Question

Consider the reaction IO−4(aq)+2H2O(l)⇌H4IO−6(aq);Kc=3.5×10−2 If you start with 25.0 mL of a 0.905 M solution of NaIO4, and then dilute it with water to 500.0 mL, what is the concentration of H4IO−6 at equilibrium?

Answer 1

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-]

IO−4(aq) + 2H2O(l) <------------------------>H4IO−6(aq)

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

-x                                                                 + x                ----------------------> changed

0.04525-x                                                       x                -----------------------> equilibrium

Kc = [H4IO−6]/[IO−4]

3.5×10^−2 = x / 0.04525-x

x = 1.584 x 10^-3 - 3.5×10^−2 x

x = 0.00153

concentration of H4IO−6 at equilibrium = x = 0.00153 M