For a concentrated solution of a weak electrolyte (K_(eq)= equilibrium constant) A₂B₃ of concentration 'c', the degree of dissociation 'α' is :

Solution

Chemistry ma’am · 3 steps

Write the ICE line for A₂B₃, and remember the stoichiometric coefficients become powers.

1. Dissociation. A₂B₃ ⇌ 2A³⁺ + 3B²⁻. Start with c of A₂B₃. If a fraction α dissociates, c(1-α) is left, and you get 2cα of A³⁺ and 3cα of B²⁻.

   K_(eq) = (2cα)²(3cα)³/c(1-α) = (4 × 27 c⁵α⁵)/c(1-α) = (108 c⁴α⁵)/(1-α)

2. Weak electrolyte. A weak electrolyte in a concentrated solution dissociates very little, so α ≪ 1 and 1-α ≈ 1. This gives K_(eq) ≈ 108 c⁴α⁵.
3. So. α = (K_(eq)/108c⁴)^(1/5), which is option (b).

The idea to keep. For AₓB_y, K ≈ xˣ yʸ cˣ⁺ʸ⁻¹αˣ⁺ʸ. Raise each coefficient to its own power, and one c always cancels with the undissociated concentration.

Answer from the NTA final answer key. Solution by PrepWiser.