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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 x10⁻3)2
  • b)
    (6.3 x 10⁻3) x (12.6 x 10⁻3)2
  • c)
    (6.3 x 10⁻3) x (12.6 x 10⁻3)
  • d)
    (12.6 x 10⁻3) x (12.6 x10⁻3)
Correct answer is option 'B'. Can you explain this answer?
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If the solubility of PbCl at 25oC is 6.3 x 103 mole/litre, its solubil...
Calculation of Solubility Product of PbCl2
The solubility product (Ksp) of a salt is the product of the concentrations of its ions in a saturated solution. In this case, the solubility of PbCl2 is given as 6.3 x 10^3 mole/litre.

Formula for Solubility Product (Ksp)
The solubility product (Ksp) of a salt AB is given by the equation:
Ksp = [A+]^m[B-]^n

Calculation for PbCl2
For PbCl2, the dissociation equation is:
PbCl2 (s) --> Pb2+ (aq) + 2Cl- (aq)
Given that the solubility of PbCl2 is 6.3 x 10^3 mole/litre, the concentration of Pb2+ ions is 6.3 x 10^3 mole/litre and the concentration of Cl- ions is twice that, i.e. 2 x 6.3 x 10^3 = 12.6 x 10^3 mole/litre.

Substitute Values into the Ksp Equation
Substitute the concentrations of Pb2+ and Cl- ions into the Ksp equation:
Ksp = [Pb2+][Cl-]^2
Ksp = (6.3 x 10^3)(12.6 x 10^3)^2
Ksp = (6.3 x 10^3)(12.6 x 10^3)(12.6 x 10^3)
Ksp = (6.3 x 10^3)(12.6 x 10^3)^2
Therefore, the correct option is (b) which represents the correct calculation for the solubility product of PbCl2 at 25oC.
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If the solubility of PbCl at 25oC is 6.3 x 103 mole/litre, its solubility product at that temperature isa)(6.3 x 103) x (6.3 x103)2b)(6.3 x 103) x (12.6 x 103)2c)(6.3 x 103) x (12.6 x 103)d)(12.6 x 103) x (12.6 x103)Correct answer is option 'B'. Can you explain this answer?
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If the solubility of PbCl at 25oC is 6.3 x 103 mole/litre, its solubility product at that temperature isa)(6.3 x 103) x (6.3 x103)2b)(6.3 x 103) x (12.6 x 103)2c)(6.3 x 103) x (12.6 x 103)d)(12.6 x 103) x (12.6 x103)Correct answer is option '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 If the solubility of PbCl at 25oC is 6.3 x 103 mole/litre, its solubility product at that temperature isa)(6.3 x 103) x (6.3 x103)2b)(6.3 x 103) x (12.6 x 103)2c)(6.3 x 103) x (12.6 x 103)d)(12.6 x 103) x (12.6 x103)Correct answer is option '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 If the solubility of PbCl at 25oC is 6.3 x 103 mole/litre, its solubility product at that temperature isa)(6.3 x 103) x (6.3 x103)2b)(6.3 x 103) x (12.6 x 103)2c)(6.3 x 103) x (12.6 x 103)d)(12.6 x 103) x (12.6 x103)Correct answer is option 'B'. Can you explain this answer?.
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