Question

0.1 mole of acetic acid has been dissolved per dm3 of the solution , percentage ionization of acetic acid will be

Answer 1

Explanation:

Acetic acid or ethanoic acid is an organic weak acid with Ka = 1.8 x 10^-5.

Ka = 1.8 x 10^-5

pKa = -log Ka

pKa = -log 1.8 x 10^-5

pKa = -(log 1.8 + log 10^-5)

pKa = 4.74

In water, acetic acid undergoes ionization to form H+ and CH3COO- ions.

CH3COOH <===> H+ + CH3COO-

The degree of ionization of the acid to form equimolar of H+ and CH3COO- is very small. In this respect, let [H+] = [CH3COO-] = A mole. Since CH3COOH is a very weak acid, then the concentration of CH3COOH after ionization is practically the same with its original concentration i.e. 0.1 M →[HA] = 0.1 M.

To determine the value of A, use the following formula :

Ka = {[H+] x [CH3COO-]}/[CH3COOH]}Solve

Ka = [A] x [A] /[CH3COOH]

1.8 x 10^-5 = A^2/0.1

A^2 = 1.8 x 10^-5 x 0.1

A^2 = 1.8 x 10^-6

Solve…. giving

[A] = 1.34 x 10^-3 M

Then, the percentage of ionization of acetic acid can be calculated using :

= {[A]/[HA]} x 100 %

= {[1.34 x 10^-3]/[0.1]} x 100 %

= [1.34 x 10^-4] x 100 %

= 1.34 x 10^-2 % or 0.0134%.....

Answer 2

Answer:

CH3COOH dissociates to a small extent:

CH3COOH ↔ CH3COO- + H+

Calculate [H+] from the Ka equation.

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

Because [H+] =[CH3COO-] and dissociation is very small, we can write the equation as

Ka = [H+]² / 0.1

Ka = 1.8*10^-5

1.8*10^-5 = [H+]² / 0.1

[H+]² = (1.8*10^-5)* 0.1

[H+]² = 1.8*10^-6

[H+] = 0.0013M

% dissociation = 0.0013/0.1*100 = 1.34%

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