JEE Chemistry · Hard

Chemical and Ionic Equilibrium: Ksp application MCQ

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Ksp applicationHardQuestion 600439

Question

The solubility of CaF₂ (Ksp = 3.4×10⁻¹¹) in 0.1 M NaF solution (common ion effect) is:
  1. A
    3.4×10⁻⁹ M
    Correct
  2. B
    1.84×10⁻⁵ M
  3. C
    3.4×10⁻⁸ M
  4. D
    3.4×10⁻¹¹ M

Correct answer

3.4×10⁻⁹ M

Explanation

CaF₂ ⇌ Ca²⁺ + 2F⁻. With [F⁻] ≈ 0.1 M (from NaF): Ksp = s × (0.1)² → s = 3.4×10⁻¹¹/0.01 = 3.4×10⁻⁹ M. Much less than in pure water (s ≈ 2×10⁻⁴ M).