Question

calculate the percentage ionisation of 0.01M acetic acid in 0.1M HCl Ka of acetic acid is 1.8×10^-5​

Answer 1

Answer:

a you for posting your question on embibe. Hope your prep is going well.

Here is the answer to your question:

CH3COOH + H2O <---> CH3COO (-) + H3O (+)

CH3COONa ---> CH3COO (-) + Na (+)

Ka (CH3COOH) = [CH3COO (-)] [H3O (+)] / [CH3COOH] = x (0.1 + x) / (0.1 - x).

If c0 / Ka > 100, then x is negligible and we get:

Ka = 0.1 x / 0.1 = x. x / c0 = alpha (percent ionization),

thus:

alpha = x / c0 = Ka / c0

= 1.8 x 10^-5 / 0.1 =

1.8 x 10^-4 =

1.8 x 10^-2 % = 0.018 %