Question

Molar conductivity of aqueous solution of HA is 200 Scm² mol⁻¹, pH of this solution is 4. Calculate the value of pKₐ(HA) at 25°C. Given: Λ°M(NaA) = 100 Scm²mol⁻¹; Λ°M(HCl) = 425 Scm²mol⁻¹; Λ°M(NaCl) = 125 Scm² mol⁻¹

Answer 1

4

Answer 2

To calculate the pKₐ(HA), we need to follow these steps:

1. Calculate the limiting molar conductivity of HA (Λ°M(HA)):
   According to Kohlrausch's Law of independent migration of ions, the limiting molar conductivity of a weak acid HA can be calculated from the limiting molar conductivities of strong electrolytes as follows: Λ°M(HA) = Λ°M(NaA) + Λ°M(HCl) - Λ°M(NaCl)
   Given:
   Λ°M(NaA) = 100 Scm²mol⁻¹
   Λ°M(HCl) = 425 Scm²mol⁻¹
   Λ°M(NaCl) = 125 Scm²mol⁻¹

   Λ°M(HA) = 100 + 425 - 125 = 400 Scm²mol⁻¹

2. Calculate the degree of dissociation (α) of HA:
   The degree of dissociation (α) of a weak electrolyte is the ratio of its molar conductivity (ΛM) at a given concentration to its limiting molar conductivity (Λ°M).
   α = ΛM(HA) / Λ°M(HA)
   Given: Molar conductivity of HA solution (ΛM(HA)) = 200 Scm²mol⁻¹
   Calculated: Λ°M(HA) = 400 Scm²mol⁻¹

   α = 200 / 400 = 0.5

3. Determine the concentration of H⁺ ions ([H⁺]):
   The pH of the solution is given as 4.
   pH = -log[H⁺]
   4 = -log[H⁺]
   [H⁺] = 10⁻⁴ M

4. Calculate the initial concentration (C) of the HA solution:
   For a weak acid HA, the concentration of H⁺ ions is given by [H⁺] = Cα, where C is the initial concentration of the acid.
   [H⁺] = Cα
   10⁻⁴ = C × 0.5
   C = 10⁻⁴ / 0.5 = 2 × 10⁻⁴ M

5. Calculate the acid dissociation constant (Kₐ):
   For the dissociation of a weak acid HA:
   HA(aq) ⇌ H⁺(aq) + A⁻(aq)

   Initial: C 0 0
   Equil: C(1-α) Cα Cα

   The acid dissociation constant Kₐ is given by:
   Kₐ = ([H⁺][A⁻]) / [HA]
   Kₐ = (Cα × Cα) / C(1-α)
   Kₐ = Cα² / (1-α)

   Substitute the values of C and α:
   Kₐ = (2 × 10⁻⁴) × (0.5)² / (1 - 0.5)
   Kₐ = (2 × 10⁻⁴) × 0.25 / 0.5
   Kₐ = (0.5 × 10⁻⁴) / 0.5
   Kₐ = 10⁻⁴

   Alternatively, using [H⁺] and C:
   Kₐ = [H⁺]² / (C - [H⁺])
   Kₐ = (10⁻⁴)² / (2 × 10⁻⁴ - 10⁻⁴)
   Kₐ = 10⁻⁸ / 10⁻⁴
   Kₐ = 10⁻⁴

6. Calculate pKₐ:
   pKₐ = -log(Kₐ)
   pKₐ = -log(10⁻⁴)
   pKₐ = 4

The final answer is $\boxed{4}$

Explanation of the solution:

1. Kohlrausch's Law: Used to find the limiting molar conductivity of the weak acid HA (Λ°M(HA)) from the given limiting molar conductivities of strong electrolytes: Λ°M(HA) = Λ°M(NaA) + Λ°M(HCl) - Λ°M(NaCl) = 100 + 425 - 125 = 400 Scm²mol⁻¹.
2. Degree of Dissociation (α): Calculated using the formula α = ΛM / Λ°M = 200 / 400 = 0.5.
3. Hydrogen Ion Concentration ([H⁺]): Determined from the given pH: [H⁺] = 10^(-pH) = 10⁻⁴ M.
4. Initial Concentration (C): Calculated from [H⁺] = Cα, so C = [H⁺] / α = 10⁻⁴ / 0.5 = 2 × 10⁻⁴ M.
5. Acid Dissociation Constant (Kₐ): Calculated using the equilibrium expression Kₐ = Cα² / (1-α) or Kₐ = [H⁺]² / (C - [H⁺]). Both yield Kₐ = 10⁻⁴.
6. pKₐ: Calculated as -log(Kₐ) = -log(10⁻⁴) = 4.