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30 public questions tagged with this topic.

CHEMISTRY

The K_{sp of Hg₂Cl₂ is 1.3 × 10⁻¹⁸. What is its solubility in water?

Given: The K_{sp of Hg₂Cl₂ is 1.3 × 10⁻¹⁸. What is its solubility in water? Formula: Hg₂Cl₂(s) Hg₂²+ + 2Cl-, K_{sp = S · (2S)² = 4S³. Substitution & Calculation: 4S³ = 1.3 × 10⁻¹⁸, S³ = 3.25 × 10⁻¹⁹, S approx 6.88 × 10⁻⁷M. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

The K_{sp of FeCO₃ is 3.1 × 10⁻¹¹. What is its solubility in water?

Given: The K_{sp of FeCO₃ is 3.1 × 10⁻¹¹. What is its solubility in water? These values define the system as per NCERT data. Formula: FeCO₃(s) Fe²+ + CO₃²-, K_{sp = S². This is standard NCERT relation. Substitution & Calculation: S² = 3.1 × 10⁻¹¹, S = sqrt3.1 × 10⁻¹¹approx 5.57 × 10⁻⁶M. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

The K_{sp of NiCO₃ is 1.3 × 10⁻⁷. What is its solubility in water?

Given: The K_{sp of NiCO₃ is 1.3 × 10⁻⁷. What is its solubility in water? These values define the system as per NCERT data. Formula: NiCO₃(s) Ni²+ + CO₃²-, K_{sp = S². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: S² = 1.3 × 10⁻⁷, S = sqrt1.3 × 10⁻⁷ approx 3.61 × 10⁻⁴ M. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

The K_{sp of PbF₂ is 3.3 × 10⁻⁸. What is its solubility in 0.01 M NaF ?

Given: The K_{sp of PbF₂ is 3.3 × 10⁻⁸. What is its solubility in 0.01 M NaF ? These values define the system as per NCERT data. Formula: K_{sp = [Pb²+][F-]², [F-] = 0.01, 3.3 × 10⁻⁸= S · (0.01)². This is the standard NCERT relation for this phenomenon. Substitution & Calculation: S = 3.3 × 10⁻⁸/ 0.0001 = 3.3 × 10⁻⁴ M. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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CHEMISTRY

The K_{sp of PbI₂ is 7.1 × 10⁻⁹. What is its solubility in water?

Given: The K_{sp of PbI₂ is 7.1 × 10⁻⁹. What is its solubility in water? Formula: PbI₂(s) Pb²+ + 2I-, K_{sp = S · (2S)² = 4S³. Substitution & Calculation: 4S³ = 7.1 × 10⁻⁹, S³ = 1.775 × 10⁻⁹, S approx 1.21 × 10⁻³M. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

The K_{sp of Ag₂CO₃ is 8.1 × 10⁻¹². What is the molar solubility in water?

Given: The K_{sp of Ag₂CO₃ is 8.1 × 10⁻¹². What is the molar solubility in water? These values define the system as per NCERT data. Formula: Ag₂CO₃(s) 2Ag+ + CO₃²-, K_{sp = (2S)² · S = 4S³. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 4S³ = 8.1 × 10⁻¹², S³ = 2.025 × 10⁻¹², S approx 1.27 × 10⁻⁴ M. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Relevant Chemistry topic covering principles and examples as per NCERT.

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Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

The Ksp of Ag₂CO₃ is 8.1 × 10⁻¹² . What is [Ag+] in a saturated solution with 0.01 M Na₂CO₃ ?

For Ag₂CO₃ 2Ag+ + CO₃²⁻ , Ksp = [Ag+]²[CO₃²⁻] = 8.1 × 10⁻¹² , [CO₃²⁻] ≈ 0.01 , [Ag+]² = (8.1 × 10⁻¹²/0.01) = 8.1 × 10⁻¹⁰ , [Ag+] = 2.85 × 10⁻⁵ M .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Relationship Between Kp Kc and Factors Affecting Equilibrium - Le Chatelier

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Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

The Ksp of CuS is 6.3 × 10⁻³⁶ . What is the pH at which [Cu²⁺] = 1.0 × 10⁻¹² M in a saturated solution, given Ka of H₂S

For CuS Cu²⁺ + S²⁻ , Ksp = [Cu²⁺][S²⁻] = 6.3 × 10⁻³⁶ , [S²⁻] = 6.3 × 10⁻²⁴ . For H₂S 2H+ + S²⁻ , K = Ka₁ × Ka₂ = 9.5 × 10⁻²⁷ , [S²⁻] = (K [H₂S]/[H+]²) , assume [H₂S] = 0.1 M , 6.3 × 10⁻²⁴ = (9.5 × 10⁻²⁷ × 0.1/[H+]²) , [H+]² = 1.51 × 10⁻⁴ , [H+] = 1.23 × 10⁻² , pH ≈ 1.91 .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Physical Equilibrium - Solid-Liquid Gas-Liquid and Henry's Law

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Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases

The Ksp of Fe(OH)2 is 4.9 × 10⁻¹⁷ . What is the pH at which [Fe²⁺] = 1.0 × 10⁻⁶ M in a saturated solution?

For Fe(OH)2 Fe²⁺ + 2OH- , Ksp = [Fe²⁺][OH-]² = 4.9 × 10⁻¹⁷ , (1.0 × 10⁻⁶)[OH-]² = 4.9 × 10⁻¹⁷ , [OH-]² = 4.9 × 10⁻¹¹ , [OH-] = 7.0 × 10⁻⁶ , pOH = 5.15 , pH = 14 - 5.15 = 8.85 .

Ref: NCERT Class 11 Chemistry > Chapter 6: Equilibrium > Topic: Ionic Equilibrium - Acids Bases and pH and Ionization of Weak Acids Bases

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Homogeneous and Heterogeneous Equilibria and Applications of Equilibrium Constant

For Hg₂Cl₂(s) Hg₂²⁺(aq) + 2Cl-(aq) , Ksp = 1.3 × 10⁻¹⁸ . What is [Hg₂²⁺] in a 0.01 M NaCl solution?

Ksp = [Hg₂²⁺][Cl-]² = 1.3 × 10⁻¹⁸ , [Cl-] ≈ 0.01 , [Hg₂²⁺] (0.01)² = 1.3 × 10⁻¹⁸ , [Hg₂²⁺] = (1.3 × 10⁻¹⁸/0.0001) = 1.3 × 10⁻¹⁴ M .

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

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