For each of the following cases calculate the equilbrium ratio of [I_{3}^{-}] to [I_{2}].

2.The solution above is diluted to 12.00 L.

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.

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