Answer:
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:
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
- % 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:
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
- % ionization ≈ (8.185 × 10⁻⁵ M / 0.010 M) × 100 ≈ 0.8185 % ≈ 0.82% (two significant figures)