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Posted by on Monday, October 29, 2012 at 8:35pm.

Given a saturated solution of CaF2, write the ionic equilibrium and use it to calculate the solubility of CaF2.

My answers for the ionic equation and ksp values are
CaF2(s)<->Ca2+(aq)+ 2F-(aq)
Ksp= [Ca2+][F-]2
ksp=2.02*10^-4

it asks what will happen to the value of ksp when a solid NaF is added to the solution?

and what would be the new solubility of CaF2 (in a 0.100 mol/L NaF)

  • chemistry - , Monday, October 29, 2012 at 10:31pm

    ...........CaF2(s) ==> Ca^2+ + 2F^-
    I..........solid........0.......0
    C..........solid........x.......2x
    E..........solid........x.......2x

    Ksp = (Ca^2+)(F^-)^2
    Substitute from the ICE chart and solve for x. The question asks for solubility of CaF2. You will calculate x = (Ca^2+) in moles/L but since 1 mol CaF2 dissolves to give 1 mol Ca&2+, then x is also solubility CaF2.

    If NaF is added to the solution what happens? Remember Le Chatelier's Principle which says that a system at equilibrium will try to undo what we do to it. So if we add F^-, the reaction will try to get rid of it. How can it do that. Only one way. That's by shifting to the left so that the solubility of CaF2 is decreased.

    If we make the (F^-) = 0.1 M, then
    Ksp = (Ca^2+)(F^-)^2
    Ksp = (x)(0.1).
    Solver for x.

  • chemistry - , Tuesday, March 12, 2013 at 7:45am

    what is the name for CaF2 in chemistry?

  • chemistry - , Wednesday, April 20, 2016 at 1:08am

    You're a cheesehead

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