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Iodine is sparingly soluble in pure water. However, it does `dissolve' in solutions containing excess iodide ion because of the following reaction: I-(aq) + I2(aq)= I3-(aq) K = 710 For each of the following cases calculate the equilbrium ratio of [I3-] to [I2]. 2. 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.
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