A certain weak acid, HA, has a Ka value of 6.7 X 10^-7.

Part A Calculate the percent ionization of HA in a 0.10 M solution. Express your answer to two significant figures and include the appropriate units.

Part B Calculate the percent ionization of HA in a 0.010 M solution.

Express your answer to two significant figures, and include the appropriate units.

Part A: 0.26%

Part B: 0.82%

Explanation:

Percent ionization is the percent of the original acid that has ionized:

%, ionization = (molar concentration of hydrogen ions at equilibrium / molar concentration of original acid) * 100

Part A:

1) dа ta:

Ka: 6.7 * 10 ⁻⁷ [HA] = 0.10 M %, ionization = ?

2) Equilibrium equation:

HA ⇄ H⁺ + A⁻

3) ICE (initial, change, equilbirium) table

Concentrations

HA H⁺ A⁻

Initial 0.10 0 0

Change - x + x + x

Equilibrium 0.10 - x x x

Equation: Ka = [H⁺] [A⁻] / [HA] =

6.7 * 10 ⁻⁷ = x² / (0.10 - x)

4) Solve the equation:

Since Ka << 1, you can assume x << 0.10 and 0.10 - x ≈ 0.10

6.7 * 10 ⁻⁷ ≈ x² / 0.10 ⇒ x² ≈ 6.7 * 10⁻⁸ ⇒ x ≈ 2.588 * 10⁻⁴

[H⁺] ≈ 2.588 * 10⁻⁴ M

% ionization ≈ (2.588 * 10⁻⁴ M / 0.1 M) * 100 ≈ 0.2588 % ≈ 0.26% (two significant figures)

Part B:

1) dа ta:

Ka: 6.7 * 10 ⁻⁷ [HA] = 0.010 M %, ionization = ?

2) Equilibrium equation:

HA ⇄ H⁺ + A⁻

3) ICE table:

Concentrations

HA H⁺ A⁻

Initial 0.010 0 0

Change - x + x + x

Equilibrium 0.010 - x x x

Equation: Ka = [H⁺] [A⁻] / [HA] =

6.7 * 10 ⁻⁷ = x² / (0.010 - x)

4) Solve the equation:

Since Ka << 1, you can assume x << 0.010 and 0.010 - x ≈ 0.010

6.7 * 10 ⁻⁷ ≈ x² / 0.010 ⇒ x² ≈ 6.7 * 10⁻⁹ ⇒ x ≈ 8.185 * 10⁻5

[H⁺] ≈ 8.185 * 10⁻⁵ M

% ionization ≈ (8.185 * 10⁻⁵ M / 0.010 M) * 100 ≈ 0.8185 % ≈ 0.82% (two significant figures)