2NaF + CaCl2 �¨ CaF2 + 2NaCl

An excess of 0.20 M NaF is used to precipitate the calcium ions from a
50.0 mL sample of a calcium chloride solution. What is the concentration
of the original calcium chloride solution?

Ya gotta have more info, like the mass of the CaF2 pptd.

In regards to DrBob222, there were 3 parts to this question; the first part said there were .1952 g and the mols were 2.504 * 10^-3 mol. But I don't know if this part refers to the first part because it doesn't sound like it does.

0.1952 g CaCl2/molar mass CaF2 = 0.002504 mole CaF2. Convert that to moles CaCl2 which is 0.002504.

Then concn CaCl2 = mole CaCl2/L soln = ? 0.002504/0.050 L = ??M

Thanks!

To find the concentration of the original calcium chloride solution, we need to use the concept of stoichiometry and the balanced chemical equation provided.

The balanced chemical equation is:
2NaF + CaCl2 → CaF2 + 2NaCl

From the equation, we can see that 2 moles of NaF react with 1 mole of CaCl2 to form 1 mole of CaF2 and 2 moles of NaCl.

We are given the excess of 0.20 M NaF, which means that the concentration of NaF is not limiting in the reaction. Therefore, we can assume that all the calcium ions from the calcium chloride (CaCl2) react with NaF, and the number of moles of NaF used is equal to the number of moles of CaCl2 in the original solution.

First, we calculate the number of moles of NaF used:
moles of NaF = concentration of NaF x volume of NaF
= 0.20 M x 50.0 mL
= 0.010 moles

According to the balanced equation, the ratio of moles of CaCl2 to moles of NaF is 1:2. Therefore, the number of moles of CaCl2 in the original solution is half of the moles of NaF used.

moles of CaCl2 = 0.010 moles / 2
= 0.005 moles

Finally, we calculate the concentration of the original calcium chloride solution:
concentration of CaCl2 = moles of CaCl2 / volume of CaCl2
= 0.005 moles / 50.0 mL
= 0.100 M

Therefore, the concentration of the original calcium chloride solution is 0.100 M.