Determine the pH of an aqueous solution that is 3.86×10-2 M in NaOCl?

Answer 1

HClO <==> H+ + ClO-, (pKa) 7.497
thus: 10^(-7.497) = [H+]*[ClO-] /[HClO].....(1)
We also know that 10^(-14) = [H+]*[OH-]..(2)
(2)/(1):
10^(7.497-14) = [OH-]*[HClO]/[ClO-]

Let [HClO] = X:
...ClO- + H2O <==> OH- + HClO
0.0386-X...................X.........X...
Therefore: 10^(7.497-14) = [OH-]*[HClO]/[ClO-]
10^(-6.503) = X^2/(0.0386-X)
3.1405x10^-7*(0.0386-X)=X^2
X^2 + X*3.1405x10^-7 - 0.0386*3.1405x10^-7 = 0
Positive root X = 0.0001099 = [OH-]
Hence pH = 14 + log(base 10)(0.0001099)
= 10.0