The Ka of a monoprotic weak acid is 7.25 × 10-3. What is the percent ionization of a 0.167 M solution of this acid?

If the concentration is more than 1000 times greater than the Ka, then you can make an assumption that simplifies the algebra needed to solve the problem. Otherwise, you need to solve a quadratic equation.

To find the percent ionization of a weak acid, we first need to determine the concentration of the H+ ions in the solution. Since the acid is monoprotic, it dissociates into H+ and the conjugate base in solution.

Let's assume that the weak acid is represented by the general formula HA. The ionization reaction is as follows:

HA ↔ H+ + A-

The equilibrium expression for the ionization of the weak acid is given by:

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

Given that the Ka value is 7.25 × 10^-3, we can set up an equation to solve for [H+].

7.25 × 10^-3 = [H+][A-]/(0.167 - [H+])

Since the concentration of the weak acid is more than 1000 times greater than the Ka (0.167 > 1000 * 7.25 × 10^-3), we can make an assumption that simplifies the algebra needed to solve the problem. The assumption is that [H+] in the denominator can be neglected when compared to 0.167.

Therefore, we can solve the equation as follows:

7.25 × 10^-3 = [H+][A-]/0.167

[H+] = (7.25 × 10^-3) * (0.167)

[H+] ≈ 1.21 × 10^-3 M

Now that we have the concentration of H+ ions, we can calculate the percent ionization using the formula:

Percent ionization = ([H+]/[HA]) * 100

Percent ionization = (1.21 × 10^-3 / 0.167) * 100

Percent ionization ≈ 0.724%

Therefore, the percent ionization of a 0.167 M solution of this acid is approximately 0.724% (rounded to three decimal places).