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Can you explain the answer of this question below:
If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature is
- A:(6.3 x 10⁻3) x (6.3 x 10⁻3)2
- B:(6.3 x 10⁻3) x (12.6 x10⁻3)2
- C:(6.3 x 10⁻3) x (12.6 x 10⁻3)
- D:(12.6 x 10⁻3) x (12.6 x 10⁻3)
- E:undefined
The answer is b.
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Can you explain the answer of this question below:If the solubility of...
Solubility product (Ksp) is the product of the concentrations of the ions in the equilibrium expression of a sparingly soluble salt at a particular temperature. It is given by the equation:
Ksp = [Pb2+][Cl-]
where [Pb2+] and [Cl-] are the concentrations of the ions.
Given that the solubility of PbCl at 25oC is 6.3 x 103 mole/litre, we can find the concentration of Pb2+ and Cl- ions as follows:
PbCl(s) ⇌ Pb2+(aq) + 2Cl-(aq)
Let x be the concentration of Pb2+ and 2x be the concentration of Cl-. Then, the solubility product can be expressed as:
Ksp = x(2x)2 = 4x3
Since the solubility of PbCl is 6.3 x 103 mole/litre, the concentration of Pb2+ and Cl- ions is also 6.3 x 103 mole/litre.
Substituting the values in the equation for Ksp, we get:
Ksp = 4(6.3 x 103)3 = (12.6 x 103)2
Therefore, the correct answer is option B, which is (6.3 x 103) x (12.6 x 103)2.
Ksp = [Pb2+][Cl-]
where [Pb2+] and [Cl-] are the concentrations of the ions.
Given that the solubility of PbCl at 25oC is 6.3 x 103 mole/litre, we can find the concentration of Pb2+ and Cl- ions as follows:
PbCl(s) ⇌ Pb2+(aq) + 2Cl-(aq)
Let x be the concentration of Pb2+ and 2x be the concentration of Cl-. Then, the solubility product can be expressed as:
Ksp = x(2x)2 = 4x3
Since the solubility of PbCl is 6.3 x 103 mole/litre, the concentration of Pb2+ and Cl- ions is also 6.3 x 103 mole/litre.
Substituting the values in the equation for Ksp, we get:
Ksp = 4(6.3 x 103)3 = (12.6 x 103)2
Therefore, the correct answer is option B, which is (6.3 x 103) x (12.6 x 103)2.
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Can you explain the answer of this question below:If the solubility of...
Yes....the right answer is b.....coz pbcl2 solubility product is given by pb2+solubility multiplied by (2cl-)^2 solubility.....hope i m right...
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Can you explain the answer of this question below:If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature isA:(6.3 x 10⁻3) x (6.3 x 10⁻3)2B:(6.3 x 10⁻3) x (12.6 x10⁻3)2C:(6.3 x 10⁻3) x (12.6 x 10⁻3)D:(12.6 x 10⁻3) x (12.6 x 10⁻3)E:undefinedThe answer is b.
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Can you explain the answer of this question below:If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature isA:(6.3 x 10⁻3) x (6.3 x 10⁻3)2B:(6.3 x 10⁻3) x (12.6 x10⁻3)2C:(6.3 x 10⁻3) x (12.6 x 10⁻3)D:(12.6 x 10⁻3) x (12.6 x 10⁻3)E:undefinedThe answer is b. for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Can you explain the answer of this question below:If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature isA:(6.3 x 10⁻3) x (6.3 x 10⁻3)2B:(6.3 x 10⁻3) x (12.6 x10⁻3)2C:(6.3 x 10⁻3) x (12.6 x 10⁻3)D:(12.6 x 10⁻3) x (12.6 x 10⁻3)E:undefinedThe answer is b. covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Can you explain the answer of this question below:If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature isA:(6.3 x 10⁻3) x (6.3 x 10⁻3)2B:(6.3 x 10⁻3) x (12.6 x10⁻3)2C:(6.3 x 10⁻3) x (12.6 x 10⁻3)D:(12.6 x 10⁻3) x (12.6 x 10⁻3)E:undefinedThe answer is b..
Can you explain the answer of this question below:If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature isA:(6.3 x 10⁻3) x (6.3 x 10⁻3)2B:(6.3 x 10⁻3) x (12.6 x10⁻3)2C:(6.3 x 10⁻3) x (12.6 x 10⁻3)D:(12.6 x 10⁻3) x (12.6 x 10⁻3)E:undefinedThe answer is b. for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Can you explain the answer of this question below:If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature isA:(6.3 x 10⁻3) x (6.3 x 10⁻3)2B:(6.3 x 10⁻3) x (12.6 x10⁻3)2C:(6.3 x 10⁻3) x (12.6 x 10⁻3)D:(12.6 x 10⁻3) x (12.6 x 10⁻3)E:undefinedThe answer is b. covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Can you explain the answer of this question below:If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature isA:(6.3 x 10⁻3) x (6.3 x 10⁻3)2B:(6.3 x 10⁻3) x (12.6 x10⁻3)2C:(6.3 x 10⁻3) x (12.6 x 10⁻3)D:(12.6 x 10⁻3) x (12.6 x 10⁻3)E:undefinedThe answer is b..
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Here you can find the meaning of Can you explain the answer of this question below:If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature isA:(6.3 x 10⁻3) x (6.3 x 10⁻3)2B:(6.3 x 10⁻3) x (12.6 x10⁻3)2C:(6.3 x 10⁻3) x (12.6 x 10⁻3)D:(12.6 x 10⁻3) x (12.6 x 10⁻3)E:undefinedThe answer is b. defined & explained in the simplest way possible. Besides giving the explanation of
Can you explain the answer of this question below:If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature isA:(6.3 x 10⁻3) x (6.3 x 10⁻3)2B:(6.3 x 10⁻3) x (12.6 x10⁻3)2C:(6.3 x 10⁻3) x (12.6 x 10⁻3)D:(12.6 x 10⁻3) x (12.6 x 10⁻3)E:undefinedThe answer is b., a detailed solution for Can you explain the answer of this question below:If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature isA:(6.3 x 10⁻3) x (6.3 x 10⁻3)2B:(6.3 x 10⁻3) x (12.6 x10⁻3)2C:(6.3 x 10⁻3) x (12.6 x 10⁻3)D:(12.6 x 10⁻3) x (12.6 x 10⁻3)E:undefinedThe answer is b. has been provided alongside types of Can you explain the answer of this question below:If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature isA:(6.3 x 10⁻3) x (6.3 x 10⁻3)2B:(6.3 x 10⁻3) x (12.6 x10⁻3)2C:(6.3 x 10⁻3) x (12.6 x 10⁻3)D:(12.6 x 10⁻3) x (12.6 x 10⁻3)E:undefinedThe answer is b. theory, EduRev gives you an
ample number of questions to practice Can you explain the answer of this question below:If the solubility of PbCl₂ at 25oC is 6.3 x 10⁻3 mole/litre, its solubility product at that temperature isA:(6.3 x 10⁻3) x (6.3 x 10⁻3)2B:(6.3 x 10⁻3) x (12.6 x10⁻3)2C:(6.3 x 10⁻3) x (12.6 x 10⁻3)D:(12.6 x 10⁻3) x (12.6 x 10⁻3)E:undefinedThe answer is b. tests, examples and also practice JEE tests.
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