The ratio of volumes of `CH_(3)C O O H`, 0.1 N to `CH_(3) C O O Na`, 0.1 N required to prepare a buffer solution of pH 5.74 is (given : `pK_(a)` of `CH_(3) C O O H` is 4+74 )

Options

A. `10:1`
B. `5:1`
C. `1:5`
D. `1:10`

Explanation

To solve the problem of finding the ratio of volumes of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa) required to prepare a buffer solution of pH 5.74, we can use the Henderson-Hasselbalch equation. Here are the steps to arrive at the solution:

### Step 1: Write the Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
where:
- \(\text{pH}\) is the desired pH of the buffer solution,
- \(\text{pK}_a\) is the acid dissociation constant of the weak acid,
- \([\text{A}^-]\) is the concentration of the conjugate base (sodium acetate),
- \([\text{HA}]\) is the concentration of the weak acid (acetic acid).

### Step 2: Substitute the Known Values
Given:
- \(\text{pH} = 5.74\)
- \(\text{pK}_a = 4.74\)

Substituting these values into the equation:
\[ 5.74 = 4.74 + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]

### Step 3: Rearrange the Equation
Rearranging the equation to isolate the logarithmic term:
\[ \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = 5.74 - 4.74 \]
\[ \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) = 1.00 \]

### Step 4: Convert from Logarithmic to Exponential Form
To eliminate the logarithm, we convert it to exponential form:
\[ \frac{[\text{A}^-]}{[\text{HA}]} = 10^{1.00} \]
\[ \frac{[\text{A}^-]}{[\text{HA}]} = 10 \]

### Step 5: Interpret the Ratio
The ratio of the concentrations of the conjugate base to the weak acid is 10:1. Since the concentrations of the solutions are the same (0.1 N), the ratio of the volumes will also be the same.

### Step 6: Conclusion
Thus, the ratio of the volumes of CH₃COOH to CH₃COONa required to prepare the buffer solution is:
\[ \text{Volume of } CH₃COOH : \text{Volume of } CH₃COONa = 1 : 10 \]

### Final Answer
The ratio of volumes of CH₃COOH (0.1 N) to CH₃COONa (0.1 N) required is **1:10**.

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