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The solubility of a sparingly soluble salt Ax By in water at 25ºC = 1.4 × 10–4 M. The solubility product is 1.1 × 10–11. The possibilities are
- a)x = 1, y = 2
- b)x = 2, y = 1
- c)x = 1, y = 3
- d)x = 3, y = 1
Correct answer is option 'A,B'. Can you explain this answer?
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The solubility of a sparingly soluble salt Ax By in water at 25ºC...
Understanding Solubility and Solubility Product
The solubility of a sparingly soluble salt AxBy is given as 1.4 × 10–4 M, and its solubility product (Ksp) is 1.1 × 10–11. We need to determine the possible values for x and y.
Determining the Dissociation Reaction
For the salt AxBy, the dissociation in water can be represented as:
AxBy ⇌ xA^(y+) + yB^(x–)
If we let the solubility (s) of AxBy be 1.4 × 10–4 M, then:
- The concentration of A^(y+) will be xs = 1.4 × 10–4 * x
- The concentration of B^(x–) will be ys = 1.4 × 10–4 * y
Calculating Ksp
The expression for the solubility product is:
Ksp = [A^(y+)]^x [B^(x–)]^y
Substituting the concentrations:
Ksp = (1.4 × 10–4 * x)^x * (1.4 × 10–4 * y)^y
Evaluating the Options
1. Option A: x = 1, y = 2
- Ksp = (1.4 × 10–4)^1 * (1.4 × 10–4)^2 = 1.4 × 10–4 * 1.96 × 10–8 = 2.744 × 10–12 (valid)
2. Option B: x = 2, y = 1
- Ksp = (1.4 × 10–4)^2 * (1.4 × 10–4)^1 = 1.96 × 10–8 * 1.4 × 10–4 = 2.744 × 10–12 (valid)
3. Option C: x = 1, y = 3
- Ksp = (1.4 × 10–4)^1 * (1.4 × 10–4)^3 = 1.4 × 10–4 * 2.744 × 10–11 (exceeds Ksp, invalid)
4. Option D: x = 3, y = 1
- Ksp = (1.4 × 10–4)^3 * (1.4 × 10–4)^1 = 2.744 × 10–11 * 1.4 × 10–4 (exceeds Ksp, invalid)
Conclusion
The valid options for the salt AxBy that satisfy the given solubility product and concentration are:
- A: x = 1, y = 2
- B: x = 2, y = 1
The solubility of a sparingly soluble salt AxBy is given as 1.4 × 10–4 M, and its solubility product (Ksp) is 1.1 × 10–11. We need to determine the possible values for x and y.
Determining the Dissociation Reaction
For the salt AxBy, the dissociation in water can be represented as:
AxBy ⇌ xA^(y+) + yB^(x–)
If we let the solubility (s) of AxBy be 1.4 × 10–4 M, then:
- The concentration of A^(y+) will be xs = 1.4 × 10–4 * x
- The concentration of B^(x–) will be ys = 1.4 × 10–4 * y
Calculating Ksp
The expression for the solubility product is:
Ksp = [A^(y+)]^x [B^(x–)]^y
Substituting the concentrations:
Ksp = (1.4 × 10–4 * x)^x * (1.4 × 10–4 * y)^y
Evaluating the Options
1. Option A: x = 1, y = 2
- Ksp = (1.4 × 10–4)^1 * (1.4 × 10–4)^2 = 1.4 × 10–4 * 1.96 × 10–8 = 2.744 × 10–12 (valid)
2. Option B: x = 2, y = 1
- Ksp = (1.4 × 10–4)^2 * (1.4 × 10–4)^1 = 1.96 × 10–8 * 1.4 × 10–4 = 2.744 × 10–12 (valid)
3. Option C: x = 1, y = 3
- Ksp = (1.4 × 10–4)^1 * (1.4 × 10–4)^3 = 1.4 × 10–4 * 2.744 × 10–11 (exceeds Ksp, invalid)
4. Option D: x = 3, y = 1
- Ksp = (1.4 × 10–4)^3 * (1.4 × 10–4)^1 = 2.744 × 10–11 * 1.4 × 10–4 (exceeds Ksp, invalid)
Conclusion
The valid options for the salt AxBy that satisfy the given solubility product and concentration are:
- A: x = 1, y = 2
- B: x = 2, y = 1
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The solubility of a sparingly soluble salt Ax By in water at 25ºC...
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The solubility of a sparingly soluble salt Ax By in water at 25ºC = 1.4 × 10–4 M. The solubility product is 1.1 × 10–11. The possibilities area)x = 1, y = 2b)x = 2, y = 1c)x = 1, y = 3 d)x = 3, y = 1Correct answer is option 'A,B'. Can you explain this answer?
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The solubility of a sparingly soluble salt Ax By in water at 25ºC = 1.4 × 10–4 M. The solubility product is 1.1 × 10–11. The possibilities area)x = 1, y = 2b)x = 2, y = 1c)x = 1, y = 3 d)x = 3, y = 1Correct answer is option 'A,B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The solubility of a sparingly soluble salt Ax By in water at 25ºC = 1.4 × 10–4 M. The solubility product is 1.1 × 10–11. The possibilities area)x = 1, y = 2b)x = 2, y = 1c)x = 1, y = 3 d)x = 3, y = 1Correct answer is option 'A,B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The solubility of a sparingly soluble salt Ax By in water at 25ºC = 1.4 × 10–4 M. The solubility product is 1.1 × 10–11. The possibilities area)x = 1, y = 2b)x = 2, y = 1c)x = 1, y = 3 d)x = 3, y = 1Correct answer is option 'A,B'. Can you explain this answer?.
The solubility of a sparingly soluble salt Ax By in water at 25ºC = 1.4 × 10–4 M. The solubility product is 1.1 × 10–11. The possibilities area)x = 1, y = 2b)x = 2, y = 1c)x = 1, y = 3 d)x = 3, y = 1Correct answer is option 'A,B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The solubility of a sparingly soluble salt Ax By in water at 25ºC = 1.4 × 10–4 M. The solubility product is 1.1 × 10–11. The possibilities area)x = 1, y = 2b)x = 2, y = 1c)x = 1, y = 3 d)x = 3, y = 1Correct answer is option 'A,B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The solubility of a sparingly soluble salt Ax By in water at 25ºC = 1.4 × 10–4 M. The solubility product is 1.1 × 10–11. The possibilities area)x = 1, y = 2b)x = 2, y = 1c)x = 1, y = 3 d)x = 3, y = 1Correct answer is option 'A,B'. Can you explain this answer?.
Solutions for The solubility of a sparingly soluble salt Ax By in water at 25ºC = 1.4 × 10–4 M. The solubility product is 1.1 × 10–11. The possibilities area)x = 1, y = 2b)x = 2, y = 1c)x = 1, y = 3 d)x = 3, y = 1Correct answer is option 'A,B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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