`20 mL` of `0.2M` sodium hydroxide is added to `50 mL` of `0.2 M `acetic acid to give `70 mL` of the solution. What is the `pH` of this solution. Calculate the additional volume of `0.2M NaOh` required to make the `pH` of the solution `4.74`. (Ionisation constant of `CH_(3)COOh` is `1.8 xx 10^(-5))`

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the number of millimoles of NaOH and acetic acid.
- **For NaOH:**
\[ \text{Millimoles of NaOH} = \text{Molarity} \times \text{Volume (mL)} = 0.2 \, \text{M} \times 20 \, \text{mL} = 4 \, \text{mmol} \]

- **For Acetic Acid (CH₃COOH):**
\[ \text{Millimoles of Acetic Acid} = \text{Molarity} \times \text{Volume (mL)} = 0.2 \, \text{M} \times 50 \, \text{mL} = 10 \, \text{mmol} \]

### Step 2: Determine the remaining acetic acid and formed sodium acetate after the reaction.
- **Reaction:**
\[ \text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]
- **Remaining Acetic Acid:**
\[ \text{Remaining Acetic Acid} = 10 \, \text{mmol} - 4 \, \text{mmol} = 6 \, \text{mmol} \]
- **Sodium Acetate Formed:**
\[ \text{Sodium Acetate} = 4 \, \text{mmol} \]

### Step 3: Calculate the pH of the solution using the Henderson-Hasselbalch equation.
- **pKa Calculation:**
\[ \text{pKa} = -\log(K_a) = -\log(1.8 \times 10^{-5}) \approx 4.74 \]
- **Using the Henderson-Hasselbalch equation:**
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Since concentrations are proportional to the number of millimoles:
\[ \text{pH} = 4.74 + \log\left(\frac{4}{6}\right) \]
\[ \text{pH} = 4.74 + \log(0.6667) \approx 4.74 - 0.1761 \approx 4.56 \]

### Step 4: Calculate the additional volume of NaOH required to reach pH 4.74.
- **Let \( V \) be the additional volume of NaOH added.**
- **Millimoles of additional NaOH:**
\[ \text{Millimoles of NaOH} = 0.2 \, V \]
- **New amounts after addition:**
\[ \text{Sodium Acetate} = 4 + 0.2V \quad \text{and} \quad \text{Acetic Acid} = 6 - 0.2V \]
- **Setting up the equation:**
\[ 4.74 = 4.74 + \log\left(\frac{4 + 0.2V}{6 - 0.2V}\right) \]
\[ 0 = \log\left(\frac{4 + 0.2V}{6 - 0.2V}\right) \]
\[ 1 = \frac{4 + 0.2V}{6 - 0.2V} \]
\[ 6 - 0.2V = 4 + 0.2V \]
\[ 2 = 0.4V \]
\[ V = \frac{2}{0.4} = 5 \, \text{mL} \]

### Final Answer:
- The pH of the solution after mixing is approximately **4.56**.
- The additional volume of `0.2 M NaOH` required to make the pH **4.74** is **5 mL**.