Step 1: Understanding the Concept:
A buffer solution is a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid) that resists changes in \( pH \) when small amounts of acid or base are added.
In this problem, acetic acid (\( CH_3COOH \)) is the weak acid and sodium acetate (\( CH_3COONa \)) provides the conjugate base (\( CH_3COO^- \)).
The common ion effect suppresses the dissociation of the weak acid, allowing for the use of the initial concentrations in equilibrium calculations.
The \( pH \) of such an acidic buffer is governed by the Henderson-Hasselbalch equation.
Step 2: Key Formula or Approach:
The Henderson-Hasselbalch equation for an acidic buffer is:
\[ pH = pK_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]
Where:
- \( pH \) is the required acidity level of the buffer.
- \( pK_a \) is the negative log of the acid dissociation constant.
- \( [\text{Salt}] \) is the molar concentration of the conjugate base.
- \( [\text{Acid}] \) is the molar concentration of the weak acid.
Step 3: Detailed Explanation:
From the problem description, we have:
- Target \( pH = 5.5 \)
- \( pK_a \) of acetic acid = 4.5
- Concentration of acid (\( [Acid] \)) = 0.1 M
Plug these values into the Henderson-Hasselbalch equation:
\[ 5.5 = 4.5 + \log \left( \frac{[Salt]}{0.1} \right) \]
Isolate the logarithmic term by subtracting 4.5 from both sides:
\[ 5.5 - 4.5 = \log \left( \frac{[Salt]}{0.1} \right) \]
\[ 1 = \log_{10} \left( \frac{[Salt]}{0.1} \right) \]
To eliminate the log, take the antilog (base 10) of both sides:
\[ 10^1 = \frac{[Salt]}{0.1} \]
\[ 10 = \frac{[Salt]}{0.1} \]
Solve for the salt concentration:
\[ [Salt] = 10 \times 0.1 \]
\[ [Salt] = 1.0 \text{ M} \]
Step 4: Final Answer:
The concentration of sodium acetate required to achieve the desired buffer \( pH \) is 1.0 M.
Therefore, option (3) is the correct answer.