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In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?
(Ksp for AgCl = 1.8 × 10–10,
Ksp for PbCl2 = 1.7 × 10–5) [2011M]
(Ksp for AgCl = 1.8 × 10–10,
Ksp for PbCl2 = 1.7 × 10–5) [2011M]
- a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 M
- b)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 M
- c)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 M
- d)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 M
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
In qualitative analysis, the metals of Group I can be separated from o...
Ksp = [Ag+] [Cl–] 1.8 × 10–10 = [Ag+] [0.1] [Ag+] = 1.8 × 10–9 M
Ksp = [Pb+2] [Cl–]2 1.7 × 10–5 = [Pb+2] [0.1]2 [Pb+2] = 1.7 × 10–3 M
Ksp = [Pb+2] [Cl–]2 1.7 × 10–5 = [Pb+2] [0.1]2 [Pb+2] = 1.7 × 10–3 M
Most Upvoted Answer
In qualitative analysis, the metals of Group I can be separated from o...
- ions are present in excess.
The balanced equation for the reaction between Ag+ and Cl- ions is:
Ag+ (aq) + Cl- (aq) → AgCl (s)
The balanced equation for the reaction between Pb2+ and Cl- ions is:
Pb2+ (aq) + 2Cl- (aq) → PbCl2 (s)
Since HCl is added in excess, it means that there will be an excess of Cl- ions in the solution.
The Ag+ ions will react with Cl- ions to form a white precipitate of AgCl.
The Pb2+ ions will react with Cl- ions to form a white precipitate of PbCl2.
The white precipitates can be separated from the solution by filtration. This will leave behind the other ions in the solution.
Overall, adding excess HCl to the solution will precipitate the Ag+ and Pb2+ ions as their chloride salts, allowing for the separation of Group I metals from other ions.
The balanced equation for the reaction between Ag+ and Cl- ions is:
Ag+ (aq) + Cl- (aq) → AgCl (s)
The balanced equation for the reaction between Pb2+ and Cl- ions is:
Pb2+ (aq) + 2Cl- (aq) → PbCl2 (s)
Since HCl is added in excess, it means that there will be an excess of Cl- ions in the solution.
The Ag+ ions will react with Cl- ions to form a white precipitate of AgCl.
The Pb2+ ions will react with Cl- ions to form a white precipitate of PbCl2.
The white precipitates can be separated from the solution by filtration. This will leave behind the other ions in the solution.
Overall, adding excess HCl to the solution will precipitate the Ag+ and Pb2+ ions as their chloride salts, allowing for the separation of Group I metals from other ions.
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In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer?
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In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer?.
In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer?.
Solutions for In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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Here you can find the meaning of In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and Pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl– concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium?(Ksp for AgCl = 1.8 × 10–10,Ksp for PbCl2 = 1.7 × 10–5) [2011M]a)[Ag+] = 1.8 × 10–7 M ; [Pb2+] = 1.7 × 10–6 Mb)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–5 Mc)[Ag+] = 1.8 × 10–9 M ; [Pb2+] = 1.7 × 10–3 Md)[Ag+] = 1.8 × 10–11 M ; [Pb2+] = 8.5 × 10–4 MCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice Class 11 tests.
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