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#PbSO₄

8 public questions tagged with this topic.

Batteries - Primary Secondary and Fuel Cells

A lead storage battery delivers 1 A for 4825 s. What is the mass of PbSO₄ (molar mass 303 g/mol) formed at the cathode?

Charge = 1 × 4825 = 4825 C . Cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O , 1 mol PbSO₄ requires 2F. Faradays = (4825/96500) = 0.05 F , Moles = (0.05/2) = 0.025 mol , Mass = 0.025 × 303 = 7.575 g .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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Batteries - Primary Secondary and Fuel Cells

A lead storage battery delivers 1.5 A for 10 hours during discharge. How many grams of PbSO₄ (molar mass 303 g/mol) are

Charge = 1.5 × 10 × 3600 = 54000 C . Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻ . 1 mol PbSO₄ requires 2F. Faradays = (54000/96500) = 0.5596 F , Moles = (0.5596/2) = 0.2798 mol , Mass = 0.2798 × 303 = 84.78 g .

Ref: NCERT Class 12 Chemistry > Chapter 2: Electrochemistry > Topic: Batteries - Primary Secondary and Fuel Cells

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Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant

For PbSO₄(s) Pb²⁺(aq) + SO₄²⁻(aq) , Ksp = 1.6 × 10⁻⁸ . In a 0.02 M Na₂SO₄ solution, what is [Pb²⁺] at equilibrium?

Ksp = [Pb²⁺][SO₄²⁻] = 1.6 × 10⁻⁸ , [SO₄²⁻] ≈ 0.02 , [Pb²⁺] = (1.6 × 10⁻⁸/0.02) = 8.0 × 10⁻⁷ M .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant

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Buffer Solutions and Solubility Product and Common Ion Effect

The Ksp of PbSO₄ is 1.6 × 10⁻⁸ . In a 0.01 M Na₂SO₄ solution, what is [Pb²⁺] in a saturated solution?

For PbSO₄ Pb²⁺ + SO₄²⁻ , Ksp = [Pb²⁺][SO₄²⁻] = 1.6 × 10⁻⁸ . [SO₄²⁻] = 0.01 + [Pb²⁺ ≈ 0.01 M (since [Pb²⁺ is small). [Pb²⁺] = (1.6 × 10⁻⁸/0.01) = 1.6 × 10⁻⁶ M .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Buffer Solutions and Solubility Product and Common Ion Effect

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Laws of Chemical Combination - Conservation Mass Definite Multiple

What is the mass of PbSO₄ produced when 20.7 g of Pb(NO₃)₂ reacts with excess H₂SO₄? (Molar masses: Pb(NO₃)₂ = 331 g/mol

Reaction: Pb(NO₃)₂ + H₂SO₄ → PbSO₄ + 2HNO₃. Moles of Pb(NO₃)₂ = 20.7/331 ≈ 0.0625 mol. 1 mol Pb(NO₃)₂ produces 1 mol PbSO₄; mass = 0.0625 × 303 ≈ 18.94 g.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Laws of Chemical Combination - Conservation Mass Definite Multiple

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Dalton's Atomic Theory and Atomic and Molecular Masses

What is the mass of PbSO₄ produced when 33.1 g of Pb(NO₃)₂ reacts with excess H₂SO₄? (Molar masses: Pb(NO₃)₂ = 331 g/mol

Reaction: Pb(NO₃)₂ + H₂SO₄ → PbSO₄ + 2HNO₃. Moles of Pb(NO₃)₂ = 33.1/331 = 0.1 mol. 1 mol Pb(NO₃)₂ produces 1 mol PbSO₄; mass = 0.1 × 303 = 30.3 g.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Dalton's Atomic Theory and Atomic and Molecular Masses

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