Description
Iodine sparingly soluble pure water dissolve'solutions containing excess iodide ion because following reaction:
| I-(aq) + I2(aq) I3-(aq) K = 710 calculate the equilbrium ratio of [I3-] to [I2]. 5.00×10-2 mol I2 1.00 L of 5.00×10-1 M KI solution |
Explanation & Answer
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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 = ??
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