Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: missUK11 on April 20, 2012, 12:59:31 PM
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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
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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.
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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 ?
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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.