How many moles of Ca(NO₃)₂ must be added to 1.0 L of a 0.10 M HF solution to begin precipitation of CaF₂(s)? For CaF₂, K_sp = 4.0 × 10^-11. A) 4.0 × 10^-9 mol B) 4.0 × 10^-10 mol C) 4.5 × 10^-7 mol D) 4.0 × 10^-12 mol E) 4.0 × 10^-13 mol

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To begin precipitation of CaF2, the product of the concentrations of Ca2+ and F- ions must equal the Ksp of CaF2. Given Ksp = 4.0 × 10^-11 and the initial concentration of HF is 0.10 M, assuming complete dissociation of HF (which is a weak acid, but for simplicity in calculation), the concentration of F- would also be 0.10 M. The equation for Ksp is [Ca2+][F-]^2 = 4.0 × 10^-11. Substituting [F-] = 0.10, we solve for [Ca2+] = 4.0 × 10^-11 / (0.10)^2 = 4.0 × 10^-9 mol/L. Since the volume is 1.0 L, the amount of Ca2+ needed is also 4.0 × 10^-9 mol.

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