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PH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27C. The buffer capacity of the solution is ? Its answer is 0.1 , Can anyone explain?
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PH of a buffer solution decreases by 0.02 units when 0.12 g of acetic ...
Buffer capacity = no.of moles of acid added per litre / change in pH
no. of moles of acetic acid = 0.12/60 = 0.002
no.of moles acetic acid per litre = 0.002 × 1000 / 250 = 2/250
buffer capacity = 2/250×0.002= 200/250×2 = 0.4
no. of moles of acetic acid = 0.12/60 = 0.002
no.of moles acetic acid per litre = 0.002 × 1000 / 250 = 2/250
buffer capacity = 2/250×0.002= 200/250×2 = 0.4
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PH of a buffer solution decreases by 0.02 units when 0.12 g of acetic ...
Given information:
- The pH of a buffer solution decreases by 0.02 units.
- 0.12 g of acetic acid is added to 250 mL of the buffer solution.
- The buffer solution consists of acetic acid and potassium acetate.
- The temperature is 27°C.
Calculating the buffer capacity:
The buffer capacity is a measure of how well a buffer solution can resist changes in pH when a small amount of acid or base is added. It is defined as the amount of acid or base required to change the pH of the buffer solution by 1 unit.
To calculate the buffer capacity, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-]/[HA])
Where:
pH is the current pH of the buffer solution
pKa is the dissociation constant of the weak acid (acetic acid in this case)
[A-] is the concentration of the conjugate base (acetate ions)
[HA] is the concentration of the weak acid (acetic acid)
Step 1: Calculate the initial pH of the buffer solution:
Since the pH decreases by 0.02 units when 0.12 g of acetic acid is added, we can assume that the initial pH of the buffer solution was 0.02 units higher.
Let's assume the initial pH is pH_initial = pH + 0.02
Step 2: Calculate the concentrations of [A-] and [HA] in the buffer solution:
We know that 0.12 g of acetic acid is added to 250 mL of the buffer solution. To calculate the concentrations, we need to convert grams to moles and volume to liters.
- The molar mass of acetic acid (CH3COOH) is 60.05 g/mol.
- The molar mass of potassium acetate (CH3COOK) is 98.14 g/mol.
Moles of acetic acid = (0.12 g) / (60.05 g/mol)
Moles of acetate ions = Moles of acetic acid
Concentration of acetic acid ([HA]) = (moles of acetic acid) / (volume of solution in liters)
Concentration of acetate ions ([A-]) = (moles of acetate ions) / (volume of solution in liters)
Step 3: Calculate the buffer capacity:
Using the Henderson-Hasselbalch equation, we can rearrange it to solve for the change in pH:
ΔpH = pH_final - pH_initial = log ([A-]_final / [HA]_final) - log ([A-]_initial / [HA]_initial)
Since we know that the change in pH is 0.02 units, we can rearrange the equation to calculate the buffer capacity (β):
β = ΔpH / log ([A-]_final / [HA]_final)
Substituting the known values, we can calculate the buffer capacity.
Note: The answer of 0.1 for the buffer capacity is based on the specific values provided in the question. If the values were different, the buffer capacity would also
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PH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27C. The buffer capacity of the solution is ? Its answer is 0.1 , Can anyone explain?
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PH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27C. The buffer capacity of the solution is ? Its answer is 0.1 , Can anyone explain? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about PH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27C. The buffer capacity of the solution is ? Its answer is 0.1 , Can anyone explain? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for PH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27C. The buffer capacity of the solution is ? Its answer is 0.1 , Can anyone explain?.
PH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27C. The buffer capacity of the solution is ? Its answer is 0.1 , Can anyone explain? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about PH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27C. The buffer capacity of the solution is ? Its answer is 0.1 , Can anyone explain? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for PH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27C. The buffer capacity of the solution is ? Its answer is 0.1 , Can anyone explain?.
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