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 process of determining the amount of NaOH required to achieve a pH of 5.00 when mixed with acetic acid (CH₃COOH).

### Step 1: Calculate the initial moles of CH₃COOH and NaOH

1. **Calculate moles of CH₃COOH:**
\[ \text{Volume of CH}_3\text{COOH} = 50 \, \text{ml} = 0.050 \, \text{L} \]
\[ \text{Concentration of CH}_3\text{COOH} = 0.2 \, \text{M} \]
\[ \text{Moles of CH}_3\text{COOH} = 0.2 \, \text{mol/L} \times 0.050 \, \text{L} = 0.01 \, \text{mol} \]

2. **Calculate moles of NaOH:**
\[ \text{Volume of NaOH} = 30 \, \text{ml} = 0.030 \, \text{L} \]
\[ \text{Concentration of NaOH} = 0.2 \, \text{M} \]
\[ \text{Moles of NaOH} = 0.2 \, \text{mol/L} \times 0.030 \, \text{L} = 0.006 \, \text{mol} \]

### Step 2: Determine the reaction between CH₃COOH and NaOH

The reaction between acetic acid and sodium hydroxide is:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

Since we have 0.01 mol of acetic acid and 0.006 mol of NaOH, NaOH is the limiting reagent.

### Step 3: Calculate remaining moles after neutralization

1. **Moles of CH₃COOH remaining:**
\[ \text{Moles of CH}_3\text{COOH remaining} = 0.01 - 0.006 = 0.004 \, \text{mol} \]

2. **Moles of CH₃COONa formed:**
\[ \text{Moles of CH}_3\text{COONa} = 0.006 \, \text{mol} \]

### Step 4: Introduce additional NaOH

Let \( V \) be the additional volume of 0.2 M NaOH added. The moles of additional NaOH will be:
\[ \text{Moles of additional NaOH} = 0.2 \times \frac{V}{1000} = 0.0002V \, \text{mol} \]

### Step 5: Calculate total moles after adding additional NaOH

After adding \( V \) ml of NaOH:
- Total moles of NaOH = \( 0.006 + 0.0002V \)
- Moles of CH₃COOH remaining = \( 0.004 \, \text{mol} \)
- Moles of CH₃COONa = \( 0.006 \, \text{mol} \)

### Step 6: Use the Henderson-Hasselbalch equation

The pH of a buffer solution is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right) \]
Where:
- \( \text{pKa} = -\log(2 \times 10^{-5}) \approx 4.7 \)
- \( [\text{A}^-] = \frac{0.006 + 0.0002V}{\text{Total Volume}} \)
- \( [\text{HA}] = \frac{0.004}{\text{Total Volume}} \)

Total volume = \( 80 + V \) ml.

### Step 7: Set up the equation for pH = 5

Substituting into the Henderson-Hasselbalch equation:
\[ 5 = 4.7 + \log\left(\frac{0.006 + 0.0002V}{0.004}\right) \]
This simplifies to:
\[ 0.3 = \log\left(\frac{0.006 + 0.0002V}{0.004}\right) \]

### Step 8: Solve for V

Convert the logarithmic equation to exponential form:
\[ 10^{0.3} = \frac{0.006 + 0.0002V}{0.004} \]
Calculating \( 10^{0.3} \approx 2 \):
\[ 2 = \frac{0.006 + 0.0002V}{0.004} \]
Cross-multiplying gives:
\[ 0.008 = 0.006 + 0.0002V \]
Rearranging yields:
\[ 0.0002V = 0.002 \implies V = \frac{0.002}{0.0002} = 10 \, \text{ml} \]

### Step 9: Find x

Given that the extra volume of 0.2 M NaOH required to make the pH 5.00 is \( \frac{10}{x} \):
\[ 10 = \frac{10}{x} \implies x = 1 \]

### Final Answer
The value of \( x \) is \( 1 \).