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: Calculate the number of moles of sodium dichloroacetate
Given:
- Volume of sodium dichloroacetate (V) = 50 mL = 0.050 L
- Molarity of sodium dichloroacetate (C) = 0.2 M

Using the formula for moles:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume} = 0.2 \, \text{mol/L} \times 0.050 \, \text{L} = 0.01 \, \text{mol} \]

### Step 2: Determine the pKa from the given pKb
Given:
- pKb of acetate ion = 9.26

Using the relationship:
\[ \text{pKa} = 14 - \text{pKb} = 14 - 9.26 = 4.74 \]

### Step 3: Set up the buffer equation
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

Since we are preparing a buffer at pH 4.74, we can substitute:
\[ 4.74 = 4.74 + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

This simplifies to:
\[ 0 = \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]

### Step 4: Solve for the ratio of salt to acid
From the logarithmic equation, we derive:
\[ \frac{[\text{Salt}]}{[\text{Acid}]} = 1 \]

This means that the number of moles of salt (sodium acetate) must equal the number of moles of acid (sodium dichloroacetate).

### Step 5: Set up the moles of salt and acid
Let \( V \) be the volume of `0.2 M NaOH` added. The moles of NaOH added will be:
\[ \text{Moles of NaOH} = 0.2 \, \text{mol/L} \times V \, \text{L} = 0.2V \, \text{mol} \]

After the reaction, the moles of sodium dichloroacetate will be:
\[ \text{Moles of sodium dichloroacetate remaining} = 0.01 - 0.2V \]

### Step 6: Set up the equation for moles
Since the moles of salt must equal the moles of acid:
\[ 0.2V = 0.01 - 0.2V \]

### Step 7: Solve for V
Rearranging the equation gives:
\[ 0.2V + 0.2V = 0.01 \]
\[ 0.4V = 0.01 \]
\[ V = \frac{0.01}{0.4} = 0.025 \, \text{L} = 25 \, \text{mL} \]

### Final Answer
The volume of `0.2 M NaOH` needed is **25 mL**.

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