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Analytical Chemistry - Systematic Treatment of Equilibrum

  1. Oct 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the molar solubility of Ag2C2O4 (silver oxalate) in a solution that has a fixed H3O+ concentration of 1.0 x 10^-6 M.

    2. Relevant equations



    3. The attempt at a solution

    I found the pertinent reactions, the charge balance and the mass balance so far, but I don't know what to do next.

    Pertinent Reactions
    Ag2C2O4(s) → 2Ag+(aq) + C2O4(2-)(aq) Ksp = 5.40 x 10^-12
    H2C2O4(aq) + H2O(l) → HC2O4-(aq) + H3O+(aq)
    HC2O4-(aq) + H2O(l) →C2O4(2-)(aq) + H30+(aq)

    Charge Balance
    [Ag+] + [H3O+] = 2[C2O4(2-)] + [HC2O4-]

    Mass Balance
    [Ag+] = 2[C2O4(2-)]
    [Ag+] = 2{[C2O4(2-)] + [HC2O4-] + [H2C2O4]}
     
  2. jcsd
  3. Oct 4, 2012 #2

    Borek

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    Staff: Mentor

    What reactions take place, what are equilibrium constants (other than Ksp you already listed) for them?
     
  4. Oct 4, 2012 #3
    I found that Ka1 = 5.60 x 10^-2 and ka2 =5.10 x 10^-5. Another site listed 5.4x10^-2 and 5.4x10^-5.

    EDIT: Actually, I finally found it in my text book. Ka1 = 5.62E-2 and ka2 = 5.42E-5
     
    Last edited: Oct 4, 2012
  5. Oct 4, 2012 #4

    Borek

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    Staff: Mentor

    Write formulas for both dissociation constants. Count your equations and your unknowns - are you ready to solve?
     
  6. Oct 4, 2012 #5
    Hey Borek, I wrote out all the formulas. My question now is what exactly am I solving for? Am I suppose to solve for all the concentrations? [Ag+], [C2O4(2-)], [HC2O4-] and [H2C2O4]?

    Also how am I suppose to solve for Ag+? Is it 5.4E-12 = [2x]^2 [x]?
     
  7. Oct 4, 2012 #6

    Borek

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    Staff: Mentor

    Technically you should have a set of n-equations in n-unknowns - yes, solve for all unknowns and you are done.

    Easier said than done, these are not linear equations, so you have to look for approximations or go numerical.

    It doesn't make sense to add another unknown x, you already have plenty of unknowns and this one is not going to make the problem much easier.
     
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