Common Ion/solubility problem =T

[hopesremix]

I'm having trouble with the solubility section of gchem. Can anyone help me out with this one problem?

The solubility of CaF2 (Ksp= 4 x 10^-11) in a 0.1 M solution of Ca(NO3)2 is approx:

answer: 1 x 10^-5 M

why is that the answer?? Help will be much appreciated.

---

Members don't see this ad.

 

[mitshi]

CaF2 --> Ca2+ + 2F-
Assume x mol of Ca2+ and 2x mol of F- dissolve in 1 liter of solution, where x denotes solubility of CaF2.

Since 0.1 M of Ca(NO3)2 gives you an additional 0.1 mol of Ca2+ per each dm3, the Ksp will therefore be:

(x + 0.1)(2x)^2 = 4 x 10^-11
Assuming x << 0.1,
(2x)^2 = 4 x 10^-10
x^2 = 1 x 10^-10
x = 1 x 10^-5 mol/liter

 

[E. coli]

I'm having trouble with the solubility section of gchem. Can anyone help me out with this one problem?

The solubility of CaF2 (Ksp= 4 x 10^-11) in a 0.1 M solution of Ca(NO3)2 is approx:

answer: 1 x 10^-5 M

why is that the answer?? Help will be much appreciated.

Ksp = [Ca] [F]^2

Since you have 0.1 M solution of Ca(NO3)2, [Ca] is going to be 0.1. This is because of common ion effect.

ksp = (0.1 M)[F]^2
4*10^-11 = (0.1 M)[F]^2
4 * 10^-10 = [F]^2

[F] = 2 * 10^-5 M

Since one CaF2 molecule has two molecules of F, the solubility of the entire CaF2 molecule will be 1* 10^-5 M.

Hope this helps !

 

[hopesremix]

thanks guys =)