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)

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Pertanyaan

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)

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Answer 1

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

Pertanyaan

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)

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