Question

calculate the concentration of H3O+ ions in a mixture of 0.02 M acetic acid

Answer 1

Ka for acetic acid is 1.8 x 10-5.
Acetic acid, CH3COOH is a weak acid which ionizes partially and sodium acetate is a strong electrolyte which ionizes completely in the solution. Let x be the number of moles of acetic acid ionized.

Then, the concentrations of various species are:
CH3COOH (aq) + H2O <-----> CH3COO– (aq) + H3O+ (aq)
CH3COONa (aq) —————-> CH3COO– (aq) + Na+(aq)
Concentration of acetate ions from acetic acid = x
Since sodium acetate is completely ionized,
The concentration of acetate ion from sodium acetate = 0.1 M
Therefore, total [CH3COO–] = 0.1 + x
Concentration of acetic acid left unionized = 0.02 – x
Now, x is very small in comparison to 0.1 so that
[CH3COO–] = 0.1 + x = 0.1 mol L-1
[CH3COOH] = 0.02 – x = 0.02 mol L-1
Ionization constant, Ka is:
Ka = [H3O+] [CH3COO–]
[CH3COOH]
Ka for acetic acid is 1.8 x 10-5 (given)
1.8 x 10-5 = [H3O+] x 0.1 / 0.02
[H3O+] = 1.8 x 10-5 x 0.02 / 0.1
[H3O+] = 3.6 x 10-6 mol L-1

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Answer 2

Answer is attached

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