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Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: organosurf on November 10, 2014, 11:45:07 AM

Title: Reaction of Calcium Fluoride ( CaF2 )
Post by: organosurf on November 10, 2014, 11:45:07 AM
Dear All,

Calcium Fluoride is ionically bonded but practically insoluble in water.
Can it undergo a double displacement reaction by reacting with a strong base such as 50% NaOH(aq) with heating so as to "solubilize" it to give NaF ?

CaF2(s)   +   2NaOH(aq)    :rarrow:  Ca(OH)2(s)  :spindown:   +   2NaF(aq)

Thanks in advance.






Title: Re: Reaction of Calcium Fluoride ( CaF2 )
Post by: unsu on November 22, 2014, 12:22:48 AM
As far as I know, CaF2 does not react with aqueous solutions of acids and bases.

CaF2 Ksp = 3.9 × 10−11
Ca(OH)2 Ksp = 4.68×10−6

CaF2  ::equil:: Ca2+ + 2F-
Ca2++2OH- ::equil:: Ca(OH)2

K(reaction)= Ksp(CaF2)/Ksp(Ca(OH)2)=8.3× 10−6 (very small equilibrium constant)

ΔG°=-RT ln(K) > 0

Thermodynamics would not favour the reaction. Maybe a few ions of fluoride would be produced.
Am I correct?
Title: Re: Reaction of Calcium Fluoride ( CaF2 )
Post by: Borek on November 22, 2014, 03:58:55 AM
Quote from: unsu on November 22, 2014, 12:22:48 AM
As far as I know, CaF2 does not react with aqueous solutions of acids and bases.

To some extent it definitely does. HF is a weak acid, so in the solutions of strong acids F- gets protonated, shifting dissolution to the right. How substantial the change is is another question.

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K(reaction)= Ksp(CaF2)/Ksp(Ca(OH)2)=8.3× 10−6 (very small equilibrium constant)

What reaction? It is not clear from what you wrote.

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Thermodynamics would not favour the reaction. Maybe a few ions of fluoride would be produced.

"Few ions" of fluoride are already present in the saturated solution of CaF2.

Quote
Am I correct?

I agree with your general conclusion that the reaction doesn't work, but I am not convinced your reasoning is a correct one.
Title: Re: Reaction of Calcium Fluoride ( CaF2 )
Post by: organosurf on November 26, 2014, 03:12:21 PM
Dear All,
Thanks for your input. CaF2 is almost insoluble in water, 0.0015g/100 mL (20 °C).
For the 0.0015g that is soluble, can this be reacted by double displacement with a soluble compound that is above Calcium in the electrochemical series, ( eg. Na2CO3 ) so as to precipitate the Calcium as another insoluble salt such as CaCO3 ?  Meaning, the reaction rate will be slow, but the equilibrium shifts to the right :

CaF2(s)  +  Na2CO3(aq)    :rarrow:  CaCO3(s) :spindown:  +  2NaF(aq)
insoluble      soluble                       insoluble                       soluble

As the CaCO3 is removed from solution, more CaF2 ionizes ( CaF is ionically bonded ) and goes into solution, reacting with the Na2CO3, thus shifting equilibrium to the right.
The ionic bonding of NaF is more stronger and stable than that of Na2CO3, again favouring formation of NaF. Is this explanation plausible ?  ???