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Ksp of a salt Ni(OH)2 is 2 x 10^-15 then, molar solubility of Ni(OH)2 in 0.01M NaOH is A) 2 x 10^-15 B) 2 x 10^-5 C) 2 x 10^-11 D) 10^-7?
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Ksp of a salt Ni(OH)2 is 2 x 10^-15 then, molar solubility of Ni(OH)2 ...
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Ksp of a salt Ni(OH)2 is 2 x 10^-15 then, molar solubility of Ni(OH)2 ...
Molar solubility is defined as the number of moles of a substance that can be dissolved in one liter of a solution. In order to determine the molar solubility of Ni(OH)2 in 0.01M NaOH, we need to consider the solubility product constant (Ksp) of Ni(OH)2.
The Ksp expression for the dissolution of Ni(OH)2 can be written as follows:
Ni(OH)2(s) ⇌ Ni2+(aq) + 2OH-(aq)
The Ksp expression is given by the product of the concentrations of the ions raised to the power of their stoichiometric coefficients:
Ksp = [Ni2+][OH-]^2
Given that the Ksp of Ni(OH)2 is 2 x 10^-15, we can set up an equilibrium expression as follows:
2 x 10^-15 = [Ni2+][OH-]^2
Since we are considering the dissolution of Ni(OH)2 in 0.01M NaOH, we can assume that the concentration of OH- is 0.01M. This is because NaOH is a strong base and dissociates completely in water to produce OH- ions.
Substituting the concentration of OH- into the Ksp expression, we get:
2 x 10^-15 = [Ni2+](0.01)^2
Simplifying the equation, we find:
[Ni2+] = (2 x 10^-15)/(0.01)^2 = 2 x 10^-11 M
Therefore, the molar solubility of Ni(OH)2 in 0.01M NaOH is 2 x 10^-11 M.
In conclusion, the correct answer is C) 2 x 10^-11.
The Ksp expression for the dissolution of Ni(OH)2 can be written as follows:
Ni(OH)2(s) ⇌ Ni2+(aq) + 2OH-(aq)
The Ksp expression is given by the product of the concentrations of the ions raised to the power of their stoichiometric coefficients:
Ksp = [Ni2+][OH-]^2
Given that the Ksp of Ni(OH)2 is 2 x 10^-15, we can set up an equilibrium expression as follows:
2 x 10^-15 = [Ni2+][OH-]^2
Since we are considering the dissolution of Ni(OH)2 in 0.01M NaOH, we can assume that the concentration of OH- is 0.01M. This is because NaOH is a strong base and dissociates completely in water to produce OH- ions.
Substituting the concentration of OH- into the Ksp expression, we get:
2 x 10^-15 = [Ni2+](0.01)^2
Simplifying the equation, we find:
[Ni2+] = (2 x 10^-15)/(0.01)^2 = 2 x 10^-11 M
Therefore, the molar solubility of Ni(OH)2 in 0.01M NaOH is 2 x 10^-11 M.
In conclusion, the correct answer is C) 2 x 10^-11.
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Ksp of a salt Ni(OH)2 is 2 x 10^-15 then, molar solubility of Ni(OH)2 in 0.01M NaOH is A) 2 x 10^-15 B) 2 x 10^-5 C) 2 x 10^-11 D) 10^-7?
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Ksp of a salt Ni(OH)2 is 2 x 10^-15 then, molar solubility of Ni(OH)2 in 0.01M NaOH is A) 2 x 10^-15 B) 2 x 10^-5 C) 2 x 10^-11 D) 10^-7? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Ksp of a salt Ni(OH)2 is 2 x 10^-15 then, molar solubility of Ni(OH)2 in 0.01M NaOH is A) 2 x 10^-15 B) 2 x 10^-5 C) 2 x 10^-11 D) 10^-7? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ksp of a salt Ni(OH)2 is 2 x 10^-15 then, molar solubility of Ni(OH)2 in 0.01M NaOH is A) 2 x 10^-15 B) 2 x 10^-5 C) 2 x 10^-11 D) 10^-7?.
Ksp of a salt Ni(OH)2 is 2 x 10^-15 then, molar solubility of Ni(OH)2 in 0.01M NaOH is A) 2 x 10^-15 B) 2 x 10^-5 C) 2 x 10^-11 D) 10^-7? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Ksp of a salt Ni(OH)2 is 2 x 10^-15 then, molar solubility of Ni(OH)2 in 0.01M NaOH is A) 2 x 10^-15 B) 2 x 10^-5 C) 2 x 10^-11 D) 10^-7? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Ksp of a salt Ni(OH)2 is 2 x 10^-15 then, molar solubility of Ni(OH)2 in 0.01M NaOH is A) 2 x 10^-15 B) 2 x 10^-5 C) 2 x 10^-11 D) 10^-7?.
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