Question

Consider the reaction IO−4(aq)+2H2O(l)⇌H4IO−6(aq);Kc=3.5×10−2
If you start with 21.0 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

The K expression

K = [H4IO6-] /[IO4-]

Total Volume: V = 21+500 = 521, recalcualte concentration

initially

[IO4-] = M1*V1/Vtotal = (0.905)*21/521 = 0.03648 M

[H4IO6-] = 0

in equilibrium, due to sotichiometry

[IO4-] = 0.03648 - x

[H4IO6-] = 0 + x

substitute in Kc

K = [H4IO6-] /[IO4-]

(3.5*10^-2) = x /(0.03648 - x)

solve for x

0.035*0.03648 -0.035x = x

x(1+0.035) = 0.0012768

x = 0.0012768/(1.035)

x= 0.0012336

substitute in H4IO6-

[H4IO6-] = 0 + x

[H4IO6-] = 0 + 0.0012336 = 0.0012336 M