Chemistry Exam  >  Chemistry Questions  >  At 298 K, 0.1 mol of ammonium acetate and 0.1... Start Learning for Free
At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water. The pH of the resulting solution is
[Given: pKa of acetic acid is 4.75]
  • a)
    4.9
  • b)
    4.6
  • c)
     4.3
  • d)
     2.3
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are ...
Calculating pH of ammonium acetate solution

Step 1: Write the dissociation reaction of ammonium acetate:

NH4C2H3O2 → NH4+ + C2H3O2-

Step 2: Write the expression for the equilibrium constant:

Ka = [NH4+][C2H3O2-]/[NH4C2H3O2]

Step 3: Calculate the concentration of NH4+ and C2H3O2-

[NH4+] = 0.1 mol/L

[C2H3O2-] = 0.14 mol/L

Step 4: Calculate the concentration of NH4C2H3O2

Since ammonium acetate is a strong electrolyte, the concentration of NH4C2H3O2 is equal to the sum of NH4+ and C2H3O2-

[NH4C2H3O2] = [NH4+] + [C2H3O2-] = 0.1 mol/L + 0.14 mol/L = 0.24 mol/L

Step 5: Substitute the values into the expression for Ka to calculate the value of Ka

Ka = [NH4+][C2H3O2-]/[NH4C2H3O2] = (0.1 mol/L)(0.14 mol/L)/(0.24 mol/L) = 5.83 × 10^-5


Calculating pH of acetic acid solution

Step 1: Write the dissociation reaction of acetic acid:

CH3COOH → H+ + CH3COO-

Step 2: Write the expression for the equilibrium constant:

Ka = [H+][CH3COO-]/[CH3COOH]

Step 3: Calculate the concentration of H+ and CH3COO-

[CH3COO-] = 0.14 mol/L

Since the solution is not neutral, we need to use the Henderson-Hasselbalch equation to calculate the concentration of H+.

pH = pKa + log([A-]/[HA])

where [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

Substituting the values, we get:

pH = 4.75 + log(0.14/0.1) = 4.6


Therefore, the pH of the solution is 4.6, which corresponds to option B.
Free Test
Community Answer
At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are ...
4.6 answer
pH =pKa+ log ( conjugate base ÷ conjugate acid )
= 4.75+log (.1÷.14) = 4.6
Explore Courses for Chemistry exam
At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water. The pH of the resulting solution is[Given: pKa of acetic acid is 4.75]a)4.9b)4.6c)4.3d)2.3Correct answer is option 'B'. Can you explain this answer?
Question Description
At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water. The pH of the resulting solution is[Given: pKa of acetic acid is 4.75]a)4.9b)4.6c)4.3d)2.3Correct answer is option 'B'. Can you explain this answer? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water. The pH of the resulting solution is[Given: pKa of acetic acid is 4.75]a)4.9b)4.6c)4.3d)2.3Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water. The pH of the resulting solution is[Given: pKa of acetic acid is 4.75]a)4.9b)4.6c)4.3d)2.3Correct answer is option 'B'. Can you explain this answer?.
Solutions for At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water. The pH of the resulting solution is[Given: pKa of acetic acid is 4.75]a)4.9b)4.6c)4.3d)2.3Correct answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Chemistry. Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free.
Here you can find the meaning of At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water. The pH of the resulting solution is[Given: pKa of acetic acid is 4.75]a)4.9b)4.6c)4.3d)2.3Correct answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water. The pH of the resulting solution is[Given: pKa of acetic acid is 4.75]a)4.9b)4.6c)4.3d)2.3Correct answer is option 'B'. Can you explain this answer?, a detailed solution for At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water. The pH of the resulting solution is[Given: pKa of acetic acid is 4.75]a)4.9b)4.6c)4.3d)2.3Correct answer is option 'B'. Can you explain this answer? has been provided alongside types of At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water. The pH of the resulting solution is[Given: pKa of acetic acid is 4.75]a)4.9b)4.6c)4.3d)2.3Correct answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice At 298 K, 0.1 mol of ammonium acetate and 0.14 mol of acetic acid are dissolved in 1 L of water. The pH of the resulting solution is[Given: pKa of acetic acid is 4.75]a)4.9b)4.6c)4.3d)2.3Correct answer is option 'B'. Can you explain this answer? tests, examples and also practice Chemistry tests.
Explore Courses for Chemistry exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev