Solubility Product, finding molar concentration

[Ace.]

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 \:\:\:\: excess \:\:\:\: 0 \:\:\:\:\:\:\:\:\:\:\:\:\: y
C \:\:\:\: -x\:\:\:\:\:\:\:\:\: +1x10-5 \:\:\:\: +2x10-5
E \:\:\:\:excess \:\:\:\: 1x10-5 \:\:\:\: y + 2x10-5

Ksp = [Cu²⁺][IO₃^1-]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).

 

[Borek]

isn't 2IO₃^- in 2:1 ratio with NaIO₃?

But you are calculating [IO₃^-], not twice that.

 

[Ace.]

So the only situation the 2:1 ratio applies is if I was trying to find moles?

 

[Borek]

Moles of what?