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When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only with
- a)10–4 M (Ag+) and 10–4 M (Cl–)
- b)10–5 M (Ag+) and 10–5 M (Cl–)
- c)10–6 M (Ag+) and 10–6 M (Cl–)
- d)10–10 M (Ag+) and 10–10 M (Cl–)
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
When equal volumes of the following solutions are mixed, precipitation...
Answer:
a) 10^-4 M (Ag⁺) and 10^-4 M (Cl⁻) is the correct answer.
Explanation:
Given data:
When equal volumes of the some solutions are mixed, precipitation of AgCl will occur
K{sp} = 1.8x10^-10
When two solutions are mixed having equal volumes, the concentrated will be reduced.
It will become half of the initial solution.
Also, AgCl will precipate only when
[Ag+][Cl-] > Ksp
Hence we will check all the options
a)
[Ag+]= 10^-4/2 M
[Cl-] = 10^-4/2 M
[Ag+][Cl-] = 10^-4/2 M*10^-4/2 M
= 10^-8/4
= 2.5*10^-9
Which is greater than Ksp
Thus (a) is the correct answer
Thus 10⁻⁴ M (Ag⁺) and 10⁻⁴ M (Cl⁻) is the correct answer.
a) 10^-4 M (Ag⁺) and 10^-4 M (Cl⁻) is the correct answer.
Explanation:
Given data:
When equal volumes of the some solutions are mixed, precipitation of AgCl will occur
K{sp} = 1.8x10^-10
When two solutions are mixed having equal volumes, the concentrated will be reduced.
It will become half of the initial solution.
Also, AgCl will precipate only when
[Ag+][Cl-] > Ksp
Hence we will check all the options
a)
[Ag+]= 10^-4/2 M
[Cl-] = 10^-4/2 M
[Ag+][Cl-] = 10^-4/2 M*10^-4/2 M
= 10^-8/4
= 2.5*10^-9
Which is greater than Ksp
Thus (a) is the correct answer
Thus 10⁻⁴ M (Ag⁺) and 10⁻⁴ M (Cl⁻) is the correct answer.
Most Upvoted Answer
When equal volumes of the following solutions are mixed, precipitation...
Understanding Precipitation of AgCl
When mixing solutions containing silver ions (Ag+) and chloride ions (Cl-), precipitation occurs when the product of their concentrations exceeds the solubility product constant (Ksp).
Key Concept: Ksp of AgCl
- Ksp of AgCl = 1.8 × 10–10
- Precipitation occurs when Q (the reaction quotient) > Ksp
Calculating Q for Various Concentrations
When equal volumes of two solutions are mixed, the concentrations of Ag+ and Cl- are halved. Therefore, for equal volume mixing:
- Concentration of Ag+ = [Ag+]initial / 2
- Concentration of Cl- = [Cl-]initial / 2
Analysis of Each Option
- Option A: 10–4 M Ag+ and 10–4 M Cl-
- After mixing: [Ag+] = [Cl-] = 5 × 10–5 M
- Q = [Ag+][Cl-] = (5 × 10–5)(5 × 10–5) = 2.5 × 10–9
- Since 2.5 × 10–9 > 1.8 × 10–10, precipitation occurs.
- Option B: 10–5 M Ag+ and 10–5 M Cl-
- After mixing: [Ag+] = [Cl-] = 5 × 10–6 M
- Q = (5 × 10–6)(5 × 10–6) = 2.5 × 10–11
- Since 2.5 × 10–11 < 1.8="" ×="" 10–10,="" no="" precipitation="" />
- Option C: 10–6 M Ag+ and 10–6 M Cl-
- After mixing: [Ag+] = [Cl-] = 5 × 10–7 M
- Q = (5 × 10–7)(5 × 10–7) = 2.5 × 10–13
- Since 2.5 × 10–13 < 1.8="" ×="" 10–10,="" no="" precipitation="" />
- Option D: 10–10 M Ag+ and 10–10 M Cl-
- After mixing: [Ag+] = [Cl-] = 5 × 10–11 M
- Q = (5 × 10–11)(5 × 10–11) = 2.5 × 10–21
- Since 2.5 × 10–21 < 1.8="" ×="" 10–10,="" no="" precipitation="" />
Conclusion
Only option A leads to precipitation of AgCl, as it exceeds the Ksp value.
When mixing solutions containing silver ions (Ag+) and chloride ions (Cl-), precipitation occurs when the product of their concentrations exceeds the solubility product constant (Ksp).
Key Concept: Ksp of AgCl
- Ksp of AgCl = 1.8 × 10–10
- Precipitation occurs when Q (the reaction quotient) > Ksp
Calculating Q for Various Concentrations
When equal volumes of two solutions are mixed, the concentrations of Ag+ and Cl- are halved. Therefore, for equal volume mixing:
- Concentration of Ag+ = [Ag+]initial / 2
- Concentration of Cl- = [Cl-]initial / 2
Analysis of Each Option
- Option A: 10–4 M Ag+ and 10–4 M Cl-
- After mixing: [Ag+] = [Cl-] = 5 × 10–5 M
- Q = [Ag+][Cl-] = (5 × 10–5)(5 × 10–5) = 2.5 × 10–9
- Since 2.5 × 10–9 > 1.8 × 10–10, precipitation occurs.
- Option B: 10–5 M Ag+ and 10–5 M Cl-
- After mixing: [Ag+] = [Cl-] = 5 × 10–6 M
- Q = (5 × 10–6)(5 × 10–6) = 2.5 × 10–11
- Since 2.5 × 10–11 < 1.8="" ×="" 10–10,="" no="" precipitation="" />
- Option C: 10–6 M Ag+ and 10–6 M Cl-
- After mixing: [Ag+] = [Cl-] = 5 × 10–7 M
- Q = (5 × 10–7)(5 × 10–7) = 2.5 × 10–13
- Since 2.5 × 10–13 < 1.8="" ×="" 10–10,="" no="" precipitation="" />
- Option D: 10–10 M Ag+ and 10–10 M Cl-
- After mixing: [Ag+] = [Cl-] = 5 × 10–11 M
- Q = (5 × 10–11)(5 × 10–11) = 2.5 × 10–21
- Since 2.5 × 10–21 < 1.8="" ×="" 10–10,="" no="" precipitation="" />
Conclusion
Only option A leads to precipitation of AgCl, as it exceeds the Ksp value.
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When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only witha)10–4 M (Ag+) and 10–4 M (Cl–)b)10–5 M (Ag+) and 10–5 M (Cl–)c)10–6 M (Ag+) and 10–6 M (Cl–)d)10–10 M (Ag+) and 10–10 M (Cl–)Correct answer is option 'A'. Can you explain this answer?
Question Description
When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only witha)10–4 M (Ag+) and 10–4 M (Cl–)b)10–5 M (Ag+) and 10–5 M (Cl–)c)10–6 M (Ag+) and 10–6 M (Cl–)d)10–10 M (Ag+) and 10–10 M (Cl–)Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only witha)10–4 M (Ag+) and 10–4 M (Cl–)b)10–5 M (Ag+) and 10–5 M (Cl–)c)10–6 M (Ag+) and 10–6 M (Cl–)d)10–10 M (Ag+) and 10–10 M (Cl–)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only witha)10–4 M (Ag+) and 10–4 M (Cl–)b)10–5 M (Ag+) and 10–5 M (Cl–)c)10–6 M (Ag+) and 10–6 M (Cl–)d)10–10 M (Ag+) and 10–10 M (Cl–)Correct answer is option 'A'. Can you explain this answer?.
When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only witha)10–4 M (Ag+) and 10–4 M (Cl–)b)10–5 M (Ag+) and 10–5 M (Cl–)c)10–6 M (Ag+) and 10–6 M (Cl–)d)10–10 M (Ag+) and 10–10 M (Cl–)Correct answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only witha)10–4 M (Ag+) and 10–4 M (Cl–)b)10–5 M (Ag+) and 10–5 M (Cl–)c)10–6 M (Ag+) and 10–6 M (Cl–)d)10–10 M (Ag+) and 10–10 M (Cl–)Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only witha)10–4 M (Ag+) and 10–4 M (Cl–)b)10–5 M (Ag+) and 10–5 M (Cl–)c)10–6 M (Ag+) and 10–6 M (Cl–)d)10–10 M (Ag+) and 10–10 M (Cl–)Correct answer is option 'A'. Can you explain this answer?.
Solutions for When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only witha)10–4 M (Ag+) and 10–4 M (Cl–)b)10–5 M (Ag+) and 10–5 M (Cl–)c)10–6 M (Ag+) and 10–6 M (Cl–)d)10–10 M (Ag+) and 10–10 M (Cl–)Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice When equal volumes of the following solutions are mixed, precipitation of AgCl (Ksp = 1.8×10–10) will occur only witha)10–4 M (Ag+) and 10–4 M (Cl–)b)10–5 M (Ag+) and 10–5 M (Cl–)c)10–6 M (Ag+) and 10–6 M (Cl–)d)10–10 M (Ag+) and 10–10 M (Cl–)Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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