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?

 

[DeeNos]

As a scientist, you are correct in questioning the use of ratios in this problem. In this case, the ratio of 2:1 for IO3- to NaIO3 does not apply because we are not looking at the overall reaction between NaIO3 and Cu(IO3)2. We are only concerned with the solubility product (Ksp) of Cu(IO3)2, which is a measure of the equilibrium between the solid Cu(IO3)2 and its dissociated ions in solution. Therefore, the ratio of 2:1 does not factor into the calculation of the molar concentration of NaIO3.

In general, ratios are used to determine the stoichiometry of a chemical reaction, but in this case, we are not looking at a reaction between two substances. We are looking at the solubility of one substance in a solution. So, we only need to consider the concentration of the ions from the dissociation of the solute (Cu(IO3)2) in the solvent (NaIO3).

I hope this clarifies why the ratio is not used in this calculation. Keep up the good work in your studies!