Question

Determine the percentage of ionisation of acetic acid at.
1. pH 3.76
2. pH 5.76

Answer 1

To calculate the percentage of ionization of acetic acid at a given pH, we need to first determine the concentration of acetic acid and its conjugate base (acetate) at that pH.

1. pH 3.76:
At pH 3.76, the concentration of H+ ions is 10^(-3.76) = 1.71 x 10^(-4) M.
Since acetic acid is a weak acid with a dissociation constant (Ka) of 1.8 x 10^(-5), we can assume that the concentration of H+ ions comes from the dissociation of acetic acid:
CH3COOH (acetic acid) --> H+ + CH3COO-

Let x be the concentration of H+ ions and acetate ions at equilibrium. Then, the equilibrium concentrations of acetic acid and acetate ions are CH3COOH = C - x and CH3COO- = x, where C is the initial concentration of acetic acid.

Using the equilibrium expression for acetic acid:
Ka = [H+][CH3COO-] / [CH3COOH]
1.8 x 10^(-5) = (x)(x) / (C - x)

Since x is much smaller than C, we can assume that C - x is approximately equal to C. Therefore:
1.8 x 10^(-5) = x^2 / C
x^2 = 1.8 x 10^(-5) * C
x = sqrt(1.8 x 10^(-5) * C)

Substitute the values and solving for x, we get:
1.71 x 10^(-4) = sqrt(1.8 x 10^(-5) * C)
C = (1.71 x 10^(-4))^2 / 1.8 x 10^(-5) = 1.62 x 10^(-3) M

Now, the percentage of ionization can be calculated using the formula:
% ionization = [CH3COO-] / [CH3COOH] * 100
% ionization = x / C * 100
% ionization = 1.71 x 10^(-4) / 1.62 x 10^(-3) * 100
% ionization ≈ 10.56%

2. pH 5.76:
Similarly, at pH 5.76, the concentration of H+ ions is 10^(-5.76) = 1.58 x 10^(-6) M.

Using the same approach as above, we can calculate the percentage of ionization:
% ionization = [CH3COO-] / [CH3COOH] * 100
% ionization = x / C * 100
% ionization = 1.58 x 10^(-6) / C * 100
% ionization ≈ 0.10%