Chemistry

posted by .

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

My answer for the ionic equation is
CaF2(s)<->Ca2+(aq)+ 2F-(aq)
Ksp= [Ca2+][F-]2

the given ksp is 3.32*10-4

how do i figure out the solubility??
i know the stoichiometry is 1:2 but i don't know how to finish this problem... please help

  • Chemistry -

    Everything you've done so far is very good.
    ..........CaF2 ==> Ca^2+ + 2F^-
    I........solid......0.......0
    C........solid.......x......2x
    E........solid......x.......2x

    Ksp = [Ca^2+][F^-]^2 = 3.32E-4
    3.32E-4 = (x)(2x)^2 and solve for x.
    x = cncn Ca^2+ in mols/L. The question asks for solubility of CaF2 but since 1 mol Ca^2+ is obtain for each CaF2 solid molecule that dissolves, then x = solubility CaF2 (in M or mols/L)

  • Chemistry -

    ksp=(x)(2x)2=3.32*10^-4
    = (x)(4x2)=3.32*10^-4
    = 4x3=3.32*10^-4
    then i took the cubed root of that and got 2.02*10^-4???

    is that correct?

  • Chemistry -

    for this same question it also asks...

    what will happen to the value of ksp?
    and
    Calculate the new solubility of CaF2 (in 0.100mol/L NaF)

    how do i go about doing the second part?

    the ksp will increase beacuse you are adding more to the solution?

  • Chemistry -

    I agree down to 4x^3 = 3.32E-4. However, when I divide by 4 and take the cube root of the result I don't get your value.
    When 0.1M NaF is added the solubility decreases because you now have a common ion of F^-
    Here 3.32E-4 = (x)(2x+0.1)
    and solve for x.

    NO. NO. NO. Ksp does NOT change, K is a constant. Why would we call it a constant if it were not constant. Why does it not change? Adding more SOLID CaF2 will not make more of it dissolve. The amount that will dissolve will dissolve regardless of whether you add another spoonful of solid or not. By the way, raising or lowering the T WILL change K because the solubility changes with changing temperature.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Chemistry

    At saturation 1.5 x 10^-4 moles caF2 have dissolved in 1L of water. What is Ksp for CaF2 at this temperature?
  2. chem/solubility

    the Ksp of CaF2 = 1.46*10^-10 and Ka of HF = 3.5*10^-4 Calculate the pH of a solution in which the solubility of CaF2 = .0100 moles/liter. so Ksp =[Ca2+][F-]^2 and Ka =[H3O+][F-]/[HF] I'm not sure how to continue...
  3. chemistry

    The mineral CaF2 has a solubility of 2.1 x 10^-4 M. What is the Ksp of CaF2?
  4. Chemistry 114

    Write down the Ksp expression for CaF2. If the molar solubility of CaF2 at 35℃ is 1.24×10^(-3)mol/dm3, calculate the value of Ksp.
  5. Chemistry

    What is the molar solubility of CaF2 if o,1o mol of NaF are added to 1.0L of a saturated solution of CaF2?
  6. chemistry

    What is the solubility of CaF2 in a buffer solution containing .30 M HCHO2 and .20 M NaCHO2?
  7. Chemistry

    Consider a saturated solution of calcium fluoride in 0.086 M potassium nitrate. Complete the following tasks, and then answer the question. a) Write the chemical equation corresponding to Ksp. b) Write the defining expression for the …
  8. chemistry

    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 …
  9. chemistry

    Given the following solubility constants, which list arranges the solutes in order of increasing solubility?
  10. Chemistry

    Calcium Fluoride, CaF2, adopts the fluorite lattice, which is described as a face-centered cubic array of Ca2+ ions with F-1 ions with Ca2+ ions in half of the cubic holes. The Radii of Ca2+ and F-1 are 126 and 117 pm, respectively. …

More Similar Questions