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Iodine sparingly soluble pure water dissolve'solutions containing excess iodide ion because following reaction:

I-(aq) + I2(aq) I3-(aq)       K = 710 
calculate the equilbrium ratio of [I3-] to [I2].
5.00×10-2 mol I2  1.00 L of 5.00×10-1  M KI solution

Feb 2nd, 2015

Thank you for the opportunity to help you with your question!

Let UNITs guide you; always USE THEM in your calculation to prevent errors

Firstly, the eqn: I- (aq) + I2 (aq) -----> I3- (aq); K = 710 L/mol

K = [i3-]/([i2][i-])

5.00E-2 mol I2 +1.00 liters 5.00E-1 M KI
[KI]init = 5.00E-1 moles/liter
[I2]init = 5.00E-2 moles/liter

[KI]eq = 5.00E-1 moles/liter - x moles/liter
[I2]eq = 5.00E-2 moles/liter - x moles/liter
[I3-]eq = x moles/liter

K= 710 liters/mole = (x moles/liter) / [5.00E-1 - x moles/liter)(5.00E-2 - x moles/liter)]

SOLVE for x

[I3-]/[I2] = x moles I3- / (5.00E-2 - x moles/liter) I2 = ?? 

Basic mathematics is a prerequisite to chemistry – I just try to help you with the methodology of solving the problem.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 13th, 2015

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