Solubility Product, finding molar concentration

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.

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+][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).