Consider the reaction IO4(aq) 2H2O(l)H4IO6(aq) Kc 3 5102 Ifyou start w...

Consider the reaction IOâˆ’4(aq)+2H2O(l)â‡ŒH4IOâˆ’6(aq);Kc=3.5Ã—10âˆ’2 If you start with 25.0 mL of a 0.905 M solution of...

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Consider the reaction IOâˆ’4(aq)+2H2O(l)â‡ŒH4IOâˆ’6(aq);Kc=3.5Ã—10âˆ’2 Ifyou start with 25.0 mL of a 0.905 M solution of NaIO4, and thendilute it with water to 500.0 mL, what is the concentration ofH4IOâˆ’6 at equilibrium?

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4.1 Ratings (852 Votes)

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) = 0.04525M = initial [IO4-]

IOâˆ’4(aq) + 2H2O(l) <------------------------>H4IOâˆ’6(aq)

0.04525Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â -----------------------> initial

-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â + xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â ----------------------> changed

0.04525-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â -----------------------> equilibrium

Kc = [H4IOâˆ’6]/[IOâˆ’4]

3.5Ã—10^âˆ’2 = x / 0.04525-x

x = 1.584 x 10^-3 - 3.5Ã—10^âˆ’2 x

x = 0.00153

concentration of H4IOâˆ’6 at equilibrium = x = 0.00153 M

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