Question

Calculate the hydroxide ion concentration in a 0.0500 M NaOCl solution. The acid dissociation constant for HOCl is 3.0 x 10^-8.

Answer 1

first from the Ka value find the Kb

Ka x Kb = 1.0 x 10^-14

Kb = 1.0 x 10^-14 / Ka = 1.0 x 10^-14 / 3.0 x 10^-8   = 3.3 x 10^-7

now construct ICE table

     NaOCl (aq) + H2O (l) <----> HOCl (aq) + NaOH (aq)

I    0.05                                                0                    0

C   -x                                                     +x                 +x

E 0.05 - x                                               +x                +x

Kb = [HOCl] [NaOH] / [NaOCl]

3.3 x 10^-7 = [x] [x] / [0.05-x]

x^2 + x 3.3 * 10^-7 - 1.65 * 10^-8 = 0

solve the quadratic rquation

x = 0.0001283

from ICE table equilibrium concentration of NaOH = x = 0.0001283

[OH-] = 0.0001283 or 1.28 x 10^-3 M