The pH of vinegar is 2.4, and the acetic acid in vinegar has a concentration of about 0.85 M. The pK_a of acetic acid is 4.75. What is the percent ionization of acetic acid in vinegar?

Question

Quizplus · Accepted Answer

To find the percent ionization, we need to use the equation for the acid dissociation constant:

Ka = [H+][A-]/[HA]

Since vinegar is primarily composed of acetic acid and water, we can assume that the initial concentration of acetic acid is equal to the concentration of the acid remaining after dissociation:

[HA]0 = 0.85 M

Let x be the concentration of H+ produced by dissociation. Then the concentration of A- is also x. Therefore, the equilibrium concentration of HA is:

[HA] = [HA]0 - x
[HA] = 0.85 M - x

Plugging these values into the Ka expression and solving for x gives:

Ka = (x^2) / (0.85 - x) = 1.8 x 10^-5
x = 0.0021 M

The percent ionization is the ratio of the concentration of H+ produced by dissociation to the initial concentration of acetic acid, multiplied by 100:

% ionization = (0.0021 / 0.85) x 100 = 0.247%

Rounding to one significant figure gives answer C: 0.47%.

The PH of Vinegar Is 2