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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

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.

Jun 13th, 2015

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Feb 1st, 2015
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Feb 1st, 2015
Sep 21st, 2017
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