Question

What volume of \(0.2M \, CH_3COOH\) needs to be added to \(100 \, ml\) of \(0.4M \, CH_3COONa\) solution to prepare a buffer of pH equal to \(4.91\)? Given that \(pK_a\) of \(CH_3COOH\) is \(4.76\).

• A. \(282.6 \, ml\)
• B. \(213.65 \, ml\)
• C. \(101.41 \, ml\)
• D. \(141.54 \, ml\)

Hint:

For acidic buffer questions, use \(pH = pK_a + \log \frac{[\text{Salt}]}{[\text{Acid}]}\). Always calculate moles first, then convert them into volume using molarity.

Answer 1

The Correct Option is D

Step 1: Identify the buffer system.
Here, \(CH_3COOH\) is a weak acid and \(CH_3COONa\) is the salt of its conjugate base.
So, this is an acidic buffer system.
For acidic buffer, we use Henderson-Hasselbalch equation:
\[ pH = pK_a + \log \frac{[\text{Salt}]}{[\text{Acid}]}. \]

Step 2: Substitute the given values.
Given:
\[ pH = 4.91, \quad pK_a = 4.76. \]
Substituting in the formula:
\[ 4.91 = 4.76 + \log \frac{[\text{Salt}]}{[\text{Acid}]}. \]

Step 3: Find the ratio of salt to acid.
Subtracting \(4.76\) from both sides:
\[ 4.91 - 4.76 = \log \frac{[\text{Salt}]}{[\text{Acid}]}. \]
\[ 0.15 = \log \frac{[\text{Salt}]}{[\text{Acid}]}. \]
Therefore:
\[ \frac{[\text{Salt}]}{[\text{Acid}]} = 10^{0.15}. \]
\[ \frac{[\text{Salt}]}{[\text{Acid}]} \approx 1.412. \]

Step 4: Calculate moles of salt present.
The salt given is \(CH_3COONa\).
Volume of \(CH_3COONa\) solution:
\[ 100 \, ml = 0.1 \, L. \]
Molarity of \(CH_3COONa\):
\[ 0.4M. \]
Moles of salt:
\[ \text{Moles of salt} = Molarity \times Volume. \]
\[ \text{Moles of salt} = 0.4 \times 0.1 = 0.04 \, mol. \]

Step 5: Calculate required moles of acid.
Using the ratio:
\[ \frac{\text{Moles of salt}}{\text{Moles of acid}} = 1.412. \]
\[ \frac{0.04}{\text{Moles of acid}} = 1.412. \]
So,
\[ \text{Moles of acid} = \frac{0.04}{1.412}. \]
\[ \text{Moles of acid} \approx 0.02833 \, mol. \]

Step 6: Calculate volume of \(CH_3COOH\) required.
Molarity of \(CH_3COOH\) is \(0.2M\).
Using the formula:
\[ Molarity = \frac{\text{Moles}}{\text{Volume in L}}. \]
So,
\[ \text{Volume in L} = \frac{\text{Moles}}{Molarity}. \]
\[ \text{Volume of } CH_3COOH = \frac{0.02833}{0.2}. \]
\[ \text{Volume of } CH_3COOH = 0.14165 \, L. \]

Step 7: Convert volume into ml and choose the correct option.
\[ 0.14165 \, L = 141.65 \, ml. \]
This is closest to \(141.54 \, ml\).
Therefore, the correct option is (D).
Final Answer:
The required volume of \(0.2M \, CH_3COOH\) is:
\[ \boxed{141.54 \, ml} \]