The volume of `0.2 M NaOH` needed to prepare a buffer of `pH 4.74` with `50 mL` fo `0.2M` sodium dicholoroacetate acid `(pK_(b) ` of `CH_(3)COO^(-)=9.26)` is `:`

Options

A. `50mL`
B. `25mL`
C. `20mL`
D. `10mL`

Explanation

To solve the problem of finding the volume of `0.2 M NaOH` needed to prepare a buffer of `pH 4.74` with `50 mL` of `0.2 M` sodium dichloroacetate, we can follow these steps:

### Step 1: Understand the Buffer System
A buffer solution consists of a weak acid and its conjugate base. In this case, sodium dichloroacetate acts as the weak acid, and the conjugate base will be formed when we add NaOH.

### Step 2: Use 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 (4.74).
- \(\text{pK}_a\) is the dissociation constant of the weak acid. Since we have the \(pK_b\) of the conjugate base, we can find \(pK_a\) using the relation:
\[ pK_a + pK_b = 14 \]
Thus,
\[ pK_a = 14 - 9.26 = 4.74 \]

### Step 3: Set Up the Equation
Now, substituting the values into the Henderson-Hasselbalch equation:
\[ 4.74 = 4.74 + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
This simplifies to:
\[ 0 = \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
This implies:
\[ \frac{[\text{A}^-]}{[\text{HA}]} = 1 \]
Therefore, the concentrations of the conjugate base and the weak acid must be equal.

### Step 4: Calculate the Amount of Acid
The concentration of the weak acid (sodium dichloroacetate) is `0.2 M` and the volume is `50 mL`. Thus, the moles of the weak acid are:
\[ \text{moles of HA} = 0.2 \, \text{M} \times 0.050 \, \text{L} = 0.01 \, \text{moles} \]

### Step 5: Determine the Moles of Base Needed
Since the ratio of the conjugate base to the weak acid is 1:1, we need `0.01 moles` of the conjugate base (which is formed by adding NaOH).

### Step 6: Calculate the Volume of NaOH Required
The concentration of NaOH is `0.2 M`. To find the volume of NaOH needed to provide `0.01 moles`:
\[ \text{Volume of NaOH} = \frac{\text{moles}}{\text{concentration}} = \frac{0.01 \, \text{moles}}{0.2 \, \text{M}} = 0.05 \, \text{L} = 50 \, \text{mL} \]

### Final Answer
The volume of `0.2 M NaOH` needed to prepare the buffer is **50 mL**.

---