The solubility of calcium chromate is 1.56 × 10^-3 g/100 mL of solution. What is the K_sp for CaCrO₄? A) 2.4 × 10^-4 B) 1.5 × 10^-5 C) 7.6 × 10^-6 D) 1.0 × 10^-8 E)

Question

The solubility of calcium chromate is 1.56 × 10^-3 g/100 mL of solution. What is the K_sp for CaCrO₄? A) 2.4 × 10^-4 B) 1.5 × 10^-5 C) 7.6 × 10^-6 D) 1.0 × 10^-8 E) < 1.0 × 10^-8

Quizplus · Accepted Answer

The solubility expression for calcium chromate is: 
CaCrO4 (s) ⇌ Ca2+ (aq) + CrO42- (aq)
The equilibrium expression for the solubility product constant is:
Ksp = [Ca2+] [CrO42-]
We are given that the solubility of calcium chromate is 1.56 × 10-3 g/100 mL of solution. We can use this information to find the concentration of Ca2+ and CrO42- in the saturated solution.
1. Convert 1.56 × 10-3 g/100 mL to mol/L:
1.56 × 10-3 g/100 mL × 1 mol/172.11 g = 9.05 × 10-6 mol/L
2. The solubility of calcium chromate gives us the concentration of both ions in the saturated solution:
[Ca2+] = 9.05 × 10-6 mol/L
[CrO42-] = 9.05 × 10-6 mol/L
3. Substitute these values into the equilibrium expression and solve for Ksp:
Ksp = [Ca2+] [CrO42-] = (9.05 × 10-6)2 = 8.20 × 10-11
Therefore, the answer is D) 1.0 × 10-8.

The Solubility of Calcium Chromate Is 1