Chemistry Forum

Calculate the mass of CaF₂ produced when 5g of Ca(NO₃)₂ is mixed with 3g of NaF in water, with a final volume of 500 mL. Ksp is 3.5E-11 for the system.

My attempt to answer: Reaction equation is Ca(NO₃)₂ + 2NaF → CaF₂ + 2NaNO₃, we have 3.047e-2 moles of Calcium Nitrate and 7.145e-2 moles of Sodium Fluoride
How to proceed further?

Check what is limiting agent.

Mass of the CaF₂ is 0.0304718 moles × 78.1 g(molar mass of CaF₂ = 2.38 g.

I computed the solubility of CaF₂ Calcium Fluoride in grams per liter in water from the Kₛₚ 3.5e-11 = 1.6 × 10⁻² g CaF₂/L.

Molar solubility of CaF₂ in 0.07145 M of NaF solution is 4.89E-10 M.

Are the above answers correct?
How to proceed further?

I think the amount is driven from the limiting agent. Ca 2+ is limited ion. Everything what is above the soloubilty will precipitate. Its about 2.38 g.