The solubility of calcium fluoride, CaF₂, in pure water is 2.15 × 10^-⁴ moles per liter. Calculate the value of K_sp for calcium fluoride from this data.

Question

Quizplus · Accepted Answer

The solubility product constant, Ksp, for CaF2 is calculated using the expression Ksp = [Ca^2+][F^-]^2. Given the solubility of CaF2 is 2.15 × 10^-4 M, [Ca^2+] = 2.15 × 10^-4 M and [F^-] = 2 × 2.15 × 10^-4 M. Therefore, Ksp = (2.15 × 10^-4)(2 × 2.15 × 10^-4)^2 = 3.98 × 10^-11.

The Solubility of Calcium Fluoride, CaF2, in Pure Water Is

The Solubility of Calcium Fluoride, CaF₂, in Pure Water Is