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#solubility

39 public questions tagged with this topic.

CHEMISTRY

The solubility product of Bi₂S₃ is 1.0 × 10⁻⁹⁷. What is its solubility in water?

Given: The solubility product of Bi₂S₃ is 1.0 × 10⁻⁹⁷. What is its solubility in water? Formula: Bi₂S₃(s) 2Bi³+ + 3S²-, K_{sp = (2S)² · (3S)³ = 108S⁵. Substitution & Calculation: 108S⁵ = 1.0 × 10⁻⁹⁷, S⁵ = 9.26 × 10⁻¹⁰⁰, S approx 1.32 × 10⁻²⁰M. 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 BaCrO₄ is 1.2 × 10⁻¹⁰. What is its solubility in 0.01 M Na₂CrO₄ ?

Given: The K_{sp of BaCrO₄ is 1.2 × 10⁻¹⁰. What is its solubility in 0.01 M Na₂CrO₄ ? These values define the system as per NCERT data. Formula: K_{sp = [Ba²+][CrO₄²-], [CrO₄²-] = 0.01, 1.2 × 10⁻¹⁰= S · 0.01. This is standard NCERT relation. Substitution & Calculation: S = 1.2 × 10⁻⁸M. 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 Fe(OH)2 is 4.9 × 10⁻¹⁷. What is its solubility in water?

Given: The K_{sp of Fe(OH)2 is 4.9 × 10⁻¹⁷. What is its solubility in water? These values define the system as per NCERT data. Formula: Fe(OH)2(s) Fe²+ + 2OH-, K_{sp = S · (2S)² = 4S³. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 4S³ = 4.9 × 10⁻¹⁷, S³ = 1.225 × 10⁻¹⁷, S approx 2.3 × 10⁻⁶ 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

For Ca(OH)2(s) Ca²+ + 2OH-, K_{sp = 5.5 × 10⁻⁶. What is [Ca²+] in 0.01 M NaOH ?

Given: For Ca(OH)2(s) Ca²+ + 2OH-, K_{sp = 5.5 × 10⁻⁶. What is [Ca²+] in 0.01 M NaOH ? These values define the system as per NCERT data. Formula: K_{sp = [Ca²+][OH-]², [OH-] = 0.01 + 2S approx 0.01. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: 5.5 × 10⁻⁶= [Ca²+] · (0.01)², [Ca²+] = 5.5 × 10⁻² 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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