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A buffer solution is prepared in which the concentration of NH3 is 0.30M and the concentration of NH4+ is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 × 10–5, what is the pH of this solution ? (log 2.7 = 0.433). [2011]
- a)9.08
- b)9.43
- c)11.72
- d)8.73
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A buffer solution is prepared in which the concentration of NH3 is 0.3...
Given [NH3] = 0.3 M, [NH4+] = 0.2 M, Kb = 1.8 × 10–5.
∴ pKb = 4.74
= 4.56
pH = 14 – 4.56 = 9.436
pH = 14 – 4.56 = 9.436
Most Upvoted Answer
A buffer solution is prepared in which the concentration of NH3 is 0.3...
To find the value of Kb for NH3, we need to set up the equilibrium expression for the reaction:
NH3 + H2O ⇌ NH4+ + OH-
The equilibrium constant expression is given by:
Kb = [NH4+][OH-] / [NH3]
We are given the concentration of NH3 as 0.30 M and NH4+ as 0.20 M. However, we are not given the concentration of OH-. In order to calculate Kb, we need to determine the concentration of OH- at equilibrium.
Since NH3 is a weak base, it reacts with water to form OH-. The reaction is as follows:
NH3 + H2O ⇌ NH4+ + OH-
We can assume that the concentration of OH- at equilibrium is x M. Therefore, the concentration of NH4+ at equilibrium is also x M.
Using the given concentrations and the assumption above, we can set up the following equation:
Kb = (x)(x) / (0.30)
Simplifying, we have:
1.8 = x^2 / 0.30
Multiplying both sides by 0.30, we get:
0.54 = x^2
Taking the square root of both sides, we find:
x ≈ 0.735
Therefore, the concentration of OH- at equilibrium is approximately 0.735 M.
Now, we can substitute the values into the equilibrium constant expression to find Kb:
Kb = (0.20)(0.735) / (0.30) = 0.490
Therefore, the value of Kb for NH3 is approximately 0.490.
NH3 + H2O ⇌ NH4+ + OH-
The equilibrium constant expression is given by:
Kb = [NH4+][OH-] / [NH3]
We are given the concentration of NH3 as 0.30 M and NH4+ as 0.20 M. However, we are not given the concentration of OH-. In order to calculate Kb, we need to determine the concentration of OH- at equilibrium.
Since NH3 is a weak base, it reacts with water to form OH-. The reaction is as follows:
NH3 + H2O ⇌ NH4+ + OH-
We can assume that the concentration of OH- at equilibrium is x M. Therefore, the concentration of NH4+ at equilibrium is also x M.
Using the given concentrations and the assumption above, we can set up the following equation:
Kb = (x)(x) / (0.30)
Simplifying, we have:
1.8 = x^2 / 0.30
Multiplying both sides by 0.30, we get:
0.54 = x^2
Taking the square root of both sides, we find:
x ≈ 0.735
Therefore, the concentration of OH- at equilibrium is approximately 0.735 M.
Now, we can substitute the values into the equilibrium constant expression to find Kb:
Kb = (0.20)(0.735) / (0.30) = 0.490
Therefore, the value of Kb for NH3 is approximately 0.490.
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A buffer solution is prepared in which the concentration of NH3 is 0.30M and the concentration of NH4+ is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 × 10–5, what is the pH of this solution ? (log 2.7 = 0.433). [2011]a)9.08b)9.43c)11.72d)8.73Correct answer is option 'B'. Can you explain this answer?
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A buffer solution is prepared in which the concentration of NH3 is 0.30M and the concentration of NH4+ is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 × 10–5, what is the pH of this solution ? (log 2.7 = 0.433). [2011]a)9.08b)9.43c)11.72d)8.73Correct answer is option 'B'. 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 A buffer solution is prepared in which the concentration of NH3 is 0.30M and the concentration of NH4+ is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 × 10–5, what is the pH of this solution ? (log 2.7 = 0.433). [2011]a)9.08b)9.43c)11.72d)8.73Correct answer is option 'B'. 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 A buffer solution is prepared in which the concentration of NH3 is 0.30M and the concentration of NH4+ is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 × 10–5, what is the pH of this solution ? (log 2.7 = 0.433). [2011]a)9.08b)9.43c)11.72d)8.73Correct answer is option 'B'. Can you explain this answer?.
A buffer solution is prepared in which the concentration of NH3 is 0.30M and the concentration of NH4+ is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 × 10–5, what is the pH of this solution ? (log 2.7 = 0.433). [2011]a)9.08b)9.43c)11.72d)8.73Correct answer is option 'B'. 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 A buffer solution is prepared in which the concentration of NH3 is 0.30M and the concentration of NH4+ is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 × 10–5, what is the pH of this solution ? (log 2.7 = 0.433). [2011]a)9.08b)9.43c)11.72d)8.73Correct answer is option 'B'. 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 A buffer solution is prepared in which the concentration of NH3 is 0.30M and the concentration of NH4+ is 0.20 M. If the equilibrium constant, Kb for NH3 equals 1.8 × 10–5, what is the pH of this solution ? (log 2.7 = 0.433). [2011]a)9.08b)9.43c)11.72d)8.73Correct answer is option 'B'. Can you explain this answer?.
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