How much water is required to dissolve 25.2 mg of CaF2?
- ChemTeamLv 72 months agoFavourite answer
I have a solution here:
Look at problem #3.
I decided to start from the Ksp of CaF2 and arrive at the mg/L value by calculation.
The value that billrussell42 started with can be looked up here:
The Ksp value I used is also given with the solubility data used.
- Anonymous2 months ago
The kid just wants the answer handed to him, not to learn it.
- hcbiochemLv 72 months ago
You can also approach this using the Ksp for CaF2. I found multiple values for the Ksp for this compound, but using 3.9X10^-11,
Ksp = [Ca2+][F-]^2 = 3.9X10^-11
Letting [Ca2+] = S, [F-] = 2S, then
3.9X10^-11 = S (2S)^2 = 4S^3
S = 2.14X10^-4 M
So, a saturated solution of CaF2 has a [CaF2] = 2.14X10^-4 M
25.2 mg = 0.0252 g
moles CaF2 = 0.0252 g / 78.07 g/mol = 3.23X10^-4 moles CaF2
Volume = 3.23X10^-4 mol / 2.14X10^-4 mol/L = 1.51 L
- billrussell42Lv 72 months ago
Calcium fluoride has a solubility in water of
0.015 g/L (18 °C)
0.016 g/L (20 °C)
taking the 20ºC number, that is 0.016 mg/mL
25.2 mg / 0.016 mg/mL = 1575 mL or 1.575 L
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- HernandoLv 62 months ago
About a cup and a half, I've found.