Determine the pH of the following solutions: a. 10 mL of 0.1molNH_(4)OH(Kb=10^-5) b. 0.01 M acetic acid (Ka=4times 10^-4) c. 0.1 M H_(2)SO_(4) d. 100 mL of 0.002 M Ca(OH)_(2)

Here's how to determine the pH of each solution:**a. 10 mL of 0.1 mol NH₄OH (Kb = 10⁻⁵)*** **Understanding the Solution:** Ammonium hydroxide (NH₄OH) is a weak base. We need to use the Kb value to calculate the hydroxide ion concentration ([OH⁻]) and then use that to find the pH.* **Setting up the Equilibrium:** NH₄OH(aq) ⇌ NH₄⁺(aq) + OH⁻(aq) Initial: 0.1 M 0 0 Change: -x +x +x Equilibrium: 0.1-x x x* **Using the Kb Expression:** Kb = [NH₄⁺][OH⁻] / [NH₄OH] = 10⁻⁵ 10⁻⁵ = x² / (0.1 - x) Since Kb is small, we can approximate (0.1 - x) ≈ 0.1 10⁻⁵ = x² / 0.1 x² = 10⁻⁶ x = [OH⁻] = 10⁻³ M* **Calculating pOH and pH:** pOH = -log[OH⁻] = -log(10⁻³) = 3 pH + pOH = 14 pH = 14 - 3 = **11****b. 0.01 M acetic acid (Ka = 4 × 10⁻⁴)*** **Understanding the Solution:** Acetic acid (CH₃COOH) is a weak acid. We'll use the Ka value to find the hydrogen ion concentration ([H⁺]) and then calculate the pH.* **Setting up the Equilibrium:** CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq) Initial: 0.01 M 0 0 Change: -x +x +x Equilibrium: 0.01-x x x* **Using the Ka Expression:** Ka = [CH₃COO⁻][H⁺] / [CH₃COOH] = 4 × 10⁻⁴ 4 × 10⁻⁴ = x² / (0.01 - x) Since Ka is small, we can approximate (0.01 - x) ≈ 0.01 4 × 10⁻⁴ = x² / 0.01 x² = 4 × 10⁻⁶ x = [H⁺] = 2 × 10⁻³ M* **Calculating pH:** pH = -log[H⁺] = -log(2 × 10⁻³) = **2.7****c. 0.1 M H₂SO₄*** **Understanding the Solution:** Sulfuric acid (H₂SO₄) is a strong acid. It completely ionizes in solution.* **Ionization:** H₂SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq) Since H₂SO₄ is a strong acid, [H⁺] = 2 × 0.1 M = 0.2 M* **Calculating pH:** pH = -log[H⁺] = -log(0.2) = **0.7****d. 100 mL of 0.002 M Ca(OH)₂*** **Understanding the Solution:** Calcium hydroxide (Ca(OH)₂) is a strong base. It completely dissociates in solution.* **Dissociation:** Ca(OH)₂(aq) → Ca²⁺(aq) + 2OH⁻(aq) Since Ca(OH)₂ is a strong base, [OH⁻] = 2 × 0.002 M = 0.004 M* **Calculating pOH and pH:** pOH = -log[OH⁻] = -log(0.004) = 2.4 pH + pOH = 14 pH = 14 - 2.4 = **11.6**