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CHM 101 CHM 101 Past Questions Objective Question
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Question

The solubility of CaF₂ is 2.0 × 10⁻⁴ M at 25°C. What is the Ksp of CaF₂?
Options
A 3.2 × 10⁻¹¹
Correct Answer
B 8.0 × 10⁻¹²
C 1.6 × 10⁻¹¹
D 4.0 × 10⁻¹²
Correct Answer

Option A is the correct answer.

Detailed Explanation

CaF₂(s) ⇌ Ca²⁺(aq) + 2F⁻(aq). If solubility s = 2.0×10⁻⁴ M, then [Ca²⁺] = s, [F⁻] = 2s. Ksp = [Ca²⁺][F⁻]² = s × (2s)² = 4s³ = 4 × (2.0×10⁻⁴)³ = 4 × 8.0×10⁻¹² = 3.2×10⁻¹¹.

Hint

Relate the ion concentrations to the molar solubility using the dissociation equation.

Question Info

Course: CHM 101
Topic: CHM 101 Past Questions
Type: Objective
Difficulty: Not specified
Question 1261 of 1872
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