# How much water is required to dissolve 25.2 mg of CaF2?

### 5 Answers

- ChemTeamLv 72 months agoFavourite answer
I have a solution here:

https://www.chemteam.info/Equilibrium/Ksp-molar-so...

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:

https://www.google.com/search?client=firefox-b-1-d...

The Ksp value I used is also given with the solubility data used.

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- Anonymous2 months ago
The kid just wants the answer handed to him, not to learn it.

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- 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

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- 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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