The solubility product of PbI₂ is 8.0 × 10⁻⁹. The solubility of lead iodide in 0.1 molar solution of lead nitrate is x × 10⁻⁶. mol/L. The value of x is __________. (Rounded off to the nearest integer) [Given √2 = 1.41]

Solution

Chemistry ma’am · 3 steps

Lead nitrate already supplies the Pb²⁺, so only the iodide carries the unknown s.

1. Set up concentrations. PbI₂ ⇌ Pb²⁺ + 2I⁻. If s mol/L of PbI₂ dissolves, [Pb²⁺] = 0.1 + s and [I⁻] = 2s. Since s is tiny, 0.1 + s ≈ 0.1.

   Kₛₚ = [Pb²⁺][I⁻]² = (0.1)(2s)² = 0.4 s²

2. Solve for s. 0.4 s² = 8.0 × 10⁻⁹, so s² = 2 × 10⁻⁸. Then s = √2 × 10⁻⁴ = 1.41 × 10⁻⁴ mol L⁻¹.
3. So. s = 141 × 10⁻⁶ mol L⁻¹, so x = 141.

The common slip: 282. Writing [I⁻] = s instead of 2s gives 0.1 s² = 8 × 10⁻⁹ and x = 282. Each PbI₂ releases two iodide ions.

The idea to keep. With a common ion, take its concentration from the added salt and let s appear only in the other ion. Never forget the stoichiometric coefficient: 2s gets squared.

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