Q.The electrical resistance of a column of $0.05\ mol\ L^{-1}$ NaOH solution of diameter $1\ cm$ and length $50\ cm$ is $5.55 \times 10^{3}$ ohm. Calculate its resistivity, conductivity and molar conductivity.

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 …

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: …

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, $E^0=0.34V$).

…

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 …

(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 …