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Equilibrium Solubility Question

  1. May 15, 2010 #1
    1. The problem statement, all variables and given/known data
    1. A 1L solution contains 0.1M of each Pb+2,Ca+2, and Sr+2. Which ion precipitates last as Na2SO4 is slowly added with no change in volume?
    2. What is the concentration of the ion that precipitates first when the second ion precipitates?

    Given:
    Ksp PbSO4=1.8x10^-8
    Ksp CaSO4=7.1x10^-5
    Ksp SrSO4=3.4x10^-7


    2. Relevant equations



    3. The attempt at a solution
    1. Pb2+ precipitates first as it has the lowest Ksp value.
    2. This is where i get stuck. I know SrSO4 is the second ion to precipitate. I get:
    ksp=[Sr2+][SO42-]
    4x10^-7=[Sr2+][SO42-]

    Don't know where to go from here. Help is appreciated.
     
  2. jcsd
  3. May 16, 2010 #2

    Borek

    User Avatar

    Staff: Mentor

    Not sure what your problem is - so far you are right, Pb will precipitate first, Sr second. Not many possibilities left at this stage.

    Think about it this way - calculate (knowing cation concentrations) at what concentration of SO42- they will precipitate. Obviously when you add sulfate concentration of SO42- can only go up starting from zero, so when you sort cations according to concentration of sulfate at which they start to precipitate, you have your answer. Not surprisingly, as concentrations of cations are identical, list you are looking for will not differ from just the sorted list of Ksp values.

    --
    methods
     
  4. May 16, 2010 #3
    So...
    ksp=[Sr2+][SO42-]
    3.4x10^-7=[Sr2+][SO42-]
    3.4x10^-7=[0.1M][SO42-]
    [SO42-]=3.4x10^-6M

    Now in Pb2+
    1.8x10^-8=[Pb2+][SO42-]
    1.8x10^-8=[Pb2+][3.4x10^-6M]
    [Pb2+]=5.29x10^-3M

    Is this how i do it, or do i have to set up an ICE table?
     
  5. May 17, 2010 #4

    Borek

    User Avatar

    Staff: Mentor

    No, you got it wrong. Strontium sulfate starts to precipitate when sulfate concentration is as you have calculated, that's OK. At this moment nothing else can precipitate. But then, when you add more sodium sulfate, strontium is removed from the solution and concentration of free SO42- goes up - till it is high enough to precipitate next cation.

    --
    methods
     
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