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At 25 °C, the solubility product (Ksp) of CaF2 in water is 3.2 × 10–11. The solubility (in mole per kg of water) of the salt at the same temperature (ignore ion pairing) is
- a)4.0 × 10-6
- b)3.2 × 10-4
- c)2.5 × 10-4
- d)2.0 × 10-4
Correct answer is option 'D'. Can you explain this answer?
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Solubility Product and Solubility of CaF2 in Water
Solubility Product (Ksp)
- At 25°C, the solubility product (Ksp) of CaF2 in water is 3.2 x 10^11.
- This means that when CaF2 is dissolved in water, it dissociates into Ca2+ and F- ions until a certain concentration is reached.
- At this concentration, the product of the concentrations of Ca2+ and F- ions is equal to the Ksp value of CaF2.
Solubility of CaF2
- The solubility of CaF2 refers to the maximum amount of CaF2 that can dissolve in a given amount of water at a certain temperature.
- It is usually expressed in moles per kg of water.
- At 25°C, the solubility of CaF2 can be calculated using the Ksp value and the ion concentrations in the solution.
Calculation
- The dissociation of CaF2 in water can be represented by the equation:
CaF2 (s) ↔ Ca2+ (aq) + 2F- (aq)
- Let x be the concentration of Ca2+ and 2x be the concentration of F- in the solution.
- Then, the product of the ion concentrations is:
x x (2x)^2 = 4x^3
- This product should be equal to the Ksp value of CaF2, which is 3.2 x 10^11.
- Therefore, we can solve for x:
4x^3 = 3.2 x 10^11
x = 2 x 10^-4
- This means that the concentration of Ca2+ in the solution is 2 x 10^-4 M, and the concentration of F- ions is 4 x 10^-4 M.
- The solubility of CaF2 is equal to the concentration of CaF2 that dissolves in the solution, which is equal to the concentration of Ca2+ ions since CaF2 dissolves stoichiometrically.
- Therefore, the solubility of CaF2 is 2 x 10^-4 mole per kg of water.
Answer
- The correct answer is option D, which is 2.0 x 10^-4.
Solubility Product (Ksp)
- At 25°C, the solubility product (Ksp) of CaF2 in water is 3.2 x 10^11.
- This means that when CaF2 is dissolved in water, it dissociates into Ca2+ and F- ions until a certain concentration is reached.
- At this concentration, the product of the concentrations of Ca2+ and F- ions is equal to the Ksp value of CaF2.
Solubility of CaF2
- The solubility of CaF2 refers to the maximum amount of CaF2 that can dissolve in a given amount of water at a certain temperature.
- It is usually expressed in moles per kg of water.
- At 25°C, the solubility of CaF2 can be calculated using the Ksp value and the ion concentrations in the solution.
Calculation
- The dissociation of CaF2 in water can be represented by the equation:
CaF2 (s) ↔ Ca2+ (aq) + 2F- (aq)
- Let x be the concentration of Ca2+ and 2x be the concentration of F- in the solution.
- Then, the product of the ion concentrations is:
x x (2x)^2 = 4x^3
- This product should be equal to the Ksp value of CaF2, which is 3.2 x 10^11.
- Therefore, we can solve for x:
4x^3 = 3.2 x 10^11
x = 2 x 10^-4
- This means that the concentration of Ca2+ in the solution is 2 x 10^-4 M, and the concentration of F- ions is 4 x 10^-4 M.
- The solubility of CaF2 is equal to the concentration of CaF2 that dissolves in the solution, which is equal to the concentration of Ca2+ ions since CaF2 dissolves stoichiometrically.
- Therefore, the solubility of CaF2 is 2 x 10^-4 mole per kg of water.
Answer
- The correct answer is option D, which is 2.0 x 10^-4.
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At 25 °C, the solubility product (Ksp) of CaF2 in water is 3.2 × 10–11. The solubility (in mole per kg of water) of the salt at the same temperature (ignore ion pairing) isa)4.0 × 10-6b)3.2 × 10-4c)2.5 × 10-4d)2.0 × 10-4Correct answer is option 'D'. Can you explain this answer?
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At 25 °C, the solubility product (Ksp) of CaF2 in water is 3.2 × 10–11. The solubility (in mole per kg of water) of the salt at the same temperature (ignore ion pairing) isa)4.0 × 10-6b)3.2 × 10-4c)2.5 × 10-4d)2.0 × 10-4Correct answer is option 'D'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about At 25 °C, the solubility product (Ksp) of CaF2 in water is 3.2 × 10–11. The solubility (in mole per kg of water) of the salt at the same temperature (ignore ion pairing) isa)4.0 × 10-6b)3.2 × 10-4c)2.5 × 10-4d)2.0 × 10-4Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 25 °C, the solubility product (Ksp) of CaF2 in water is 3.2 × 10–11. The solubility (in mole per kg of water) of the salt at the same temperature (ignore ion pairing) isa)4.0 × 10-6b)3.2 × 10-4c)2.5 × 10-4d)2.0 × 10-4Correct answer is option 'D'. Can you explain this answer?.
At 25 °C, the solubility product (Ksp) of CaF2 in water is 3.2 × 10–11. The solubility (in mole per kg of water) of the salt at the same temperature (ignore ion pairing) isa)4.0 × 10-6b)3.2 × 10-4c)2.5 × 10-4d)2.0 × 10-4Correct answer is option 'D'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about At 25 °C, the solubility product (Ksp) of CaF2 in water is 3.2 × 10–11. The solubility (in mole per kg of water) of the salt at the same temperature (ignore ion pairing) isa)4.0 × 10-6b)3.2 × 10-4c)2.5 × 10-4d)2.0 × 10-4Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 25 °C, the solubility product (Ksp) of CaF2 in water is 3.2 × 10–11. The solubility (in mole per kg of water) of the salt at the same temperature (ignore ion pairing) isa)4.0 × 10-6b)3.2 × 10-4c)2.5 × 10-4d)2.0 × 10-4Correct answer is option 'D'. Can you explain this answer?.
Solutions for At 25 °C, the solubility product (Ksp) of CaF2 in water is 3.2 × 10–11. The solubility (in mole per kg of water) of the salt at the same temperature (ignore ion pairing) isa)4.0 × 10-6b)3.2 × 10-4c)2.5 × 10-4d)2.0 × 10-4Correct answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry.
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