Question

In: Chemistry

What is the pH of a 0.250 M NaOCl solution? Kb = 3.4 x 10-7

What is the pH of a 0.250 M NaOCl solution? Kb = 3.4 x 10-7

Solutions

Expert Solution

construct the ICE table

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

I    0.25                                             0                  0

C    -x                                                +x               +x

E 0.25-x                                             +x              +x

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

3.4 x 10-7 = [x] [x] / [0.25-x]

x2 + x 3.4 * 10-7 - 8.5 x 10-8 = 0

solve the quadratic equation

x = 2.91 x 10-4 M = [NaOH]

pOH = -log(OH-) = -log(2.91 x 10-4) = 3.53

pH + pOH = 14

pH = 14-pOH = 14-3.53 = 10.46M


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