0 + \lambda_{NO_3^-}^0$

$\Lambda_{m(NaOH)}^0 = \lambda_{Na^+}^0 + \lambda_{OH^-}^0$

$\Lambda_{m(NaNO_3)}^0 = \lambda_{Na^+}^0 + \lambda_{NO_3^-}^0$ …

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The limiting molar conductivity of water, $\Lambda^0_{m(H_2O)}$ , is found by applying Kohlrausch’s law of independent migration of ions. It equals the sum of the limiting conductivities of its constituent ions, $H^+$ and $OH^-$ , which can be obtained by combining the conductivities of strong electrolytes. The correct expressions are (i) and (iii).

The key idea here is Kohlrausch’s law: at infinite dilution, each ion contributes a fixed amount to the molar conductivity of an electrolyte, independent of the other ion it travels with. So $\Lambda^0_m$ for any electrolyte is simply the sum of the limiting conductivities of its cation and anion.

For water, which dissociates as $H_2O \rightleftharpoons H^+ + OH^-$ , its limiting molar conductivity is:

\Lambda^0_{m(H_2O)} = \lambda^0_{H^+} + \lambda^0_{OH^-}

We don’t know these individual ionic conductivities directly, but we can get them by combining data from strong electrolytes that contain these ions — strong electrolytes are the ones whose $\Lambda_m^0$ can actually be measured directly, by extrapolating $\Lambda_m$ vs $\sqrt{c}$ to zero concentration.

Let’s check each option step by step.

Option (i): $\Lambda^0_{m(HCl)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaCl)}$

Write each in terms of ionic conductivities:
- $\Lambda^0_{m(HCl)} = \lambda^0_{H^+} + \lambda^0_{Cl^-}$
- $\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}$
- $\Lambda^0_{m(NaCl)} = \lambda^0_{Na^+} + \lambda^0_{Cl^-}$
Adding the first two and subtracting the third:

(\lambda^0_{H^+} + \lambda^0_{Cl^-}) + (\lambda^0_{Na^+} + \lambda^0_{OH^-}) - (\lambda^0_{Na^+} + \lambda^0_{Cl^-})

The $\lambda^0_{Na^+}$ and $\lambda^0_{Cl^-}$ cancel, leaving $\lambda^0_{H^+} + \lambda^0_{OH^-}$ , which is exactly $\Lambda^0_{m(H_2O)}$ . HCl, NaOH and NaCl are all strong electrolytes, so this is a legitimate calculation. (i) is correct.

Option (ii): $\Lambda^0_{m(HNO_3)} + \Lambda^0_{m(NaNO_3)} - \Lambda^0_{m(NaOH)}$

Write them out:
- $\Lambda^0_{m(HNO_3)} = \lambda^0_{H^+} + \lambda^0_{NO_3^-}$
- $\Lambda^0_{m(NaNO_3)} = \lambda^0_{Na^+} + \lambda^0_{NO_3^-}$
- $\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}$
Sum the first two and subtract the third:

(\lambda^0_{H^+} + \lambda^0_{NO_3^-}) + (\lambda^0_{Na^+} + \lambda^0_{NO_3^-}) - (\lambda^0_{Na^+} + \lambda^0_{OH^-})

The $\lambda^0_{Na^+}$ cancels, but we get $\lambda^0_{H^+} + 2\lambda^0_{NO_3^-} - \lambda^0_{OH^-}$ , which is not $\lambda^0_{H^+} + \lambda^0_{OH^-}$ . So (ii) is incorrect.

Option (iii): $\Lambda^0_{m(HNO_3)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaNO_3)}$

Write:
- $\Lambda^0_{m(HNO_3)} = \lambda^0_{H^+} + \lambda^0_{NO_3^-}$
- $\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}$
- $\Lambda^0_{m(NaNO_3)} = \lambda^0_{Na^+} + \lambda^0_{NO_3^-}$
Adding the first two and subtracting the third:

(\lambda^0_{H^+} + \lambda^0_{NO_3^-}) + (\lambda^0_{Na^+} + \lambda^0_{OH^-}) - (\lambda^0_{Na^+} + \lambda^0_{NO_3^-})

The $\lambda^0_{Na^+}$ and $\lambda^0_{NO_3^-}$ cancel, leaving $\lambda^0_{H^+} + \lambda^0_{OH^-}$ . HNO₃, NaOH and NaNO₃ are all strong electrolytes, so this is also legitimate. (iii) is correct.

Option (iv): $\Lambda^0_{m(NH_4OH)} + \Lambda^0_{m(HCl)} - \Lambda^0_{m(NH_4Cl)}$ $Λ_{m (N H_{4} O H)}^{0} + Λ_{m (H C l)}^{0} - Λ_{m (N H_{4} C l)}^{0}$ Write:
- $\Lambda^0_{m(NH_4OH)} = \lambda^0_{NH_4^+} + \lambda^0_{OH^-}$
- $\Lambda^0_{m(HCl)} = \lambda^0_{H^+} + \lambda^0_{Cl^-}$
- $\Lambda^0_{m(NH_4Cl)} = \lambda^0_{NH_4^+} + \lambda^0_{Cl^-}$ …

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Method: Kohlrausch’s Law of Independent Migration of Ions

Concept: 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. This lets you build $\Lambda_m^0$ of one substance from other substances' $\Lambda_m^0$ values — provided every substance used is a strong electrolyte (only strong-electrolyte $\Lambda_m^0$ can be measured directly, by extrapolating $\Lambda_m$ vs $\sqrt{c}$ to $c=0$ ).

Steps:

Write the expression for $\Lambda^0_m$ of water Water dissociates as:

H_2O \rightleftharpoons H^+ + OH^-

So,

\Lambda^0_{m(H_2O)} = \lambda^0_{H^+} + \lambda^0_{OH^-}

Express each given electrolyte in terms of ionic conductivities

For example:
- $\Lambda^0_{m(HCl)} = \lambda^0_{H^+} + \lambda^0_{Cl^-}$
- $\Lambda^0_{m(NaOH)} = \lambda^0_{Na^+} + \lambda^0_{OH^-}$
- $\Lambda^0_{m(NaCl)} = \lambda^0_{Na^+} + \lambda^0_{Cl^-}$
Combine to isolate $\lambda^0_{H^+} + \lambda^0_{OH^-}$

Take option (i):

\Lambda^0_{m(HCl)} + \Lambda^0_{m(NaOH)} - \Lambda^0_{m(NaCl)}

Substitute:

= (\lambda^0_{H^+} + \lambda^0_{Cl^-}) + (\lambda^0_{Na^+} + \lambda^0_{OH^-}) - (\lambda^0_{Na^+} + \lambda^0_{Cl^-})

Cancel $\lambda^0_{Na^+}$ and $\lambda^0_{Cl^-}$ :

= \lambda^0_{H^+} + \lambda^0_{OH^-} = \Lambda^0_{m(H_2O)}

HCl, NaOH, NaCl are all strong electrolytes — valid.

Check other options similarly
- Option (iii) also works, using only strong electrolytes HNO₃, NaOH, NaNO₃: …

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Common Mistakes & How to Avoid Them

Mistake 1: Not recognising Kohlrausch’s Law of independent migration of ions

Students often try to memorise the formula without understanding the logic behind it.

How to avoid:

Kohlrausch’s law says:

\Lambda_m^0 = \lambda_+^0 + \lambda_-^0

For water ( $H_2O$ ), the ions are $H^+$ and $OH^-$ . So:

\Lambda_{m(H_2O)}^0 = \lambda_{H^+}^0 + \lambda_{OH^-}^0

Now, any combination of strong electrolytes that gives you $\lambda_{H^+}^0 + \lambda_{OH^-}^0$ is correct.

Mistake 2: Confusing addition/subtraction signs

Students often misplace the signs when combining electrolytes.

How to avoid:

Write each electrolyte’s ionic contributions explicitly:

$\Lambda_{m(HCl)}^0 = \lambda_{H^+}^0 + \lambda_{Cl^-}^0$
$\Lambda_{m(NaOH)}^0 = \lambda_{Na^+}^0 + \lambda_{OH^-}^0$
$\Lambda_{m(NaCl)}^0 = \lambda_{Na^+}^0 + \lambda_{Cl^-}^0$

Now compute:

\Lambda_{m(HCl)}^0 + \Lambda_{m(NaOH)}^0 - \Lambda_{m(NaCl)}^0 = (\lambda_{H^+}^0 + \lambda_{Cl^-}^0) + (\lambda_{Na^+}^0 + \lambda_{OH^-}^0) - (\lambda_{Na^+}^0 + \lambda_{Cl^-}^0)

Cancel $\lambda_{Na^+}^0$ and $\lambda_{Cl^-}^0$ → $\lambda_{H^+}^0 + \lambda_{OH^-}^0$ ✓

Mistake 3: Assuming only one combination is correct

The question says two or more options may be correct, but students often stop after finding one.

How to avoid:

Check every option using the same method. For option (iii):

\Lambda_{m(HNO_3)}^0 + \Lambda_{m(NaOH)}^0 - \Lambda_{m(NaNO_3)}^0

$\Lambda_{m(HNO_3)}^0 = \lambda_{H^+}^0 + \lambda_{NO_3^-}^0$
$\Lambda_{m(NaOH)}^0 = \lambda_{Na^+}^0 + \lambda_{OH^-}^0$
$\Lambda_{m(NaNO_3)}^0 = \lambda_{Na^+}^0 + \lambda_{NO_3^-}^0$

Cancel $\lambda_{Na^+}^0$ and $\lambda_{NO_3^-}^0$ → $\lambda_{H^+}^0 + \lambda_{OH^-}^0$ ✓

So both (i) and (iii) are correct.

Mistake 4: Wrongly accepting option (iv) because the ion algebra 'cancels'

Option (iv) is $\Lambda_{m(NH_4OH)}^0 + \Lambda_{m(HCl)}^0 - \Lambda_{m(NH_4Cl)}^0$ . If you write it out formally:

(\lambda_{NH_4^+}^0 + \lambda_{OH^-}^0) + (\lambda_{H^+}^0 + \lambda_{Cl^-}^0) - (\lambda_{NH_4^+}^0 + \lambda_{Cl^-}^0) = \lambda_{H^+}^0 + \lambda_{OH^-}^0

This LOOKS identical to (i) and (iii) — but it is not accepted as a valid answer, and it's a genuine trap.

Why it's wrong: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AHSEC Higher Secondary (HS) Final Examination 2026Set ANNUAL2 marks
$Q. State the variation of conductivity and molar conductivity of acetic acid with dilution.$
›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 …

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AHSEC Higher Secondary (HS) Final Examination 2024Set ANNUAL2 marks
$Q. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.$
›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: …
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AHSEC Higher Secondary (HS) Final Examination 2023Set ANNUAL2 marks
$Q. The following limiting molar conductivities are given as: lambda-m-zero(H2SO4) = x S cm^2 mol^-1, lambda-m-zero(K2SO4) = y S cm^2 mol^-1, lambda-m-zero(CH3COOK) = z S cm^2 mol^-1. Calculate limiting molar conductivity of acetic acid. OR The cell potential for the following cell is 0.576V at 298K. Calculate the pH of the solution: Pt | H2(g) | H+(aq) || Cu2+(0.01M) | Cu(s). Given, E-zero(Cu2+/Cu) = 0.34V.$
›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:

$2\lambda^0(H^+) + \lambda^0(SO_4^{2-}) = x \quad \text{...(i), from } H_2SO_4$

$2\lambda^0(K^+) + \lambda^0(SO_4^{2-}) = y \quad \text{...(ii), from } K_2SO_4$

$\lambda^0(CH_3COO^-) + \lambda^0(K^+) = z \quad \text{...(iii), from } CH_3COOK$

Subtracting (ii) from (i): $2\lambda^0(H^+) - 2\lambda^0(K^+) = x - y \Rightarrow \lambda^0(H^+) = \lambda^0(K^+) + \dfrac{x-y}{2}$

We want $\lambda_m^0(CH_3COOH) = \lambda^0(CH_3COO^-) + \lambda^0(H^+)$ . From (iii), $\lambda^0(CH_3COO^-) = z - \lambda^0(K^+)$ , so:

$\lambda_m^0(CH_3COOH) = \left[z - \lambda^0(K^+)\right] + \left[\lambda^0(K^+) + \frac{x-y}{2}\right] = z + \frac{x-y}{2}$

(Equivalently, $\lambda_m^0(CH_3COOH) = \lambda_m^0(CH_3COOK) + \lambda_m^0(\tfrac12H_2SO_4) - \lambda_m^0(\tfrac12K_2SO_4)$ .)

Option 2 — pH from cell potential:

Cell: $Pt|H_2(g)|H^+(aq)\ ||\ Cu^{2+}(0.01M)|Cu(s)$ , with anode = H₂/H⁺ (oxidation, $E^0=0V$ ) and cathode = Cu²⁺/Cu (reduction, $E^0=0.34V$ ).

…

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AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL2 marks
$Q. Define molar conductivity of a solution. Explain how molar conductivity changes with change in concentration of solution for a weak and a strong electrolyte. (1+1=2)$
›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 …

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AHSEC Higher Secondary (HS) Final Examination 2020Set ANNUAL2 marks
$Q. (a) Following reactions occur at cathode during the electrolysis of aqueous silver chloride: Ag+(aq) + e- -> Ag(s), E° = +0.80V; H+(aq) + e- -> 1/2 H2(g), E° = 0.00V. On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode and why?$
(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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Q.Λm(H2O)0\Lambda^0_{m(H_2O)}Λm(H2​O)0​ is equal to _______________. (Two or more than two options may be correct.)

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Q. $\Lambda^0_{m(H_2O)}$ is equal to _______________. (Two or more than two options may be correct.)

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