calculate molar solubility and g/L?

Question

The concentration of calcium ions in a saturated solution of calcium fluoride is 2 x 10^-4 M. What is the solubility of the calcium fluoride in moles/L (molar solubility) and g/L.

I don't understand what to do with the 2 x 10^-4. Can someone show me how to do this question? Thankyou!

Answer 1

Calcium fluoride is CaF2. In water this will dissociate :

CaF2 ↔ Ca2+ + 2F-

From this equation you can see that for every mol Ca 2+ you have, you start with 1mol CaF2

You are told that you have 2*10^-4M Ca 2+ ions in the solution. That is 2*10^-4 moles of Ca 2+ ions in 1 litre of solution. You can now conclude that the concentration of CaF2 must be equal to this , or 2*10^-4 mol CaF2 per litre. Answer to first part



What is this as a mass, g per litre?

Molar mass CaF2 = 40.1 + 2*19 = 78.1g/mol

2*10^-4 mol = 2*10^-4*78.1 = 0.0156g/litre. Answer to second part.

Answer 2

CaF2 = > Ca2+ + 2F-



for each mole of CaF2 that dissolves 1 mole of Ca2+ is formed

if we have 2 x 10^-4 Ca2+ in solution 2 x 10^-4 moles of CaF2 has dissolved



molar solubility of CaF2 is 2 x 10^-4 moles per liter



mass of CaF2 that dissolves per liter = 2 x 10^-4 x 78 = 0.0156 g/ liter

Answer 3

Lead iodide is almost insoluble. The equilibrium in a saturated solution of PbI2 is:

PbI2(s) ⇌ Pb2+(aq) + 2I–(aq)

the solubility of lead iodide was found to be 1.53×10−3 M.



What is this concentration expressed in g/L?