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 I-(aq) + I2(aq) I3-(aq)       K = 710

For each of the following cases calculate the equilbrium ratio of [I3-] to [I2].

5.00×10-2 mol of I2 is added to 1.00 L of 5.00×10-1 M KI solution.

2.    The solution above is diluted to 12.00 L.

Feb 1st, 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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Feb 1st, 2015
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Feb 1st, 2015
May 30th, 2017
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