A 50.0-mL sample of 0.100 M Ca(NO3)2 is mixed with 50.00 mL of 0.200 M NaF. When the system has come to equil?

Question

A 50.0-mL sample of 0.100 M Ca(NO3)2 is mixed with 50.00 mL of 0.200 M NaF. When the system has come to equilibrium, which of the following sets of conditions will hold? The Ksp for CaF2 is 4.0 x 10-11.

Moles Solid CaF2 Formed [Ca2+] (M) [F-] (M)


A.
5.0 x 10-3 3.5 x 10-4 M 7.0 x 10-4 M

B. 5.0 x 10-3 3.4 x 10-9 M 0.05 M

C. 5.0 x 10-3 2.2 x 10-4 M 4.3 x 10-4 M

D. 5.0 x 10-3 3.5 x 10-4 M 4.3 x 10-4 M

E. 10.0 x 10-3 1.3 x 10-5 M 1.3 x 10-5 M

Answer 1

In the 50 mL solution you have
mole Ca+2 = M1*V1
when you mix the solution you initially have the sam mole Ca+2 but since the volume changes, the concentration will also change.

thus [Ca+2]o= M1V1/(V1+V2) =0.1*50/(50+50) =0.05 M
Note that I didn't convert mL into L since the convertion factor is simplified by the ratio.

Similarly [F-]o = M2V2/(V1+V2)= 0.2*50/(50+50) =0.1 M

.. .. .. .. .. CaF2 <=> Ca+2 +2F-
Initial .. .. .. .. .. .. .. .. 0.05 .. .0.1
Precipitate. .. .. .. .. .. x .. .. .. 2x
Produce .. .. x
At equil .. .. .. .. .. .. .. 0.05-x .. 0.1-2x

Ksp =[CA+2][F-]^2 = (0.05-x)(0.1-2x)^2 =4*10^-11 Do the calculations and you get a cubic equation.
The only acceptable solution is
x= 0.04978

so the moles of CaF2 formed are = x*V= 0.04978*0.1= 5.0 *10^-3
[Ca+2]= 0.05-0.04978= 0.00022 =2.2*10^-4
[F-]= 0.1-2*0.04978 = 4.4*10^-4

So the answer is C