For this problem, we’re being asked to calculate the solubility of CaF2(s) in 0.300 M NaF(aq). Since the compounds are ionic compounds, they form ions when dissociating in water. The dissociation of CaF2 and NaF in water are as follows:
Calcium is in Group 2A so it’s charge is +2. Fluorine is in Group 7A so its charge is –1:
CaF2(s) ⇌ Ca2+(aq) + 2 F–(aq)
Sodium is in Group 1A so it’s charge is +1. Fluorine is in Group 7A so its charge is –1:
NaF(s) → Na+(aq) + F–(aq)
Notice that there is a common ion present, F–. The common ion effect states that the solubility of a salt is lower in the presence of a common ion.
We can construct an ICE table for the dissociation of CaF2. Remember that solids are ignored in the ICE table.
What is the solubility of CaF2 (assume Ksp = 4.0 x 10-12 ) in 0.030 M NaF?
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