Calculate the pH of a .44M solution of NaOCl with a congjugate acid Ka of 3.5x10-8.?

Answer 1

Ka (HOCl) = 3.5 * 10^-8, so Kb (OCl-) = 2.86 * 10^-7.

Kb = [HOCl][OH-}/[OCl-]
2.86 *10^-7 = x^2/0.44-x x<<<<0.44, so it drops out of denominator

2.86*10^-7 = x^2/0.44
1.26*10^-7 = x^2
x = [OH-] = 3.55*10^-4
[H+] = 2.82*10^-11, pH = 10.55