Question

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.

Answer 1

Answer:

• 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) Data:

• 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) Data:

• 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)

Answer 2

The percent ionization of HA is 2.6%.

The ICE table is as follows;

      HA(aq) ⇄ H^+(aq)     +   A^-(aq)

I      0.1            0                     0

C     -x              +x                   +x

E    0.1 - x          x                     x

If the Ka = 6.7 X 10^-7

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

6.7 X 10^-7= [x] [x]/[0.1 - x]

6.7 X 10^-7([0.1 - x) = x^2

6.7  X 10^-8 - 6.7 X 10^-7x =  x^2

x^2 + 6.7 X 10^-7x - 6.7  X 10^-8 = 0

x = 0.00026 M

The percent ionization is obtained from; [A^-]/[HA] X  100/1

=  0.00026 M/ 0.010 M  X  100/1 = 2.6%

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