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Consider the reaction: IO4- (aq) + 2H20 (l) <----> H4IO6- (aq) If you start with 25.0...

Consider the reaction: IO4- (aq) + 2H20 (l) <----> H4IO6- (aq) If you start with 25.0 mL of a 0.905M solution of NaIO4, and then dilute it with water to 500.0mL, what is the concentration of H4IO6- at equilbrium? Assume Kc = 3.5 x 10^-2

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Answer #1

C1V1 = C2V2

C1 = original solution concentration = 0.905 M
V1 = volume of original solution = 25 mL
C2 = concentration of diluted solution = ??
V2 = volume of diluted solution = 500 mL

C2 = C1V1 / V2 = (0.905)(25 mL) / (500 mL) = 0.04525 M = initial [IO4-]

Molarity . . . . . . . .IO4- + 2H2O <==> H4IO6-
Initial . . . . . . . . .0.04525 . . . . . . . . . . . . . .0
Change . . . . . . . . .-x . . . . . . . . . . . . . . . .x
Final . . . . . . . . 0.04525-x . . . . . . . . . . . . . x

Kc = [H4IO6-] / [IO4-] = 0.035 . . .note that H2O does not appear in the Kc expression since it is a pure liquid.

(x) / (0.04525-x) = 0.035
x = (0.035)(0.04525-x)
x = -0.035x + 0.00158
1.035x = 0.00158
x = 0.00152 = [H4IO6-] at equilibrium

So the [H4IO6-] at equilibrium = 0.00152 M

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