Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at 100.15°C. When 0.2 mol of NaCl is added to the resulting solution, it was observed that the solution froze at -0.8° C. The solubility product of PbCl₂ formed is __________ × 10⁻⁶ at 298 K. (Nearest integer)
Given : K_b=0.5 K kg mol⁻¹ and K_f=1.8 K kg mol⁻¹. Assume molality to the equal to molarity in all cases.

Solution

Chemistry ma’am · 5 steps

Count particles from each colligative reading; the missing ones are the PbCl₂ that precipitated.

1. Find lead nitrate. Pb(NO₃)₂ gives 3 ions, so Δ T_b = 0.15 = K_b × i × m = 0.5 × 3 × m. That gives m = 0.1, so there is 0.1 mol of Pb²⁺ and 0.2 mol of NO₃⁻.
2. Particles from freezing. After NaCl is added, the total particle molality is 0.8/1.8 = 4/9 = 0.444. Without any precipitate it would be 0.1 + 0.2 + 0.2 + 0.2 = 0.7, from Pb²⁺, NO₃⁻, Na⁺ and Cl⁻.
3. Amount precipitated. Each x mol of PbCl₂ that precipitates removes x of Pb²⁺ and 2x of Cl⁻. That is 3x particles in all. So 0.7 - 3x = 0.444, which gives x = 0.0852.
4. Ions left dissolved. [Pb²⁺] = 0.1 - 0.0852 = 0.0148 M and [Cl⁻] = 0.2 - 0.1704 = 0.0296 M. These are the concentrations in the saturated solution.

   Kₛₚ = [Pb²⁺][Cl⁻]² = 0.0148 × (0.0296)² = 1.30 × 10⁻⁵

5. So. Kₛₚ = 13.0 × 10⁻⁶, which rounds to 13.

The common slip: 438. Writing Kₛₚ = [Pb²⁺][Cl⁻] without squaring the chloride term gives about 4.4 × 10⁻⁴. Written as a multiple of 10⁻⁶, that is 438.

The idea to keep. When a precipitate forms, colligative data counts only the particles still dissolved. The drop in the particle count tells you how much came out of solution.

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