Given that the ka for hocl is 3.5 * 10-8, calculate the k value for the reaction of hocl with oh-.

a. 3.5 * 10-22

b. 3.5 * 10-8

c. 2.9 * 10-7

d. 3.5 * 106

e. none of these

Following reaction is involved in above system

HOCl (aq) ↔ H + (aq) + OCl - (aq)

OCl - (aq) + H2O (l) ↔ HOCl (aq) + OH - (aq)

Now, if the system is obeys 1st order kinetics we have

K = [OCl-][H + ]/[HOCl] ... (1)

∴ [HOCl-] / [OCl-] = [H+] (1 / 3.0 * 10-8) ... (2)

and now considering that system is obeying 2nd order kinetics, we have

K = [OH-][HOCl-] / [OCl-] ... (3)

Subs 2 in 3 we get

K = [OH-][H+] (1 / 3.0 * 10-8)

we know that, [OH-][H+] = 10 - 14

∴K = 3.3 * 10 - 7

Thus, correct answer is e i. e none of these