If the solubility of CaF2 (s) in water is 5.0 x 10^-3. The solubility will be
A. The same in 0.10 M NaF
B. less than 0.10 M NaF
C. Higher in 0.10 M NaF
D.None of the above
Why can't you use the same screen name. It takes longer to answer when we must surf through different names. At least you're consistent;ie., I see all of the names start with the letter M.
........CaF2 ==> Ca^2+ + 2F^-
Ksp = (Ca^2+)(F^-)^2 = 5E-3
Then NaF --> Na^+ + F^-
Note NaF is 100^ ionized and 0.1M NaF gives 0.1M Na^+ and 0.1M F^-. So F^- is a common ion, that will increase F^- in the CaF2 equilibrium, the reaction will shift to the left and ........
To determine the effect of NaF on the solubility of CaF2 in water, we need to understand the concept of common ion effect.
When a solid compound, such as CaF2, is dissolved in water, it dissociates into its respective ions. In this case, CaF2 dissolves to form Ca2+ and 2F- ions. The solubility of CaF2 can be expressed as the concentration of the Ca2+ and F- ions in the solution.
Now, when we add NaF to the solution, it also dissociates into Na+ and F- ions. Since F- ions are already present in the CaF2 solution, the addition of NaF increases the concentration of F- ions in the solution. This increase in F- ions concentration can affect the equilibrium of the dissolution process.
According to Le Chatelier's principle, adding more of a product (in this case, F- ions) will shift the equilibrium towards the reactants (CaF2(s)) side in order to relieve the stress caused by the increased concentration. As a result, the solubility of CaF2 will decrease in the presence of NaF.
Therefore, the correct answer is B. The solubility of CaF2 will be less than 0.10 M NaF.