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What is the solubility of AgCl in 0.20 M NH3 ?
Given : Ksp(AgCl) = 1.7 × 10-10 M2, K1 = [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1 and
K2 = [Ag(NH3)2+] / [Ag (NH3)+ [NH3] = 7.14 × 103 M-1
- a)6.66 × 10-3
- b)9.66 × 10-5
- c)9.66 × 10-3
- d)6.66 × 10-5
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 &...
If x be the concentration of AgCl in the solution, then [Cl–] = x
From the Ksp for AgCl, we derive
If we answer that the majority of the dissolved
Ag+ goes into solution as Ag(NH3)2+ then
Ag(NH3)2+ then
Ag(NH3)2+ = x
Since two molecules of NH3 are required for every Ag(NH3)2+ ion formed, we have
∴ x = [Ag(NH3)2+] = 9.6 × 10–3 M, which is the solubility of AgCl in 0.2 M NH3
From the Ksp for AgCl, we derive
If we answer that the majority of the dissolved
Ag+ goes into solution as Ag(NH3)2+ then
Ag(NH3)2+ then
Ag(NH3)2+ = x
Since two molecules of NH3 are required for every Ag(NH3)2+ ion formed, we have
∴ x = [Ag(NH3)2+] = 9.6 × 10–3 M, which is the solubility of AgCl in 0.2 M NH3
Most Upvoted Answer
What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 &...
Solubility of AgCl in 0.20 M NH3
Given:
Ksp(AgCl) = 1.7 × 10^-10 M^2
K1 = [Ag(NH3)] / [Ag][NH3] = 2.33 × 10^3 M^-1
K2 = [Ag(NH3)2] / [Ag(NH3)][NH3] = 7.14 × 10^3 M^-1
To find: Solubility of AgCl in 0.20 M NH3
Solution:
Step 1: Write the chemical equation for the dissolution of AgCl in NH3.
AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl-(aq)
Step 2: Write the equilibrium expressions for the reaction.
Ksp = [Ag+][Cl-] = 1.7 × 10^-10 M^2
Kf1 = [Ag(NH3)2+]/[Ag+][NH3] = 2.33 × 10^3 M^-1
Kf2 = [Ag(NH3)2+]/[Ag(NH3)][NH3] = 7.14 × 10^3 M^-1
Step 3: Use the equilibrium expressions to calculate the concentration of Ag+ and Cl- ions in solution.
Let x be the solubility of AgCl in NH3. Then, the concentration of Ag+ and Cl- ions can be expressed as follows:
[Ag+] = [Ag(NH3)2+] = Kf2[x][NH3]
[Cl-] = Ksp / [Ag+] = (1.7 × 10^-10 M^2) / (Kf2[x][NH3])
Step 4: Use the equilibrium expressions to calculate the concentration of NH3.
Since NH3 is a weak base, we can assume that the concentration of NH3 is equal to the initial concentration of NH3, which is 0.20 M.
[NH3] = 0.20 M
Step 5: Use the concentration of Ag+, Cl-, and NH3 to calculate the value of x, which is the solubility of AgCl in NH3.
Substituting the values of [Ag+], [Cl-], and [NH3] in the equilibrium expression for the dissolution of AgCl in NH3, we get:
Ksp = [Ag+][Cl-] = Kf2[x][NH3] × (1.7 × 10^-10 M^2) = 2.33 × 10^3 M^-1 × x × (0.20 M)^2 / (7.14 × 10^3 M^-1)
Solving for x, we get:
x = 9.66 × 10^-3 M
Therefore, the solubility of AgCl in 0.20 M NH3 is 9.66 × 10^-3 M.
Answer: Option C (9.66 × 10^-3 M)
Given:
Ksp(AgCl) = 1.7 × 10^-10 M^2
K1 = [Ag(NH3)] / [Ag][NH3] = 2.33 × 10^3 M^-1
K2 = [Ag(NH3)2] / [Ag(NH3)][NH3] = 7.14 × 10^3 M^-1
To find: Solubility of AgCl in 0.20 M NH3
Solution:
Step 1: Write the chemical equation for the dissolution of AgCl in NH3.
AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2+(aq) + Cl-(aq)
Step 2: Write the equilibrium expressions for the reaction.
Ksp = [Ag+][Cl-] = 1.7 × 10^-10 M^2
Kf1 = [Ag(NH3)2+]/[Ag+][NH3] = 2.33 × 10^3 M^-1
Kf2 = [Ag(NH3)2+]/[Ag(NH3)][NH3] = 7.14 × 10^3 M^-1
Step 3: Use the equilibrium expressions to calculate the concentration of Ag+ and Cl- ions in solution.
Let x be the solubility of AgCl in NH3. Then, the concentration of Ag+ and Cl- ions can be expressed as follows:
[Ag+] = [Ag(NH3)2+] = Kf2[x][NH3]
[Cl-] = Ksp / [Ag+] = (1.7 × 10^-10 M^2) / (Kf2[x][NH3])
Step 4: Use the equilibrium expressions to calculate the concentration of NH3.
Since NH3 is a weak base, we can assume that the concentration of NH3 is equal to the initial concentration of NH3, which is 0.20 M.
[NH3] = 0.20 M
Step 5: Use the concentration of Ag+, Cl-, and NH3 to calculate the value of x, which is the solubility of AgCl in NH3.
Substituting the values of [Ag+], [Cl-], and [NH3] in the equilibrium expression for the dissolution of AgCl in NH3, we get:
Ksp = [Ag+][Cl-] = Kf2[x][NH3] × (1.7 × 10^-10 M^2) = 2.33 × 10^3 M^-1 × x × (0.20 M)^2 / (7.14 × 10^3 M^-1)
Solving for x, we get:
x = 9.66 × 10^-3 M
Therefore, the solubility of AgCl in 0.20 M NH3 is 9.66 × 10^-3 M.
Answer: Option C (9.66 × 10^-3 M)
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What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 × 10-10M2, K1= [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1andK2= [Ag(NH3)2+] / [Ag (NH3)+[NH3] = 7.14 × 103M-1a)6.66 × 10-3b)9.66 × 10-5c)9.66 × 10-3d)6.66 × 10-5Correct answer is option 'C'. Can you explain this answer?
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What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 × 10-10M2, K1= [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1andK2= [Ag(NH3)2+] / [Ag (NH3)+[NH3] = 7.14 × 103M-1a)6.66 × 10-3b)9.66 × 10-5c)9.66 × 10-3d)6.66 × 10-5Correct answer is option 'C'. 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 What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 × 10-10M2, K1= [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1andK2= [Ag(NH3)2+] / [Ag (NH3)+[NH3] = 7.14 × 103M-1a)6.66 × 10-3b)9.66 × 10-5c)9.66 × 10-3d)6.66 × 10-5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 × 10-10M2, K1= [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1andK2= [Ag(NH3)2+] / [Ag (NH3)+[NH3] = 7.14 × 103M-1a)6.66 × 10-3b)9.66 × 10-5c)9.66 × 10-3d)6.66 × 10-5Correct answer is option 'C'. Can you explain this answer?.
What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 × 10-10M2, K1= [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1andK2= [Ag(NH3)2+] / [Ag (NH3)+[NH3] = 7.14 × 103M-1a)6.66 × 10-3b)9.66 × 10-5c)9.66 × 10-3d)6.66 × 10-5Correct answer is option 'C'. 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 What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 × 10-10M2, K1= [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1andK2= [Ag(NH3)2+] / [Ag (NH3)+[NH3] = 7.14 × 103M-1a)6.66 × 10-3b)9.66 × 10-5c)9.66 × 10-3d)6.66 × 10-5Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 × 10-10M2, K1= [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1andK2= [Ag(NH3)2+] / [Ag (NH3)+[NH3] = 7.14 × 103M-1a)6.66 × 10-3b)9.66 × 10-5c)9.66 × 10-3d)6.66 × 10-5Correct answer is option 'C'. Can you explain this answer?.
Solutions for What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 × 10-10M2, K1= [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1andK2= [Ag(NH3)2+] / [Ag (NH3)+[NH3] = 7.14 × 103M-1a)6.66 × 10-3b)9.66 × 10-5c)9.66 × 10-3d)6.66 × 10-5Correct answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 × 10-10M2, K1= [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1andK2= [Ag(NH3)2+] / [Ag (NH3)+[NH3] = 7.14 × 103M-1a)6.66 × 10-3b)9.66 × 10-5c)9.66 × 10-3d)6.66 × 10-5Correct answer is option 'C'. Can you explain this answer?, a detailed solution for What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 × 10-10M2, K1= [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1andK2= [Ag(NH3)2+] / [Ag (NH3)+[NH3] = 7.14 × 103M-1a)6.66 × 10-3b)9.66 × 10-5c)9.66 × 10-3d)6.66 × 10-5Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 × 10-10M2, K1= [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1andK2= [Ag(NH3)2+] / [Ag (NH3)+[NH3] = 7.14 × 103M-1a)6.66 × 10-3b)9.66 × 10-5c)9.66 × 10-3d)6.66 × 10-5Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice What is the solubility of AgCl in 0.20 M NH3?Given : Ksp(AgCl) = 1.7 × 10-10M2, K1= [Ag(NH3)+] / [Ag+] [NH3] = 2.33 × 103M-1andK2= [Ag(NH3)2+] / [Ag (NH3)+[NH3] = 7.14 × 103M-1a)6.66 × 10-3b)9.66 × 10-5c)9.66 × 10-3d)6.66 × 10-5Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
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