Start learning →

Solubility of CaF2 in water (Ksp = 4.0 × 10⁻¹¹):

A10⁻⁶ M
B10⁻³ M
C4 × 10⁻¹¹ M
D2.15 × 10⁻⁴ M
Answer & Solution
Correct answer: D. 2.15 × 10⁻⁴ M
CaF2 ⇌ Ca²+ + 2F-. If s is solubility, [Ca²+] = s and [F-] = 2s. Ksp = s(2s)² = 4s³. **So s = (Ksp/4)^(1/3) = (10⁻¹¹)^(1/3) ≈ 2.15 × 10⁻⁴ M.**
Practise this question with Quill — JEE Advanced →

Keep practising Unit 7A1: Equilibrium Constant and Dynamic Equilibrium

Browse all Unit 7A1: Equilibrium Constant and Dynamic Equilibrium questions for JEE Advanced, or explore more Chemistry topics.

Start a Unit 7A1: Equilibrium Constant and Dynamic Equilibrium lesson →