Solubility Product, finding molar concentration

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Discussion Overview

The discussion revolves around a homework problem involving the solubility product (Ksp) of Cu(IO3)2 and the calculation of molar concentration of NaIO3 in solution. Participants explore the relationships between the ions in solution and the application of stoichiometric ratios in their calculations.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant presents a calculation for the molar concentration of NaIO3 based on the dissolution of Cu(IO3)2 and the Ksp value.
  • Another participant questions the application of the stoichiometric ratio, specifically whether the 2:1 ratio between IO3- and NaIO3 should affect the calculation of [IO3-].
  • A third participant seeks clarification on when the 2:1 ratio is relevant, suggesting it may only apply to moles rather than concentrations.
  • There is an ongoing confusion regarding the correct interpretation of stoichiometric ratios in the context of the problem.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the application of stoichiometric ratios in their calculations, indicating that multiple competing views remain on how to approach the problem.

Contextual Notes

There is a lack of consensus on when to apply the stoichiometric ratios in calculations, particularly in transitioning from moles to concentrations. This may depend on the specific context of the problem and the definitions used.

Who May Find This Useful

Students working on solubility product problems, particularly those involving stoichiometry and concentration calculations in chemistry.

Ace.
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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 \:\:\:\: 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).
 
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Ace. said:
isn't 2IO3- in 2:1 ratio with NaIO3?

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

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