Weak Acids, Weak Bases, and K_a/K_b

Understand weak acid/base equilibria, acid and base dissociation constants (K_a and K_b), and percent ionization.

Weak Acids, Weak Bases, and K_a/K_b

Weak Acids

Definition: Partially ionizes in water (< 100%)

General equilibrium:

$\ce{HA(aq) + H2O(l) <=> H3O^+(aq) + A^-(aq)}$

Simplified: HA ⇌ H⁺ + A⁻

Examples: CH₃COOH, HF, HNO₂, H₃PO₄

Acid Dissociation Constant (K_a)

Equilibrium expression:

$K_a = \frac{[H^+][A^-]}{[HA]}$

K_a magnitude:

Larger K_a = stronger acid
Typical range: 10⁻² to 10⁻¹⁴
Strong acids: K_a >> 1

Common weak acids:

| Acid | K_a | |------|-----| | HF | 6.8 × 10⁻⁴ | | HNO₂ | 4.5 × 10⁻⁴ | | CH₃COOH | 1.8 × 10⁻⁵ | | H₂CO₃ | 4.3 × 10⁻⁷ |

Weak Acid pH Calculation

Use ICE table:

Example: HA ⇌ H⁺ + A⁻

| | HA | H⁺ | A⁻ | |-|---------|-------|-------| | I | [HA]₀ | 0 | 0 | | C | -x | +x | +x | | E | [HA]₀-x | x | x |

K_a expression:

$K_a = \frac{x^2}{[HA]_0 - x}$

Solve for x = [H⁺], then pH = -log x

Small x Approximation (5% Rule)

When [HA]₀/K_a > 100:

Assume: [HA]₀ - x ≈ [HA]₀

$K_a \approx \frac{x^2}{[HA]_0}$

$x = \sqrt{K_a \cdot [HA]_0}$

Check validity:

Calculate x
If x/[HA]₀ < 5%, approximation valid
If x/[HA]₀ > 5%, use quadratic

Weak Bases

Definition: Partially ionize by accepting H⁺

General equilibrium:

$\ce{B(aq) + H2O(l) <=> BH^+(aq) + OH^-(aq)}$

Examples: NH₃, CH₃NH₂, pyridine

Base Dissociation Constant (K_b)

Equilibrium expression:

$K_b = \frac{[BH^+][OH^-]}{[B]}$

Common weak bases:

| Base | K_b | |------|-----| | NH₃ | 1.8 × 10⁻⁵ | | CH₃NH₂ | 4.4 × 10⁻⁴ | | C₅H₅N (pyridine) | 1.7 × 10⁻⁹ |

Weak Base pH Calculation

ICE table: B + H₂O ⇌ BH⁺ + OH⁻

| | B | BH⁺ | OH⁻ | |-|---------|-------|-------| | I | [B]₀ | 0 | 0 | | C | -x | +x | +x | | E | [B]₀-x | x | x |

K_b expression:

$K_b = \frac{x^2}{[B]_0 - x}$

Solve for x = [OH⁻]:

Calculate x from K_b
pOH = -log[OH⁻]
pH = 14.00 - pOH

Relationship between K_a and K_b

For conjugate acid-base pair:

$K_a \times K_b = K_w = 1.0 \times 10^{-14}$

Example: NH₃/NH₄⁺

K_b(NH₃) = 1.8 × 10⁻⁵
K_a(NH₄⁺) = K_w/K_b = 5.6 × 10⁻¹⁰

Stronger acid → weaker conjugate base Stronger base → weaker conjugate acid

Percent Ionization

Measure of acid/base strength:

$\text{% ionization} = \frac{[H^+]_{\text{eq}}}{[HA]_0} \times 100\%$

For weak acids:

Typically < 5%
Increases with dilution
Larger K_a → larger % ionization

Example: 0.10 M acetic acid

K_a = 1.8 × 10⁻⁵
[H⁺] = 1.3 × 10⁻³ M
% ionization = (1.3 × 10⁻³/0.10) × 100% = 1.3%

pK_a and pK_b

Analogous to pH:

$\text{pK}_a = -\log K_a$

$\text{pK}_b = -\log K_b$

Relationship:

$\text{pK}_a + \text{pK}_b = 14.00$

Interpretation:

Smaller pK_a = stronger acid
Smaller pK_b = stronger base

Polyprotic Acids

Multiple ionizable protons:

Example: H₂SO₃

First ionization: H₂SO₃ ⇌ H⁺ + HSO₃⁻

K_a1 (larger)

Second ionization: HSO₃⁻ ⇌ H⁺ + SO₃²⁻

K_a2 (smaller)

Pattern: K_a1 >> K_a2 >> K_a3

For pH: Usually only first ionization matters (K_a1 >> K_a2)

📚 Practice Problems

1Problem 1easy

❓ Question:

Calculate the pH of a 0.10 M solution of acetic acid (CH₃COOH). K_a = 1.8 × 10⁻⁵.

💡 Show Solution

Given:

[CH₃COOH]₀ = 0.10 M
K_a = 1.8 × 10⁻⁵

Equilibrium: CH₃COOH ⇌ H⁺ + CH₃COO⁻

Set up ICE table:

| | CH₃COOH | H⁺ | CH₃COO⁻ | |-|---------|-----|---------| | I | 0.10 | 0 | 0 | | C | -x | +x | +x | | E | 0.10-x | x | x |

Write K_a expression:

$K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{x^2}{0.10-x}$

Check for approximation:

$\frac{[HA]_0}{K_a} = \frac{0.10}{1.8 \times 10^{-5}} = 5.6 \times 10^3$

Since ratio > 100, try small x approximation:

$1.8 \times 10^{-5} = \frac{x^2}{0.10}$

$x^2 = (1.8 \times 10^{-5})(0.10)$

$x^2 = 1.8 \times 10^{-6}$

$x = 1.34 \times 10^{-3}$

Check validity:

$\frac{x}{[HA]_0} = \frac{1.34 \times 10^{-3}}{0.10} = 0.0134 = 1.34\%$

1.34% < 5% ✓ Approximation valid!

Calculate pH:

[H⁺] = x = 1.34 × 10⁻³ M

$\text{pH} = -\log(1.34 \times 10^{-3})$

$\text{pH} = 2.87$

Answer: pH = 2.87

Percent ionization:

$\% = \frac{1.34 \times 10^{-3}}{0.10} \times 100\% = 1.3\%$

Only 1.3% of acetic acid ionized (weak acid!)

2Problem 2hard

❓ Question:

Calculate the pH of a 0.10 M solution of acetic acid (CH₃COOH) given K_a = 1.8 × 10⁻⁵.

💡 Show Solution

Solution:

Equilibrium: CH₃COOH ⇌ H⁺ + CH₃COO⁻

ICE table: | | CH₃COOH | H⁺ | CH₃COO⁻ | |----------|---------|-----|---------| | Initial | 0.10 | 0 | 0 | | Change | -x | +x | +x | | Equil. | 0.10-x | x | x |

K_a expression: K_a = [H⁺][CH₃COO⁻] / [CH₃COOH] 1.8 × 10⁻⁵ = x² / (0.10 - x)

Simplification: Assume x << 0.10, so 0.10 - x ≈ 0.10 1.8 × 10⁻⁵ = x² / 0.10 x² = 1.8 × 10⁻⁶ x = 1.34 × 10⁻³ M = [H⁺]

Check assumption: 1.34 × 10⁻³ / 0.10 = 0.0134 = 1.34% < 5% ✓

Calculate pH: pH = -log(1.34 × 10⁻³) = 2.87

3Problem 3medium

❓ Question:

Calculate the pH of a 0.25 M ammonia (NH₃) solution. K_b = 1.8 × 10⁻⁵.

💡 Show Solution

Given:

[NH₃]₀ = 0.25 M
K_b = 1.8 × 10⁻⁵

Equilibrium: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

Set up ICE table:

| | NH₃ | NH₄⁺ | OH⁻ | |-|---------|-------|-------| | I | 0.25 | 0 | 0 | | C | -x | +x | +x | | E | 0.25-x | x | x |

Write K_b expression:

$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = \frac{x^2}{0.25-x}$

Check approximation:

$\frac{[B]_0}{K_b} = \frac{0.25}{1.8 \times 10^{-5}} = 1.4 \times 10^4 > 100$ ✓

Use approximation:

$1.8 \times 10^{-5} = \frac{x^2}{0.25}$

$x^2 = (1.8 \times 10^{-5})(0.25)$

$x^2 = 4.5 \times 10^{-6}$

$x = 2.12 \times 10^{-3}$

Check: x/[B]₀ = 2.12×10⁻³/0.25 = 0.85% < 5% ✓

Calculate pOH:

[OH⁻] = x = 2.12 × 10⁻³ M

$\text{pOH} = -\log(2.12 \times 10^{-3})$

$\text{pOH} = 2.67$

Calculate pH:

$\text{pH} = 14.00 - \text{pOH}$

$\text{pH} = 14.00 - 2.67$

$\text{pH} = 11.33$

Answer: pH = 11.33

Interpretation:

pH > 7: basic solution ✓
Weak base partially ionizes
Only ~0.85% ionized

4Problem 4hard

❓ Question:

Calculate the pH of a 0.50 M solution of ammonia (NH₃) given K_b = 1.8 × 10⁻⁵.

💡 Show Solution

Solution:

Equilibrium: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

ICE table: | | NH₃ | NH₄⁺ | OH⁻ | |----------|-------|------|------| | Initial | 0.50 | 0 | 0 | | Change | -x | +x | +x | | Equil. | 0.50-x| x | x |

K_b expression: K_b = [NH₄⁺][OH⁻] / [NH₃] 1.8 × 10⁻⁵ = x² / (0.50 - x)

Assume x << 0.50: 1.8 × 10⁻⁵ = x² / 0.50 x² = 9.0 × 10⁻⁶ x = 3.0 × 10⁻³ M = [OH⁻]

Check: 3.0 × 10⁻³ / 0.50 = 0.006 = 0.6% < 5% ✓

Calculate pOH: pOH = -log(3.0 × 10⁻³) = 2.52

Calculate pH: pH = 14.00 - 2.52 = 11.48

5Problem 5hard

❓ Question:

The pH of a 0.50 M solution of a weak acid (HA) is 2.68. Calculate: (a) K_a, (b) percent ionization, (c) K_b for the conjugate base A⁻.

💡 Show Solution

Given:

[HA]₀ = 0.50 M
pH = 2.68

Equilibrium: HA ⇌ H⁺ + A⁻

(a) Calculate K_a

Find [H⁺] from pH:

$[H^+] = 10^{-\text{pH}} = 10^{-2.68}$

$[H^+] = 2.09 \times 10^{-3} \text{ M}$

ICE table:

| | HA | H⁺ | A⁻ | |-|-----------|---------------|---------------| | I | 0.50 | 0 | 0 | | C | -2.09×10⁻³ | +2.09×10⁻³ | +2.09×10⁻³ | | E | 0.498 | 2.09×10⁻³ | 2.09×10⁻³ |

At equilibrium:

[HA] = 0.50 - 0.00209 = 0.498 M
[H⁺] = [A⁻] = 2.09 × 10⁻³ M

Calculate K_a:

$K_a = \frac{[H^+][A^-]}{[HA]}$

$K_a = \frac{(2.09 \times 10^{-3})(2.09 \times 10^{-3})}{0.498}$

$K_a = \frac{4.37 \times 10^{-6}}{0.498}$

$K_a = 8.8 \times 10^{-6}$

Answer (a): K_a = 8.8 × 10⁻⁶

(b) Calculate percent ionization

$\% \text{ ionization} = \frac{[H^+]_{\text{eq}}}{[HA]_0} \times 100\%$

$\% \text{ ionization} = \frac{2.09 \times 10^{-3}}{0.50} \times 100\%$

$\% \text{ ionization} = 0.418\%$

Answer (b): 0.42% ionization

(c) Calculate K_b for conjugate base A⁻

Use relationship:

$K_a \times K_b = K_w$

$K_b = \frac{K_w}{K_a}$

$K_b = \frac{1.0 \times 10^{-14}}{8.8 \times 10^{-6}}$

$K_b = 1.1 \times 10^{-9}$

Answer (c): K_b = 1.1 × 10⁻⁹

Summary:

| Property | Value | |----------|-------| | K_a (HA) | 8.8 × 10⁻⁶ | | % ionization | 0.42% | | K_b (A⁻) | 1.1 × 10⁻⁹ |

Note: HA is moderately weak acid; A⁻ is very weak base (small K_b)

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