[missUK11]

Q. Given the Ksp of CaF₂ is 4.0x10^-11, calculate the concentration of Fluoride ions in 1.2 x 10^-3 mol kg^-1 of  KNO₃


KNO3--> k⁺ + NO3^-1

so intial conc: KNO₃= 1.2 X 10^-3 K⁺=0 NO3^-1=0

    change: KNO₃= -1.2x10^-3     K⁺= + 1.2X10-3    NO3^-1= +1.2X10-3

 
CaF₂--> Ca2+ 2F-

initial conc: CaF₂= x   Ca²⁺= 0      F-=0

change : CaF₂= - x  ca²⁺=+x     F-= +2x

So... [ca2+][F-]²=ksp
       
(x)(2x)²=4.0x10-11
x=2.15 x10-4 so does the concentration of fluoride ions =4.3x10-4 mol kg-1 ?? does the KNO3 have any effect??

thanks

[Borek]

initial conc: CaF₂= x

No, initial concentration of CaF₂ is not x, it is zero. CaF₂ is not dissolved at all, and you can assume it dissolves only by dissociation.

[ca2+][F-]²=ksp
       
(x)(2x)²=4.0x10-11

Ironically, this is correct - as thankfully Ksp doesn't contain term related to CaF₂ presence.

KNO₃ modifies the ionic strength of the solution, so yes, it changes the solubility.

[missUK11]

initial conc: CaF₂= x

No, initial concentration of CaF₂ is not x, it is zero. CaF₂ is not dissolved at all, and you can assume it dissolves only by dissociation.

[ca2+][F-]²=ksp
       
(x)(2x)²=4.0x10-11

Ironically, this is correct - as thankfully Ksp doesn't contain term related to CaF₂ presence.

KNO₃ modifies the ionic strength of the solution, so yes, it changes the solubility.

So how would i calculate the fluoride concentration by considering the ionic strength of KNO3 ?

[Borek]

See for example:

http://www.chembuddy.com/?left=pH-calculation&right=ionic-strength-activity-coefficients

While it is mostly about pH calculation, it applies here as well. Calculate ionic strength, calculate activity coefficients, and do the calculations remembering that

[tex]K_{sp} = a_{Ca^{2+}} a_{F^-}^2 = f_2 [Ca^{2+}] (f_1 [F^-])^2[/tex]

where a is the activity, and f₁ and f₂ are activity coefficients respectively for single and double charged ion.