Question

HClO is a weak acid (Ka = 4.0 × 10–8) and so the salt NaClO acts as a weak base. What is the pH of a solution that is 0.048 M in NaClO at 25 °C?

Answer 1

Lets see

K_b = Kw/Ka =1 x10^-14/4.0x10^-8

= 2.5 x 10^-7

Now for reaction

ClO^- + H₂O    HClO + OH^- ............at equilibrium , lets use the I.C.E.
I 0.048M 0 0

C -x +x +x
E 0.048 - x x x

Now, a equilibrium

K_b=  [HClO][OH^-]/[ClO^-] = x²/(0.048M- x)

value of x in (0.048M- x) is negligible

so above equation becomes

K_b= x²/0.048M

2.5 x 10^-7 = x²/0.048M

so

x² = 1.2 x 10^-8

x = 1.09 x 10^-4

so

p^OH= -log [OH^-]

= -log [1.09 x 10^-4] ............................................. as x = [OH^-]

= 3.96

so p^H   = 14 - 3.96

p^H = 10.04