Answer
The concentration of $F^-$ ions will decrease.
Work Step by Step
As we discussed in 59a, the equilibrium will move to increase the consume of the added $Ca^{2+}$, and it will produce more $CaF_2(s)$. That follows the Le Chatelier's principle.
This is the equilibrium:
$CaF_2(s) \lt -- \gt Ca^{2+}(aq) + 2F^-(aq)$
- When the equilibrium moves to the left, more $F^-$ ions will be consumed, decreasing its concentration.
