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Calculating Ksp from Molar Solubilty

  1. May 20, 2013 #1
    1. The problem statement, all variables and given/known data
    The molar solubility of magnesium fluoride, MgF2 is 1 x 10-3 in pure water. What is the Ksp for MgF2?

    a)4 x 10-3
    b)4 x 10-6
    c)4 x 10-9
    d)2 x 10-3
    e)1 x 10-3

    My book says the correct answer is C in bold.

    2. Relevant equations
    Ksp= [Mg2+][F-][F-]


    3. The attempt at a solution

    Ksp= [Mg2+][F-][F-]=[1 x 10-3][1 x 10-3][1 x 10-3]=

    1 x 10-9



    I think this should be the correct answer. For some reason the book multiplies by four. It says:

    "The relationship between the solubility product constant and molar solubility for a compound that produces 3 moles of ions for every mole of solid dissolved is as follows:

    Ksp=4x3= 4(1 x 10-3)3= 4 x 10-9 "

    I don't understand why to multiply by four.
     
  2. jcsd
  3. May 20, 2013 #2

    Borek

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    Staff: Mentor

    Write dissolution reaction and think about stoichiometry.
     
  4. May 20, 2013 #3
    MgF2 (s) ----> Mg2+ + 2F-

    There is 1 x 10-3 moles/liter of each ion. What did I do wrong? :(
     
  5. May 20, 2013 #4

    Borek

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    Staff: Mentor

    That would mean exactly the same number of fluoride and magnesium ions. Take a look at the reaction equation - are they produced in 1:1 ratio?
     
  6. May 20, 2013 #5
    Oh, I think I see. So I shouldn't have divided up the fluoride ions into individual concentrations:

    The total fluoride ion concentration is 2 x (1 x 10-3)= 2 x 10-3

    So, the Ksp= [Mg2+][F-]

    [1 x 10-3][(2 x 10-3]= 2 x 10-6
     
  7. May 20, 2013 #6
    ...Oops I forgot to include, but that's not an answer choice.
     
  8. May 20, 2013 #7

    Borek

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    Staff: Mentor

    Compare your two posts:

     
  9. May 20, 2013 #8
    The only way I could see arriving at an answer of 4 x 10-9 is by squaring (2 x 10-3) before multiplying it by (1 x 10-3), which would mean that the concentration of [F-] is squared.

    It doesn't make sense to me square the concentration of F-. If there are two moles of F- ions, per MgF2, why not just multiply 2 times the [MgF2] to get [F-].

    Where does [F-]2 come into all this or am I not supposed to square it at all?
     
  10. May 20, 2013 #9

    Borek

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    That puts us several squares back. What is Ksp and how is it written for a XaYb type salt?

    Or perhaps I should ask more general question - do you know what the reaction quotient is? Ksp is a just a specific case.
     
  11. May 20, 2013 #10
    Yes. I know about both solubility product constants and equilibrium constants.

    I guess what's happening here is something I've been trying to avoid all along: understanding why reaction orders match the coefficients in a balanced chemical equation. I've kind of just accepted it without really knowing why. I've heard that it has something to do with number of collisions or something like that, but I've never really gotten into depth.
     
  12. May 20, 2013 #11

    Borek

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    Staff: Mentor

    No, you don't have to dig into kinetics, it is enough that you learn the definition of the reaction quotient.

    (Actually reaction quotient can be easily derived from the thermodynamics, but it is still not necessary, definition is perfectly enough).
     
  13. May 20, 2013 #12
    Borek, why do I have to square the concentration of [F-] once I've already multiplied it by two to account for the double F- ions?
     
  14. May 20, 2013 #13

    Borek

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    Staff: Mentor

    There is no "why" here, simply apply the definition. Concentration calculation is one thing, properly written Ksp is another.
     
  15. May 20, 2013 #14
    Ksp = [Mg2+][F-]2=

    [1 x 10-3][2 x 10-3]2= 4 x 10-9


    Okay, thanks..... I understand how to write a Ksp, just not why it is written as such, but oh well....
     
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