How many moles of CaF₂ will dissolve in 3.0 L of 0.050 M NaF solution? (K_sp for CaF₂ = 4.0 × 10^-11) A) 4.8 × 10^-8 mol B) 8.9 × 10^-9 mol C) 8.0 × 10^-8 mol D) 2.7 × 10^-8 mol E) none of these

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The correct answer is A.The presence of NaF in the solution will cause the dissolution of CaF2 to be less than it would be in pure water. This is because the common ion effect will reduce the solubility of CaF2. The Ksp expression for CaF2 is:Ksp = [Ca2+][F-]^2The concentration of F- in the solution is 0.050 M. Substituting this value into the Ksp expression, we get:4.0 × 10^-11 = [Ca2+][0.050]^2[Ca2+] = 1.6 × 10^-6 MThe number of moles of CaF2 that will dissolve in 3.0 L of solution is:moles CaF2 = [Ca2+] × volume of solutionmoles CaF2 = 1.6 × 10^-6 M × 3.0 Lmoles CaF2 = 4.8 × 10^-8 mol

How Many Moles of CaF2 Will Dissolve in 3

How Many Moles of CaF₂ Will Dissolve in 3