Calculate the solubility (in g/L)of calcium fluoride in water at 25°C if the K_sp for CaF₂ is 3.5 × . A)9.6 × 10^-4 g/L B)2.1 × 10^-4g/L C)3.3 × 10^-2 g/L D)4.1 × 10^-2 g/L

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Using the given equilibrium constant (K_sp), we can write the equation for the dissolution of calcium fluoride as follows: CaF2 (s) ⇌ Ca2+ (aq) + 2F- (aq) The equilibrium constant expression for this equation is: K_sp = [Ca2+][F-]2 We are given that K_sp = 3.5 × 10^-11. At equilibrium, let the concentration of Ca2+ be x M, and the concentration of F- be 2x M. Then: K_sp = x(2x)^2 = 4x^3 3.5 × 10^-11 = 4x^3 x = 9.6 × 10^-5 M The solubility of calcium fluoride is equal to the concentration of Ca2+ ions at equilibrium, which is 9.6 × 10^-5 M. To convert this to g/L, we need to multiply by the molar mass of CaF2 and the conversion factor between L and m^3: solubility = (9.6 × 10^-5 mol/L) × (78.08 g/mol) × (1000 cm^3/L) × (1 m^3/10^6 cm^3) solubility = 2.1 × 10^-4 g/L or solubility = 2.1 × 10^-2 g/L

Calculate the Solubility (In G/L)of Calcium Fluoride in Water at 25°C