Question

Given that the K sp for CaF2 at 25°C is 3.9 × 10–11, will a precipitate form when 0.10 L of 2.3 × 10–4 M CaCl2 is added to 0.50 L of 5.0 × 10–3 M NaF? Assume volumes are additive. Pick one:

Yes, because Q = 5.8 × 10–9 and therefore Q > K sp.

Yes, because Q = 6.6 × 10–10 and therefore Q > K sp.

Yes, because Q = 1.6 × 10–7 and therefore Q > K sp.

No, because Q = 6.6 × 10–10 and therefore Q > K sp. Please Explain

Answer 1

The precipitant

CaF2 <--> Ca+2 and 2F-

Ksp = [Ca+][F-]^2

Ksp = 3.9*10^-11

VT = V1+V2 = 0.1+0.5 = 0.6

mol Ca+2 = MV = 0.1*(2.3*10^-4)= 0.000023

[Ca+2] = 0.000023/0.6 = 0.00003833333 M

mol F- = MV = 0.5*5*10^-3 = 0.0025 mol

[F-] = mo/V =0.0025 /0.6 = 0.0041666

then

Q = [Ca+][F-]^2

Q = 0.00003833333 *(0.0041666^2) = 6.654879*10^-10

Since Q > Ksp

expect precipitation, since this is supersaturated (i.e. plenty of ions in solution, which is not possible)