If the solubility of calcium fluoride is 2.0 × 10^-4 M, the solubility product value is ________.

Question

Quizplus · Accepted Answer

The solubility product value (Ksp) is calculated by multiplying the concentrations of the ions raised to their stoichiometric coefficients. For the chemical equation CaF2(s) ⇌ Ca2+(aq) + 2F-(aq), the Ksp = [Ca2+][F-]2. Plugging in the given solubility of 2.0 × 10^-4 M for CaF2, we get [Ca2+] = 2.0 × 10^-4 M and [F-] = 4.0 × 10^-4 M (since there are 2 moles of F- for every mole of Ca2+). Multiplying these values together and rounding to the appropriate significant figures gives a Ksp of 3.2 × 10^-11. Therefore, the correct choice is B.

If the Solubility of Calcium Fluoride Is 2