0.2 M NaOH are added to 50 ml of 0.2 M acetic acid `( K_(a) = 1.8 xx 10^(-5))` . The volume of 0.2 M NaOH required to make the pH of solution 4.74 `( -log 1.8 xx 10^(-5) = 4.74 ) `

Options

A. 25ml
B. 50 ml
C. 30 ml
D. 15ml

Answer

To solve the problem, we need to determine the volume of 0.2 M NaOH required to achieve a pH of 4.74 when added to 50 ml of 0.2 M acetic acid.

### Step-by-Step Solution:

1. **Identify the given values:**
- Concentration of acetic acid (HA) = 0.2 M
- Volume of acetic acid = 50 ml = 0.050 L
- Concentration of NaOH = 0.2 M
- Desired pH = 4.74
- \( K_a \) of acetic acid = \( 1.8 \times 10^{-5} \)

2. **Calculate the number of moles of acetic acid:**
\[ \text{Moles of acetic acid} = \text{Concentration} \times \text{Volume} = 0.2 \, \text{mol/L} \times 0.050 \, \text{L} = 0.01 \, \text{mol} = 10 \, \text{mmol} \]

3. **Determine the pKa from the given pH:**
Since the pH is given as 4.74, we know:
\[ pK_a = 4.74 \]

4. **Set up the equilibrium expression for the buffer solution:**
The buffer solution consists of acetic acid (HA) and its conjugate base (A⁻). The pH of a buffer solution can be expressed using the Henderson-Hasselbalch equation:
\[ \text{pH} = pK_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Rearranging gives:
\[ \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) = \text{pH} - pK_a = 4.74 - 4.74 = 0 \]
This implies:
\[ \frac{[\text{A}^-]}{[\text{HA}]} = 1 \]

5. **Establish the relationship between moles:**
Let \( V \) be the volume of NaOH added in liters. The moles of NaOH added will be:
\[ \text{Moles of NaOH} = 0.2 \, \text{mol/L} \times V \, \text{L} = 0.2V \, \text{mol} \]
The moles of acetic acid remaining after reaction with NaOH will be:
\[ \text{Remaining moles of HA} = 10 \, \text{mmol} - 0.2V \, \text{mmol} \]
The moles of the conjugate base (A⁻) formed will be equal to the moles of NaOH added:
\[ [\text{A}^-] = 0.2V \, \text{mmol} \]

6. **Set up the concentration expression:**
The total volume after adding NaOH will be \( 50 \, \text{ml} + V \, \text{ml} \). Therefore, the concentrations are:
\[ [\text{HA}] = \frac{10 - 0.2V}{50 + V} \quad \text{and} \quad [\text{A}^-] = \frac{0.2V}{50 + V} \]

7. **Use the ratio from the Henderson-Hasselbalch equation:**
Since \( \frac{[\text{A}^-]}{[\text{HA}]} = 1 \):
\[ \frac{0.2V}{10 - 0.2V} = 1 \]

8. **Solve for V:**
Cross-multiplying gives:
\[ 0.2V = 10 - 0.2V \]
\[ 0.4V = 10 \]
\[ V = \frac{10}{0.4} = 25 \, \text{ml} \]

### Final Answer:
The volume of 0.2 M NaOH required to make the pH of the solution 4.74 is **25 ml**.