Initial concentration of IO4- = moles of NaIO4/final volume = 0.025 x 0.915/0.250 = 0.0915 M IO4- (aq) + 2 H2O (l) <=> H4IO6- (aq) I 0.0915 0 C -a +a E 0.0915-a a Kc = [H4IO6-]/[IO4-] = a/(0.0915 - a) = 3.5 x 10-2 [H4IO6-] = a = 3.09 x 10-3 M Solution Initial concentration of IO4- = moles of NaIO4/final volume = 0.025 x 0.915/0.250 = 0.0915 M IO4- (aq) + 2 H2O (l) <=> H4IO6- (aq) I 0.0915 0 C -a +a E 0.0915-a a Kc = [H4IO6-]/[IO4-] = a/(0.0915 - a) = 3.5 x 10-2 [H4IO6-] = a = 3.09 x 10-3 M.