If 30 mL of 5.0 × 10^-4 M Ca(NO₃)₂ is added to 70 mL of 2.0 × 10^-4 M NaF, will a precipitate form? (K_sp of CaF₂ = 4.0 × 10^-11) A) No, because the ion product is less than K_sp. B) No, because the ion product is greater than K_sp. C) Not enough information is given. D) Yes, because the ion product is less than K_sp. E) Yes, because the ion product is greater than K_sp.

Question

Quizplus · Accepted Answer

The ion product (IP) is calculated by multiplying the concentrations of the ions involved in the potential precipitate. In this case, the potential precipitate is CaF2. The concentration of Ca2+ ions from Ca(NO3)2 is 5.0 × 10^-4 M in 30 mL, and the concentration of F- ions from NaF is 2.0 × 10^-4 M in 70 mL. When mixed, the total volume is 100 mL, diluting both ion concentrations. The diluted concentration of Ca2+ is (5.0 × 10^-4 M)(30 mL / 100 mL) = 1.5 × 10^-4 M, and the diluted concentration of F- is (2.0 × 10^-4 M)(70 mL / 100 mL) = 1.4 × 10^-4 M. The IP for CaF2 is calculated as [Ca2+][F-]^2. However, since the stoichiometry of CaF2 involves one Ca2+ ion and two F- ions, and given the Ksp of CaF2 is 4.0 × 10^-11, the actual IP will be less than this Ksp due to the dilution effect and the initial concentrations provided, indicating no precipitate will form.

If 30 ML of 5