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The Ksp of Ag2CrO4 is 1.1 × 10-12 at 298 K. The solubility (in mol/L) of Ag2CrO4 in a 0.1 M AgNO3 solution is [IIT-2013]
- a)1.1 × 10-11
- b)1.1 × 10-10
- c)1.1 × 10-12
- d)1.1 × 10-9
Correct answer is option 'B'. Can you explain this answer?
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The Ksp of Ag2CrO4 is 1.1 × 10-12 at 298 K. The solubility (in m...
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The Ksp of Ag2CrO4 is 1.1 × 10-12 at 298 K. The solubility (in m...
Understanding Ksp and Solubility
The solubility product constant (Ksp) is crucial for understanding the solubility of a compound in a solution. For Ag2CrO4, Ksp is given as 1.1 × 10^-12 at 298 K.
Equilibrium Expression
The dissolution of Ag2CrO4 can be represented as:
Ag2CrO4 (s) ⇌ 2 Ag+ (aq) + CrO4^2- (aq)
From this reaction, the Ksp expression is:
Ksp = [Ag+]^2 [CrO4^2-]
Calculating Solubility in 0.1 M AgNO3
In a 0.1 M AgNO3 solution, the concentration of Ag+ ions is already 0.1 M. Let the solubility of Ag2CrO4 in this solution be 's'.
Thus, the concentration of Ag+ from the dissolution of Ag2CrO4 will be:
[Ag+] = 0.1 + 2s ≈ 0.1 (since s is small compared to 0.1)
The concentration of CrO4^2- will simply be 's'.
Substituting into Ksp Expression
Now, substitute these concentrations into the Ksp expression:
1.1 × 10^-12 = (0.1)^2 * s
Solving for 's'
Now, we can rearrange this equation to solve for 's':
s = Ksp / (0.1)^2
s = 1.1 × 10^-12 / 0.01
s = 1.1 × 10^-10
Conclusion
The solubility of Ag2CrO4 in a 0.1 M AgNO3 solution is 1.1 × 10^-10 mol/L, confirming that the correct answer is option 'B'.
The solubility product constant (Ksp) is crucial for understanding the solubility of a compound in a solution. For Ag2CrO4, Ksp is given as 1.1 × 10^-12 at 298 K.
Equilibrium Expression
The dissolution of Ag2CrO4 can be represented as:
Ag2CrO4 (s) ⇌ 2 Ag+ (aq) + CrO4^2- (aq)
From this reaction, the Ksp expression is:
Ksp = [Ag+]^2 [CrO4^2-]
Calculating Solubility in 0.1 M AgNO3
In a 0.1 M AgNO3 solution, the concentration of Ag+ ions is already 0.1 M. Let the solubility of Ag2CrO4 in this solution be 's'.
Thus, the concentration of Ag+ from the dissolution of Ag2CrO4 will be:
[Ag+] = 0.1 + 2s ≈ 0.1 (since s is small compared to 0.1)
The concentration of CrO4^2- will simply be 's'.
Substituting into Ksp Expression
Now, substitute these concentrations into the Ksp expression:
1.1 × 10^-12 = (0.1)^2 * s
Solving for 's'
Now, we can rearrange this equation to solve for 's':
s = Ksp / (0.1)^2
s = 1.1 × 10^-12 / 0.01
s = 1.1 × 10^-10
Conclusion
The solubility of Ag2CrO4 in a 0.1 M AgNO3 solution is 1.1 × 10^-10 mol/L, confirming that the correct answer is option 'B'.
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The Ksp of Ag2CrO4 is 1.1 × 10-12 at 298 K. The solubility (in mol/L) of Ag2CrO4 in a 0.1 M AgNO3 solution is [IIT-2013] a)1.1 × 10-11b)1.1 × 10-10c)1.1 × 10-12d)1.1 × 10-9Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The Ksp of Ag2CrO4 is 1.1 × 10-12 at 298 K. The solubility (in mol/L) of Ag2CrO4 in a 0.1 M AgNO3 solution is [IIT-2013] a)1.1 × 10-11b)1.1 × 10-10c)1.1 × 10-12d)1.1 × 10-9Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The Ksp of Ag2CrO4 is 1.1 × 10-12 at 298 K. The solubility (in mol/L) of Ag2CrO4 in a 0.1 M AgNO3 solution is [IIT-2013] a)1.1 × 10-11b)1.1 × 10-10c)1.1 × 10-12d)1.1 × 10-9Correct answer is option 'B'. Can you explain this answer?.
The Ksp of Ag2CrO4 is 1.1 × 10-12 at 298 K. The solubility (in mol/L) of Ag2CrO4 in a 0.1 M AgNO3 solution is [IIT-2013] a)1.1 × 10-11b)1.1 × 10-10c)1.1 × 10-12d)1.1 × 10-9Correct answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The Ksp of Ag2CrO4 is 1.1 × 10-12 at 298 K. The solubility (in mol/L) of Ag2CrO4 in a 0.1 M AgNO3 solution is [IIT-2013] a)1.1 × 10-11b)1.1 × 10-10c)1.1 × 10-12d)1.1 × 10-9Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The Ksp of Ag2CrO4 is 1.1 × 10-12 at 298 K. The solubility (in mol/L) of Ag2CrO4 in a 0.1 M AgNO3 solution is [IIT-2013] a)1.1 × 10-11b)1.1 × 10-10c)1.1 × 10-12d)1.1 × 10-9Correct answer is option 'B'. Can you explain this answer?.
Solutions for The Ksp of Ag2CrO4 is 1.1 × 10-12 at 298 K. The solubility (in mol/L) of Ag2CrO4 in a 0.1 M AgNO3 solution is [IIT-2013] a)1.1 × 10-11b)1.1 × 10-10c)1.1 × 10-12d)1.1 × 10-9Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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