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If the value of the solubility product for AgBr is 4.0 x 10-12 at 25°C, calculate the solubility of AgBr(s) in water.
- a)2.5 x 10-13
- b)2.5 x 10-6
- c)2 x 10-6
- d)5.0 x10-7
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
If the value of the solubility product for AgBr is 4.0 x 10-12at 25&de...
Ksp = 4.0 x 10-12
For AB type salt, Solubility = Ksp1/2
= (4.0 x 10-12)1/2
= 2 x 10-6
For AB type salt, Solubility = Ksp1/2
= (4.0 x 10-12)1/2
= 2 x 10-6
Most Upvoted Answer
If the value of the solubility product for AgBr is 4.0 x 10-12at 25&de...
The solubility product constant (Ksp) is a measure of the extent to which a compound dissociates into its constituent ions in a solution. In the case of AgBr, the solubility product constant (Ksp) is given as 4.0 x 10^-12 at 25°C.
The expression for the solubility product constant (Ksp) of AgBr is:
Ksp = [Ag+][Br-]
where [Ag+] is the concentration of silver ions (Ag+) in the solution and [Br-] is the concentration of bromide ions (Br-) in the solution.
Since AgBr dissociates into Ag+ and Br- ions in a 1:1 ratio, the concentration of Ag+ is equal to the concentration of Br-. Therefore, both [Ag+] and [Br-] are denoted by 'x'.
Substituting the values into the expression for Ksp:
Ksp = x * x
Ksp = x^2
Given that Ksp = 4.0 x 10^-12, we can solve for 'x':
4.0 x 10^-12 = x^2
Taking the square root of both sides:
x = √(4.0 x 10^-12)
x ≈ 2.0 x 10^-6
Therefore, the solubility of AgBr at 25°C is approximately 2.0 x 10^-6 M (molar concentration).
The expression for the solubility product constant (Ksp) of AgBr is:
Ksp = [Ag+][Br-]
where [Ag+] is the concentration of silver ions (Ag+) in the solution and [Br-] is the concentration of bromide ions (Br-) in the solution.
Since AgBr dissociates into Ag+ and Br- ions in a 1:1 ratio, the concentration of Ag+ is equal to the concentration of Br-. Therefore, both [Ag+] and [Br-] are denoted by 'x'.
Substituting the values into the expression for Ksp:
Ksp = x * x
Ksp = x^2
Given that Ksp = 4.0 x 10^-12, we can solve for 'x':
4.0 x 10^-12 = x^2
Taking the square root of both sides:
x = √(4.0 x 10^-12)
x ≈ 2.0 x 10^-6
Therefore, the solubility of AgBr at 25°C is approximately 2.0 x 10^-6 M (molar concentration).
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Community Answer
If the value of the solubility product for AgBr is 4.0 x 10-12at 25&de...
Ksp=√s
so,Ksp=√4.0×10–12
i.e,Ksp=2×10-6...
so,the answer is option C.
so,Ksp=√4.0×10–12
i.e,Ksp=2×10-6...
so,the answer is option C.
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If the value of the solubility product for AgBr is 4.0 x 10-12at 25°C, calculate the solubility of AgBr(s) in water.a)2.5 x 10-13b)2.5 x 10-6c)2 x 10-6d)5.0 x10-7Correct answer is option 'C'. Can you explain this answer?
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If the value of the solubility product for AgBr is 4.0 x 10-12at 25°C, calculate the solubility of AgBr(s) in water.a)2.5 x 10-13b)2.5 x 10-6c)2 x 10-6d)5.0 x10-7Correct answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about If the value of the solubility product for AgBr is 4.0 x 10-12at 25°C, calculate the solubility of AgBr(s) in water.a)2.5 x 10-13b)2.5 x 10-6c)2 x 10-6d)5.0 x10-7Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If the value of the solubility product for AgBr is 4.0 x 10-12at 25°C, calculate the solubility of AgBr(s) in water.a)2.5 x 10-13b)2.5 x 10-6c)2 x 10-6d)5.0 x10-7Correct answer is option 'C'. Can you explain this answer?.
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