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Find the solubility of Ni(OH)2 in 0.1 M NaOH.given that the ionic product of Ni(OH)2 is 2×10^-15 a)2×10^-13 M b)2×10^-8 M c) 1×10^-13 M d) 1×10^-8 M?
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Find the solubility of Ni(OH)2 in 0.1 M NaOH.given that the ionic prod...
Solubility of Ni(OH)2 in 0.1 M NaOH
Given:
Ionic product of Ni(OH)2 = 2×10^-15
Molarity of NaOH = 0.1 M
Step 1: Writing the balanced equation
The balanced equation for the dissociation of Ni(OH)2 is:
Ni(OH)2 (s) ⇌ Ni2+ (aq) + 2OH- (aq)
Step 2: Writing the solubility product expression
The solubility product expression for Ni(OH)2 is:
Ksp = [Ni2+][OH-]^2
Step 3: Calculating the solubility of Ni(OH)2
Let's assume the solubility of Ni(OH)2 is 's' mol/L.
Using the balanced equation, we can write the equilibrium concentrations of Ni2+ and OH- ions as 's' mol/L and '2s' mol/L, respectively.
Substituting these values into the solubility product expression, we get:
Ksp = (s)(2s)^2
2×10^-15 = 4s^3
Solving for 's', we find:
s = (2×10^-15/4)^(1/3)
s ≈ 0.316 × 10^-5 M
Step 4: Considering the presence of NaOH
In the presence of NaOH, the OH- concentration will increase due to the dissociation of NaOH.
Since NaOH is a strong base, it completely dissociates to give Na+ and OH- ions.
Given that the molarity of NaOH is 0.1 M, the concentration of OH- ions will be 0.1 M.
Step 5: Comparing the solubility of Ni(OH)2 and OH- concentration from NaOH
Since the solubility of Ni(OH)2 is less than the concentration of OH- ions from NaOH, the solubility is limited by the solubility product.
Therefore, the solubility of Ni(OH)2 in 0.1 M NaOH is approximately 0.316 × 10^-5 M.
Answer: c) 1×10^-13 M
Given:
Ionic product of Ni(OH)2 = 2×10^-15
Molarity of NaOH = 0.1 M
Step 1: Writing the balanced equation
The balanced equation for the dissociation of Ni(OH)2 is:
Ni(OH)2 (s) ⇌ Ni2+ (aq) + 2OH- (aq)
Step 2: Writing the solubility product expression
The solubility product expression for Ni(OH)2 is:
Ksp = [Ni2+][OH-]^2
Step 3: Calculating the solubility of Ni(OH)2
Let's assume the solubility of Ni(OH)2 is 's' mol/L.
Using the balanced equation, we can write the equilibrium concentrations of Ni2+ and OH- ions as 's' mol/L and '2s' mol/L, respectively.
Substituting these values into the solubility product expression, we get:
Ksp = (s)(2s)^2
2×10^-15 = 4s^3
Solving for 's', we find:
s = (2×10^-15/4)^(1/3)
s ≈ 0.316 × 10^-5 M
Step 4: Considering the presence of NaOH
In the presence of NaOH, the OH- concentration will increase due to the dissociation of NaOH.
Since NaOH is a strong base, it completely dissociates to give Na+ and OH- ions.
Given that the molarity of NaOH is 0.1 M, the concentration of OH- ions will be 0.1 M.
Step 5: Comparing the solubility of Ni(OH)2 and OH- concentration from NaOH
Since the solubility of Ni(OH)2 is less than the concentration of OH- ions from NaOH, the solubility is limited by the solubility product.
Therefore, the solubility of Ni(OH)2 in 0.1 M NaOH is approximately 0.316 × 10^-5 M.
Answer: c) 1×10^-13 M
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Find the solubility of Ni(OH)2 in 0.1 M NaOH.given that the ionic product of Ni(OH)2 is 2×10^-15 a)2×10^-13 M b)2×10^-8 M c) 1×10^-13 M d) 1×10^-8 M?
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Find the solubility of Ni(OH)2 in 0.1 M NaOH.given that the ionic product of Ni(OH)2 is 2×10^-15 a)2×10^-13 M b)2×10^-8 M c) 1×10^-13 M d) 1×10^-8 M? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Find the solubility of Ni(OH)2 in 0.1 M NaOH.given that the ionic product of Ni(OH)2 is 2×10^-15 a)2×10^-13 M b)2×10^-8 M c) 1×10^-13 M d) 1×10^-8 M? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the solubility of Ni(OH)2 in 0.1 M NaOH.given that the ionic product of Ni(OH)2 is 2×10^-15 a)2×10^-13 M b)2×10^-8 M c) 1×10^-13 M d) 1×10^-8 M?.
Find the solubility of Ni(OH)2 in 0.1 M NaOH.given that the ionic product of Ni(OH)2 is 2×10^-15 a)2×10^-13 M b)2×10^-8 M c) 1×10^-13 M d) 1×10^-8 M? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Find the solubility of Ni(OH)2 in 0.1 M NaOH.given that the ionic product of Ni(OH)2 is 2×10^-15 a)2×10^-13 M b)2×10^-8 M c) 1×10^-13 M d) 1×10^-8 M? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the solubility of Ni(OH)2 in 0.1 M NaOH.given that the ionic product of Ni(OH)2 is 2×10^-15 a)2×10^-13 M b)2×10^-8 M c) 1×10^-13 M d) 1×10^-8 M?.
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