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Solubility Product, finding molar concentration

  1. Jan 2, 2014 #1
    1. The problem statement, all variables and given/known data
    If the 2.0 x 10-5 mol of Cu(IO3)2 can dissolve in 2 L of NaIO3, find the molar concentration of the NaIO3 solution. Ksp = 1.4 x 10-7 for Cu(IO3)2.

    2. Relevant equations

    3. The attempt at a solution
    Let y = [IO3-(aq)] present in the solution from NaIO3

    Cu(IO3)2(s) ↔ Cu2+(aq) + 2IO3-(aq)
    I [itex]\:\:\:\:[/itex] excess [itex]\:\:\:\:[/itex] 0 [itex]\:\:\:\:\:\:\:\:\:\:\:\:\:[/itex] y
    C [itex]\:\:\:\:[/itex] -x[itex]\:\:\:\:\:\:\:\:\:[/itex] +1x10-5 [itex]\:\:\:\:[/itex] +2x10-5
    E [itex]\:\:\:\:[/itex]excess [itex]\:\:\:\:[/itex] 1x10-5 [itex]\:\:\:\:[/itex] y + 2x10-5

    Ksp = [Cu2+][IO31-]2
    1.4 * 10-7 = [1.0 * 10-5][y + 2.0 * 10-5]
    y = 0.118M

    That is the correct answer but I am wondering why you don't divide by 2, since isn't 2IO3- in 2:1 ratio with NaIO3? I'm just totally confused when to use the ratios and when not to (like in this case).
  2. jcsd
  3. Jan 2, 2014 #2


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    Staff: Mentor

    But you are calculating [IO3-], not twice that.
  4. Jan 2, 2014 #3
    So the only situation the 2:1 ratio applies is if I was trying to find moles?
  5. Jan 2, 2014 #4


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    Staff: Mentor

    Moles of what?
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