In: Chemistry
What is the common ion effect and how may the common ion effect be used to prevent the dissolution of certain sparingly soluble salts into water solution? Calculate the concentration of CaF2 in 0.002 M KF.
1)
Common ion effect is decrease in solubility of a sparingly soluble
salt in presence of a common ion
For example when we mix CaF2 with KF
CaF2 dissolves as:
CaF2 <----> Ca2+ + 2 F-
when we add KF, we are adding more of F-. This is adding
a prduct in equilibrium. So, equilibrium shifts to reactant side
hence decreasing the solubility
2)
KF here is Strong electrolyte
It will dissociate completely to give [F-] = 0.002 M
At equilibrium:
CaF2 <----> Ca2+ + 2 F-
s 2*10^-3 + 2s
Ksp = [Ca2+][F-]^2
3.45*10^-11=(s)*(2*10^-3+ 2s)^2
Since Ksp is small, s can be ignored as compared to 2*10^-3
Above expression thus becomes:
3.45*10^-11=(s)*(2*10^-3)^2
3.45*10^-11= (s) * 4*10^-6
s = 8.625*10^-6 M
Answer: 8.6*10^-6 M