Solid sodium fluoride is slowly added to 150 mL of a barium nitrate solution until the concentration of fluoride ion is 0.0479 M. The maximum amount of barium ion remaining in solution is _____ M

Answer 1

Sodium fluoride = NaF

Barium nitrate = Ba(NO₃)₂

Reaction:

2NaF(aq) + Ba(NO₃)₂(aq) ==> 2NaNO₃(aq) + BaF₂(s)

BaF₂(s) <==> Ba²⁺(aq) + 2F^-(aq) .. Ksp = 1.84x10^-7 (Wikipedia)

Ksp = 1.84x10^-7 = [Ba²⁺][F^-]²

1.84x10^-7 = [Ba²⁺][0.0479]²

[Ba²⁺] = 1.84x10^-7 / (0.0479)²

[Ba²⁺] = 8.02x10^-5 M