Question

A solution containing 0.100 M HF also contains 0.0100 M Ca(NO3)2. Hydrochloric acid is added slowly till the solution has a pH of 2.00. Will a precipitate of calcium fluoride,CaF2 occur? for the HF, Ka=6.46 x 10^-4 and for the CaF2, Ksp=2.69 x10^-11

Answer 1

CaF₂ + 2H₃O⁺ = Ca²⁺ + 2HF + 2H₂O

[H₃O⁺] = 10^-pH = 10^-2

a_F- = K_{a, HF} / ([H₃O⁺] + K_{a, HF}) = 6.46 х 10^-4 / (10^-2 + 6.46 х 10^-4) = 6.1 х 10^-2

[F^-] = C(HF) x a_F- = 0.100 x 6.1 х 10^-2 = 6.1 х 10^{-3 M}

[Ca²⁺] x [F^-] = 0.0100 x 6.1 х 10^-3 = 6.1 х 10^-5 > K_sp = 2.69 х 10^-11