What is the calcium ion concentration in a solution prepared by mixing 447 mL of 0.407 M calcium nitrate with 467 mL of 0.363 M sodium fluoride? The 𝐾sp of calcium fluoride is 3.45×10−11. [Ca2+] =?

Question

What is the calcium ion concentration in a solution prepared by mixing 447 mL of 0.407 M calcium nitrate with 467 mL of 0.363 M sodium fluoride? The 𝐾sp of calcium fluoride is 3.45×10−11. [Ca2+] =?

Answer 1

Ca(NO₃)₂(aq) + 2NaF(aq) ==> CaF₂(s) + 2NaNO₃(aq) .. balanced equation

Ca²⁺(aq) + 2F^-(aq) ==> CaF₂(s) .. net ionic equation

moles Ca²⁺ present = 447 mls x 1 L / 1000 ml x 0.407 mol / L = 0.1819 moles Ca²⁺

moles F^- present = 467 ml x 1 L / 1000 ml x 0.363 mol / L = 0.1695 moles F^-

Volume = 447 ml + 467 ml = 914 mls = 0.914 L

[Ca²⁺] = 0.1819 mol / 0.914 L = 0.199 M

[F^-] = 0.1695 mol / 0.914 L = 0.185 M

Q = [Ca²⁺][F^-]² = (0.199)(0.185)² = 6.81x10^-3

Since Q > Ksp, a precipitate will form

moles Ca²⁺ reacted = 0.1695 mols F^- x 1 mol Ca²⁺ / 2 mol F^- = 0.08475 mols

moles Ca²⁺ left in solution = 0.1819 mols - 0.08475 = 0.0972 mols Ca²⁺ in solution

Final volume = 0.914 L

Final [Ca²⁺] = 0.0972 mols / 0.914 L = 0.106 M