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An aqueous solution contains 0.01 M RNH2 (Kb= 2 × 10-6) & 10-4 M NaOH.
The concentration of OH- is nearly:
- a)2.414 ×10-4 M
- b)10-4 M
- c)1.414×10-4 M
- d)2×10-4 M
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
An aqueous solution contains 0.01 M RNH2(Kb= 2 × 10-6) & 10-4...
x2 + 10–4 x – 2 × 10–8 = 0
x = 10–4
[OH–] = x + 10–4
= 2 × 10–4
Most Upvoted Answer
An aqueous solution contains 0.01 M RNH2(Kb= 2 × 10-6) & 10-4...
Given:
- Concentration of RNH2 (weak base) = 0.01 M
- Concentration of NaOH (strong base) = 10-4 M
- Kb (base dissociation constant) for RNH2 = 2 * 10-6
To find:
- Concentration of OH- in the solution
Formula:
- Kb = [OH-][RNH2]/[RNH3+]
Solution:
Step 1: Write down the balanced equation for the reaction between RNH2 and OH-:
RNH2 + OH- → RNH3+ + OH-
Step 2: Calculate the concentration of RNH3+ formed:
Since RNH2 is a weak base, it partially ionizes to form RNH3+ and OH-. Let x be the concentration of RNH3+ formed. Since the concentration of RNH2 is much greater than the concentration of OH-, we can assume that the amount of RNH2 reacting is not significantly changed, so the concentration of RNH3+ formed is approximately equal to x.
Step 3: Calculate the concentration of OH-:
Since the concentration of OH- formed is equal to the concentration of RNH3+ formed, we can write:
[OH-] = x
Step 4: Write the expression for Kb and substitute the known values:
Kb = [OH-][RNH2]/[RNH3+]
2 * 10-6 = x * 0.01 / x
Step 5: Solve for x:
Cross-multiplying the equation:
2 * 10-6 * x = x * 0.01
Simplifying the equation:
2 * 10-6 = 0.01
Dividing both sides by 2 * 10-6:
x = 0.01 / 2 * 10-6
x = 5 * 103
Therefore, the concentration of OH- is 5 * 10-3 M, which is equal to 2 * 10-4 M.
Hence, the correct answer is option D: 2 * 10-4 M.
- Concentration of RNH2 (weak base) = 0.01 M
- Concentration of NaOH (strong base) = 10-4 M
- Kb (base dissociation constant) for RNH2 = 2 * 10-6
To find:
- Concentration of OH- in the solution
Formula:
- Kb = [OH-][RNH2]/[RNH3+]
Solution:
Step 1: Write down the balanced equation for the reaction between RNH2 and OH-:
RNH2 + OH- → RNH3+ + OH-
Step 2: Calculate the concentration of RNH3+ formed:
Since RNH2 is a weak base, it partially ionizes to form RNH3+ and OH-. Let x be the concentration of RNH3+ formed. Since the concentration of RNH2 is much greater than the concentration of OH-, we can assume that the amount of RNH2 reacting is not significantly changed, so the concentration of RNH3+ formed is approximately equal to x.
Step 3: Calculate the concentration of OH-:
Since the concentration of OH- formed is equal to the concentration of RNH3+ formed, we can write:
[OH-] = x
Step 4: Write the expression for Kb and substitute the known values:
Kb = [OH-][RNH2]/[RNH3+]
2 * 10-6 = x * 0.01 / x
Step 5: Solve for x:
Cross-multiplying the equation:
2 * 10-6 * x = x * 0.01
Simplifying the equation:
2 * 10-6 = 0.01
Dividing both sides by 2 * 10-6:
x = 0.01 / 2 * 10-6
x = 5 * 103
Therefore, the concentration of OH- is 5 * 10-3 M, which is equal to 2 * 10-4 M.
Hence, the correct answer is option D: 2 * 10-4 M.
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An aqueous solution contains 0.01 M RNH2(Kb= 2 × 10-6) & 10-4M NaOH.The concentration of OH-is nearly:a)2.414 ×10-4Mb)10-4Mc)1.414×10-4Md)2×10-4MCorrect answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about An aqueous solution contains 0.01 M RNH2(Kb= 2 × 10-6) & 10-4M NaOH.The concentration of OH-is nearly:a)2.414 ×10-4Mb)10-4Mc)1.414×10-4Md)2×10-4MCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous solution contains 0.01 M RNH2(Kb= 2 × 10-6) & 10-4M NaOH.The concentration of OH-is nearly:a)2.414 ×10-4Mb)10-4Mc)1.414×10-4Md)2×10-4MCorrect answer is option 'D'. Can you explain this answer?.
An aqueous solution contains 0.01 M RNH2(Kb= 2 × 10-6) & 10-4M NaOH.The concentration of OH-is nearly:a)2.414 ×10-4Mb)10-4Mc)1.414×10-4Md)2×10-4MCorrect answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about An aqueous solution contains 0.01 M RNH2(Kb= 2 × 10-6) & 10-4M NaOH.The concentration of OH-is nearly:a)2.414 ×10-4Mb)10-4Mc)1.414×10-4Md)2×10-4MCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An aqueous solution contains 0.01 M RNH2(Kb= 2 × 10-6) & 10-4M NaOH.The concentration of OH-is nearly:a)2.414 ×10-4Mb)10-4Mc)1.414×10-4Md)2×10-4MCorrect answer is option 'D'. Can you explain this answer?.
Solutions for An aqueous solution contains 0.01 M RNH2(Kb= 2 × 10-6) & 10-4M NaOH.The concentration of OH-is nearly:a)2.414 ×10-4Mb)10-4Mc)1.414×10-4Md)2×10-4MCorrect answer is option 'D'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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An aqueous solution contains 0.01 M RNH2(Kb= 2 × 10-6) & 10-4M NaOH.The concentration of OH-is nearly:a)2.414 ×10-4Mb)10-4Mc)1.414×10-4Md)2×10-4MCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for An aqueous solution contains 0.01 M RNH2(Kb= 2 × 10-6) & 10-4M NaOH.The concentration of OH-is nearly:a)2.414 ×10-4Mb)10-4Mc)1.414×10-4Md)2×10-4MCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of An aqueous solution contains 0.01 M RNH2(Kb= 2 × 10-6) & 10-4M NaOH.The concentration of OH-is nearly:a)2.414 ×10-4Mb)10-4Mc)1.414×10-4Md)2×10-4MCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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