30 ml of 0.2 M NaOH is added with 50 ml `"0.2 M "CH_(3)COOH` solution. The extra volume of 0.2 M NaOH required to make the pH of the solution 5.00 is `(10)/(x).` The value of x is. The ionisation constant of `CH_(3)COOH=2xx10^(-5)`.

Answer

To solve the problem step by step, we will follow the outlined process and calculations based on the information provided.

### Step 1: Calculate the initial millimoles of NaOH and CH₃COOH
- **For NaOH:**
\[ \text{Millimoles of NaOH} = \text{Molarity} \times \text{Volume (ml)} = 0.2 \, \text{M} \times 30 \, \text{ml} = 6 \, \text{mmol} \]

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

### Step 2: Determine the remaining millimoles after the reaction
- The reaction between NaOH and CH₃COOH is:
\[ \text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]
- Since 6 mmol of NaOH reacts with 6 mmol of CH₃COOH:
- Remaining CH₃COOH = \(10 \, \text{mmol} - 6 \, \text{mmol} = 4 \, \text{mmol}\)
- Produced CH₃COONa = 6 mmol

### Step 3: Use the Henderson-Hasselbalch equation to find the initial pH
- The Henderson-Hasselbalch equation is:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
- Given \(K_a = 2 \times 10^{-5}\), we find pKa:
\[ \text{pKa} = -\log(2 \times 10^{-5}) \approx 4.7 \]
- Substitute the values into the equation:
\[ \text{pH} = 4.7 + \log\left(\frac{6}{4}\right) = 4.7 + \log(1.5) \approx 4.7 + 0.176 = 4.88 \]

### Step 4: Set up the equation to find the additional NaOH needed to reach pH 5
- Let \(V\) be the additional volume of 0.2 M NaOH required.
- The additional millimoles of NaOH added:
\[ \text{Additional millimoles} = 0.2 \times V \]
- New amounts after adding \(V\) ml of NaOH:
- New CH₃COOH = \(4 - 0.2V\) mmol
- New CH₃COONa = \(6 + 0.2V\) mmol

### Step 5: Set up the pH equation for the new conditions
- We want the pH to be 5:
\[ 5 = 4.7 + \log\left(\frac{6 + 0.2V}{4 - 0.2V}\right) \]
- Rearranging gives:
\[ 0.3 = \log\left(\frac{6 + 0.2V}{4 - 0.2V}\right) \]
- Taking the antilog:
\[ 10^{0.3} \approx 2 = \frac{6 + 0.2V}{4 - 0.2V} \]

### Step 6: Solve for \(V\)
- Cross-multiplying gives:
\[ 2(4 - 0.2V) = 6 + 0.2V \]
- Expanding and simplifying:
\[ 8 - 0.4V = 6 + 0.2V \]
\[ 8 - 6 = 0.4V + 0.2V \]
\[ 2 = 0.6V \implies V = \frac{2}{0.6} \approx 3.33 \, \text{ml} \]

### Step 7: Find the value of \(x\)
- Given that the extra volume of 0.2 M NaOH required to make the pH 5.00 is \(\frac{10}{x}\):
\[ 3.33 = \frac{10}{x} \implies x = \frac{10}{3.33} \approx 3 \]

### Final Answer
The value of \(x\) is approximately **3**.