Solubility Product, finding molar concentration

  • Thread starter Ace.
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  • #1
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Homework Statement


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.


Homework Equations





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).
 

Answers and Replies

  • #2
Borek
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isn't 2IO3- in 2:1 ratio with NaIO3?

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

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