Calculating pOH value

Question

What is the pOH value of a 0.01 M

2-methylpropanoicacid solution? 

(Ka= 1.6 .10–5)

Answer 1

HA = 2-methyl propionic acid

Ka = 1.6x10^-5

HA ==> H⁺ + A^-

Ka = 1.6x10^-5 = [H⁺][A^-]/[HA]

1.6x10^-5 = (x)(x)/0.01-x (assume x is small relative to 0.01 M and neglect it in denominator)

x² = 1.6x10^-7

x = [H⁺] = 4x10^-4 M --> this is small relative to 0.01 so assumption was valid

pH = -log 4x10^-4 = 3.40

pOH = 14 - 3.40 = 10.6

Answer 2

C4H8O2 + H2O → H3O+ + C4H7O2-

Ka = [H3O+][C4H7O2-] = x^2 = 1.6 x 10^–5

[C4H8O2] .01

x^2 = 1.6 x 10^-7

x = 4 x 10^-4 This is the H3O+ concentration.

-log 4 x 10^-4 = pH = 3.4

pOH = 14 - 3.4 = 10.6

Hopefully a better chemist will check this... (Made a sq root error...)