A student needs to prepare a buffer solution of propanoic acid and its sodium salt with pH 4.
The ratio of `([CH_3CH_2COO^(-)])/([CH_3CH_2COOH])` required to make buffer is ________ .
Given : `K_a(CH_3CH_2COOH)=1.3 xx 10^(-5)`

Options

A. 0.03
B. 0.13
C. 0.23
D. 0.33

Explanation

To solve the problem of preparing a buffer solution of propanoic acid and its sodium salt with a pH of 4, we will use the Henderson-Hasselbalch equation. Here are the steps to find the required ratio of \([CH_3CH_2COO^-]\) (the salt) to \([CH_3CH_2COOH]\) (the acid):

### Step 1: Calculate \(pK_a\)
The dissociation constant \(K_a\) for propanoic acid is given as \(1.3 \times 10^{-5}\). To find \(pK_a\), we use the formula:
\[ pK_a = -\log(K_a) \]
Substituting the value:
\[ pK_a = -\log(1.3 \times 10^{-5}) = 5 - \log(1.3) \]
Calculating \(\log(1.3)\) (approximately 0.113):
\[ pK_a \approx 5 - 0.113 = 4.887 \]

### Step 2: Apply the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation for a buffer solution is given by:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Where:
- \(pH\) is the desired pH of the buffer (4 in this case),
- \(pK_a\) is the calculated \(pK_a\) (approximately 4.887),
- \([A^-]\) is the concentration of the salt (\(CH_3CH_2COO^-\)),
- \([HA]\) is the concentration of the acid (\(CH_3CH_2COOH\)).

### Step 3: Rearranging the equation
Substituting the known values into the equation:
\[ 4 = 4.887 + \log\left(\frac{[CH_3CH_2COO^-]}{[CH_3CH_2COOH]}\right) \]
Rearranging gives:
\[ \log\left(\frac{[CH_3CH_2COO^-]}{[CH_3CH_2COOH]}\right) = 4 - 4.887 \]
\[ \log\left(\frac{[CH_3CH_2COO^-]}{[CH_3CH_2COOH]}\right) = -0.887 \]

### Step 4: Exponentiating to find the ratio
To find the ratio, we exponentiate both sides:
\[ \frac{[CH_3CH_2COO^-]}{[CH_3CH_2COOH]} = 10^{-0.887} \]
Calculating \(10^{-0.887}\) (approximately 0.13):
\[ \frac{[CH_3CH_2COO^-]}{[CH_3CH_2COOH]} \approx 0.13 \]

### Conclusion
The ratio of \([CH_3CH_2COO^-]\) to \([CH_3CH_2COOH]\) required to make the buffer solution is approximately **0.13**.

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