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