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## Homework Statement

If the 2.0 x 10

^{-5}mol of Cu(IO

_{3})

_{2}can dissolve in 2 L of NaIO

_{3}, find the molar concentration of the NaIO

_{3}solution. Ksp = 1.4 x 10

^{-7}for Cu(IO

_{3})

_{2}.

## Homework Equations

## The Attempt at a Solution

Let y = [IO3-(aq)] present in the solution from NaIO3

Cu(IO

_{3})

_{2}(s) ↔ Cu

^{2+}(aq) + 2IO

_{3}

^{-}(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 = [Cu

^{2+}][IO

_{3}

^{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 2IO

_{3}

^{-}in 2:1 ratio with NaIO

_{3}? I'm just totally confused when to use the ratios and when not to (like in this case).