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)
0.04525Â 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
-x                                                               Â
+
x              Â
----------------------> changed
0.04525-x                                                     Â
x              Â
-----------------------> 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
