Acid and Base Worksheet (weak and strong) featuring a table for calculating ion concentrations and pH/pOH values, along with five problems on ionization and dissociation constants.

Let’s solve each problem step by step.

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Problem 1: Fill out the table

We need to find [H⁺], [OH⁻], pH, and pOH for each solution.

Remember:
- For strong acids (like HCl), they fully dissociate → [H⁺] = concentration of acid.
- For strong bases (like NaOH), they fully dissociate → [OH⁻] = concentration of base.
- For weak acids/bases, we need Ka or Kb values. Since the worksheet says “see the chart with dissociation constants,” but no chart is provided here, I’ll use standard known values:

Standard Ka/Kb values (at 25°C):
- Benzoic acid (C₆H₅COOH): Ka = 6.3 × 10⁻⁵
- Methylamine (CH₃NH₂): Kb = 4.4 × 10⁻⁴
- Propionic acid (CH₃CH₂COOH): Ka = 1.3 × 10⁻⁵
- Ammonia (NH₃): Kb = 1.8 × 10⁻⁵

Also remember:
- pH = -log[H⁺]
- pOH = -log[OH⁻]
- pH + pOH = 14
- [H⁺][OH] = 1.0 × 10⁻¹⁴

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Row 1: .010 M HCl (strong acid)
→ Fully dissociates → [H⁺] = 0.010 M
→ [OH⁻] = 1.0×10⁻¹⁴ / 0.010 = 1.0×10⁻¹² M
→ pH = -log(0.010) = 2.00
→ pOH = 14 - 2.00 = 12.00

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Row 2: .0010 M NaOH (strong base)
→ Fully dissociates → [OH⁻] = 0.0010 M
→ [H⁺] = 1.0×10⁻¹⁴ / 0.0010 = 1.0×10⁻¹¹ M
→ pOH = -log(0.0010) = 3.00
→ pH = 14 - 3.00 = 11.00

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Row 3: .25 M benzoic acid (weak acid, Ka = 6.3×10⁻⁵)
Use approximation for weak acid:
HA ⇌ H⁺ + A⁻
Ka = x² / (initial - x) ≈ x² / initial (if x << initial)

x = [H⁺] = (Ka × C) = (6.3×10⁻⁵ × 0.25) = √(1.575×10⁻⁵) ≈ 0.00397 M

Check: 0.00397 / 0.25 = 1.6% < 5% → OK to approximate.

→ [H⁺] ≈ 0.00397 M
→ [OH⁻] = 1.0×10⁻¹⁴ / 0.00397 ≈ 2.52×10⁻¹² M
→ pH = -log(0.00397) ≈ 2.40
→ pOH = 14 - 2.40 = 11.60

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Row 4: .50 M methylamine (weak base, Kb = 4.4×10⁻⁴)
B + H₂O ⇌ BH⁺ + OH⁻
Kb = x² / (initial - x) ≈ x² / initial

x = [OH⁻] = √(Kb × C) = √(4.4×10⁻⁴ × 0.50) = √(2.2×10⁻⁴) ≈ 0.0148 M

Check: 0.0148 / 0.50 = 2.96% < 5% → OK

→ [OH⁻] ≈ 0.0148 M
→ [H⁺] = 1.0×10⁻¹⁴ / 0.0148 ≈ 6.76×10⁻¹³ M
→ pOH = -log(0.0148) ≈ 1.83
→ pH = 14 - 1.83 = 12.17

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Row 5: .12 M propionic acid (weak acid, Ka = 1.3×10⁻⁵)
x = [H⁺] = √(Ka × C) = √(1.3×10⁻⁵ × 0.12) = √(1.56×10⁻⁶) ≈ 0.00125 M

Check: 0.00125 / 0.12 = 1.04% < 5% → OK

→ [H⁺] ≈ 0.00125 M
→ [OH⁻] = 1.0×10⁻¹⁴ / 0.00125 = 8.0×10⁻¹² M
→ pH = -log(0.00125) ≈ 2.90
→ pOH = 14 - 2.90 = 11.10

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Row 6: .085 M ammonia (weak base, Kb = 1.8×10⁻⁵)
x = [OH⁻] = (Kb × C) = √(1.8×10⁻⁵ × 0.085) = √(1.53×10⁻⁶) ≈ 0.00124 M

Check: 0.00124 / 0.085 = 1.46% < 5% → OK

→ [OH⁻] ≈ 0.00124 M
→ [H⁺] = 1.0×10⁻¹⁴ / 0.00124 ≈ 8.06×10⁻¹² M
→ pOH = -log(0.00124) ≈ 2.91
→ pH = 14 - 2.91 = 11.09

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Now let’s fill in the table:

| Acid or base | [H⁺] | [OH⁻] | pH | pOH |
|----------------------|----------------|----------------|------|------|
| .010 M HCl | 0.010 | 1.0×10⁻¹² | 2.00 | 12.00|
| .0010 M NaOH | 1.0×10⁻¹¹ | 0.0010 | 11.00| 3.00 |
| .25 M benzoic acid | 0.00397 | 2.52×10⁻¹² | 2.40 | 11.60|
| .50 M methylamine | 6.76×10⁻¹³ | 0.0148 | 12.17| 1.83 |
| .12 M propionic acid | 0.00125 | 8.0×10⁻¹² | 2.90 | 11.10|
| .085 M ammonia | 8.06×10⁻¹² | 0.00124 | 11.09| 2.91 |

*(Note: Values rounded to 2–3 significant figures as appropriate.)*

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Problem 2: A .20 M solution of a weak acid is 5.0% ionized at equilibrium. What is this acid’s Ka?

Percent ionization = ([H⁺] at eq / initial [HA]) × 100%

So, [H⁺] = 5.0% of 0.20 M = 0.050 × 0.20 = 0.010 M

For weak acid HA ⇌ H⁺ + A⁻
At equilibrium:
[H⁺] = [A⁻] = 0.010 M
[HA] = 0.20 - 0.010 = 0.19 M

Ka = [H⁺][A⁻] / [HA] = (0.010)(0.010) / 0.19 = 0.0001 / 0.19 ≈ 5.26×10⁻⁴

Answer: Ka ≈ 5.3 × 10⁻⁴

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Problem 3: A .15 M solution of a weak base is 2.9% ionized at equilibrium. What is this base’s Kb?

Percent ionization = ([OH⁻] at eq / initial [B]) × 100%

[OH⁻] = 2.9% of 0.15 = 0.029 × 0.15 = 0.00435 M

For weak base B + H₂O ⇌ BH⁺ + OH⁻
At equilibrium:
[OH⁻] = [BH⁺] = 0.00435 M
[B] = 0.15 - 0.00435 = 0.14565 M

Kb = [BH⁺][OH⁻] / [B] = (0.00435)(0.00435) / 0.14565 ≈ 0.0000189225 / 0.14565 ≈ 1.30×10⁻⁴

Answer: Kb ≈ 1.3 × 10⁻⁴

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Problem 4: A .10 M solution of acid has a pH of 2.5. What is its Ka?

pH = 2.5 → [H⁺] = 10^(-2.5) = 0.003162 M

Assume it’s a weak acid: HA ⇌ H⁺ + A⁻
At equilibrium:
[H⁺] = [A⁻] = 0.003162 M
[HA] = 0.10 - 0.003162 ≈ 0.0968 M

Ka = (0.003162)² / 0.0968 ≈ 0.00001000 / 0.0968 ≈ 1.03×10⁻⁴

Answer: Ka ≈ 1.0 × 10⁻⁴

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Problem 5: A .20 M solution of acid has a pH of 4.2. What is its Ka?

pH = 4.2 → [H⁺] = 10^(-4.2) ≈ 6.31×10⁻⁵ M

Weak acid: HA ⇌ H⁺ + A⁻
[H⁺] = [A⁻] = 6.31×10⁻⁵ M
[HA] = 0.20 - 6.31×10⁻⁵ ≈ 0.20 M (since very small change)

Ka = (6.31×10⁻⁵)² / 0.20 ≈ 3.98×10⁻⁹ / 0.20 ≈ 1.99×10⁻⁸

Answer: Ka ≈ 2.0 × 10⁻⁸

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Final Answer:

Table filled as above.

2) Ka = 5.3 × 10⁻⁴
3) Kb = 1.3 × 10⁻⁴
4) Ka = 1.0 × 10⁻⁴
5) Ka = 2.0 × 10⁻⁸