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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].

1.  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.


Oct 16th, 2017

K = [I3-] / [I-][I2] = 710

                        I-(aq)         + I2(aq)          I3-(aq)

Initial               5e-1M             5e-2M          0

Change             -x                   -x               +x

Equilibrium      5e-1 -x           5e-2 -x         x


710 = x / (0.025 - 0.55x +x^2)

17.75 - 390.5x + 710x^2 = x

17.75 - 391.5x + 710x^2 =0

x = 0.04984404763809308 or x =0.5015644030661324

The first value is the only solution that works for the problem (concentraitons cannot be negative)

Ask for [I3-] / [I2] = 0.0498 / (5e-2 - 0.0498) = 0.0498/0.0002 = 249


For the second problem, just calculate the diluted concentrations and follow the same procedure.


Feb 3rd, 2015

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