First, we need to write the balanced chemical equation for the dissociation of acetic acid (CH3COOH):

CH3COOH + H2O ⇌ CH3COO- + H3O+

The Ka expression for this reaction is:

Ka = [CH3COO-][H3O+] / [CH3COOH]

We can assume that initially, before adding the CH3COONa, the acetic acid solution is a weak acid solution and can be approximated by the concentration of undissociated acid [CH3COOH] since the dissociation is small.

So, initially, [CH3COOH] = 0.01 M and [CH3COO-] and [H3O+] are negligible.

When we add 0.01 mol of CH3COONa, it will dissociate completely into CH3COO- and Na+.

So, the final concentration of CH3COO- will be:

[CH3COO-] = 0.01 mol / 1 L = 0.01 M

The Na+ ion will not affect the pH because it is the conjugate base of a strong acid (NaOH).

Now we need to calculate the new concentration of H3O+ ions. Since CH3COO- is the conjugate base of acetic acid, it will react with water to form some H3O+ ions and some undissociated CH3COOH:

CH3COO- + H2O ⇌ CH3COOH + OH-

The equilibrium constant for this reaction is:

Kb = [CH3COOH][OH-] / [CH3COO-]

Since Kb = Kw / Ka, where Kw is the ion product constant for water (1.0 × 10^-14), we can find Kb:

Kb = Kw / Ka = 1.0 × 10^-14 / 1.8 = 5.56 × 10^-15

At equilibrium, [OH-] = [H3O+] because water is neutral. So, we can write:

Kb = [CH3COOH][H3O+] / [CH3COO-]

[CH3COOH] = [CH3COOH]initial - [CH3COO-] = 0.01 M - 0.01 M = 0 M (since the dissociation is small)

[H3O+] = Kb * [CH3COO-] / [CH3COOH] = 5.56 × 10^-15 * 0.01 M / 0 M = infinity

This means that all of the CH3COO- ions reacted with water to form H3O+ ions and undissociated CH3COOH, resulting in an infinite concentration of H3O+ ions.

Therefore, the new pH can be approximated as:

pH = -log[H3O+] = -log(infinity) = 0

So, the pH change is:

pHfinal - pHinitial = 0 - (-log(0.01)) = 2

Therefore, the pH increases by 2 when 0.01 mol CH3COONa solution is added to one litre of 0.01 CH3COOH solution.