Question

Solution 1 is made by dissolving cadmium fluoride (CdF2, Ksp = 6.44 x 10-3) in 100.0 mL of water until excess solid is present. Solution 2 is prepared by dissolving lithium fluoride (LiF, Ksp = 1.84 x 10-3) in 200.0 mL of water until excess solid is present. Which solution has a higher fluoride ion concentration?

Answer 1

@amistre64 @Destinymasha @ganeshie8 HELP

Answer 2

do you know how to calculate the solubility of the salt knowing the Ksp?

Answer 3

Aah , no

Answer 4

@Cuanchi

Answer 5

the CdF2 is an low solubility salt in water

CdF2(s) -> Cd2(aq)+ + 2 F-(aq)

if you write the Ksp as the equilibrium constant for the reaction

Ksp= [Cd2+] [F-]^2= 6.44 x 10-3

if we said that the solubility of ;the salt is equal to the moles of salt dissolved
the [Cd2+] is going to be equal to the solubility of the salt because the soichimetry is one to one. [Cd2+] = S
but the [F-] is going to be twice the solubility of the salt [F-]= 2S
then we can replace the [Cd2+] [F-] by the expression as a solubility to have only one variable "S"
Ksp= [Cd2+] [F-]^2= 6.44 x 10-3
Ksp= S x (2S)^2 = 6.44 x 10-3
Ksp = 4S^3=6.44 x 10-3
\[S=\sqrt[3]{ \frac{ Ksp }{4 }}\]
it is a cubic root!!!
Then the [F-]= 2S => \[[F ^{-}]= 2S =2 \times\sqrt[3]{ \frac{ Ksp }{4 }}\]
You can do a similar approach with the LiF, Ksp = 1.84 x 10-3

Answer 6

https://drive.google.com/file/d/0B0iV-eKggI19dEdlRWpRVlpyUFE/view?usp=sharing

Answer 7

Thanks