1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Borek

    User Avatar

    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

    Borek

    User Avatar

    Staff: Mentor

    Moles of what?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Solubility Product, finding molar concentration
  1. Molar Concentration (Replies: 4)

  2. Molar concentration (Replies: 1)

  3. Molar concentration (Replies: 5)

  4. Molar concentrations (Replies: 3)

Loading...