Ksp Equation for CaF2: Coefficients and Stoichiometry Explained

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The discussion clarifies the correct expression for the solubility product constant (Ksp) of calcium fluoride (CaF2). The correct formula is Ksp = [Ca][F]^2, where the coefficients from the dissociation reaction are used as exponents. Some confusion arises from the misinterpretation of the formula Ksp = [Ca][2F]^2, which is incorrect. The correct understanding involves recognizing that if x moles of CaF2 dissolve, it produces x moles of Ca2+ and 2x moles of F-. This highlights the importance of stoichiometry in deriving the Ksp expression.
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CaF2⇔Ca+2F.
Many places state that Ksp=[Ca][2F]^2; Ksp=[x]*[2x]^2.
I thought the coefficient go into Ksp equation as an exponent. So, Ksp=[Ca][F]^2.

Can anyone explain which is correct, or why [2F]^2 (if that's the correct one)?
 
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staka said:
Many places state that Ksp=[Ca][2F]^2

That's wrong. Are you sure you are not misreading:

Ksp=[x]*[2x]^2.

which is correct - assuming x is molar solubility of CaF2.

Ksp=[Ca][F]^2.

That's correct.
 
Why do we get Ksp=[x]*[2x]^2?
(Ksp=[Ca][2F]^2 is from this.. misread wrong)
 
Simple stoichiometry. If x moles of CaF2 are dissolved - how many moles of Ca2+? How many moles of F-?
 
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