Consider the dissociation of the weak acid HX as given below
HX(aq) ⇌ H⁺(aq)+X⁻(aq), Ka=1.2 × 10⁻⁵
[Kₐ : dissociation constant]
The osmotic pressure of 0.03 M aqueous solution of HX at 300 K is _________ × 10⁻² bar (nearest integer).
[Given : R=0.083 L bar mol⁻¹ K⁻¹]

Solution

Chemistry ma’am · 3 steps

A weak acid adds a few extra ions, so find α from Kₐ first, then put i = 1 + α into π = iCRT.

1. Degree of dissociation. For a weak acid with small α, Kₐ ≈ Cα². So α = √((1.2 × 10⁻⁵)/0.03) = √(4 × 10⁻⁴) = 0.02.
2. Van 't Hoff factor. Each dissociated HX gives two particles. The total concentration of particles is C(1-α) + Cα + Cα = C(1+α), so i = 1 + 0.02 = 1.02.

   π = iCRT = 1.02 × 0.03 × 0.083 × 300 = 0.762 bar

3. So. π ≈ 76.2 × 10⁻² bar, which rounds to 76.

The common slip: 75. Ignoring the dissociation and taking i = 1 gives π = 0.747 bar, which is 75, not 76.

The idea to keep. For a weak electrolyte, colligative properties need i. Get α from Kₐ = Cα², then use i = 1 + (n-1)α, where n is the number of ions formed per formula unit.

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