A 0.010 M aqueous solution of a weak acid HA has a pH of 4.0.What is the degree of ionization of HA in the solution?

Question

Quizplus · Accepted Answer

From the pH value of 4.0, we can calculate the concentration of H+ ions in the solution as:

pH = -log [H+]
4.0 = -log [H+]
[H+] = 10^-4 M

Since HA is a weak acid, it undergoes partial ionization in the solution:

HA + H2O ⇌ H3O+ + A-

Let's assume that the degree of ionization of HA is α. Then, at equilibrium:

[H3O+] = α[C]initial
[A-] = α[C]initial

where [C]initial is the initial concentration of HA.

The acid dissociation constant (Ka) of HA is defined by:

Ka = [H3O+][A-]/[HA]

Substituting the equilibrium concentrations, we get:

Ka = (α[C]initial)^2 / [(1-α)[C]initial]

Simplifying, we get:

α^2 Ka = α[C]initial
α Ka = [H3O+]

Substituting the given values, we get:

α Ka = 10^-4 M

Now, we need to solve for α. We can assume that α << 1, i.e., the degree of ionization is very small. In this case, we can neglect α in the denominator of the equation and solve for α:

α Ka = [H3O+]
α x 10^-5.7 = 10^-4
α = 0.011

Therefore, the degree of ionization of HA is 1.1%, which is closest to choice D (1%).

A 0010 M Aqueous Solution of a Weak Acid HA Has