What volume of 0.1 M `Ca(OH)_(2)` will be required neutralise 200 " mL of " 0.2 M `H_(2)SO_(3)` using methyl red indicator to change the colour form pihnk (acidic medium) to yellow (basic medium)?

Options

A. 300 mL
B. 200 mL
C. 100 mL
D. 30 mL

Explanation

To solve the problem of determining the volume of 0.1 M `Ca(OH)_(2)` required to neutralize 200 mL of 0.2 M `H_(2)SO_(3)`, we can follow these steps:

### Step 1: Write the Neutralization Reaction
The neutralization reaction between calcium hydroxide and sulfurous acid can be represented as follows:

\[ \text{H}_2\text{SO}_3 + \text{Ca(OH)}_2 \rightarrow \text{CaSO}_3 + 2\text{H}_2\text{O} \]

### Step 2: Determine the n-factor for `H_(2)SO_(3)`
The n-factor is the number of moles of hydrogen ions (H⁺) that can be provided by one mole of the acid. For `H_(2)SO_(3)`, it can donate 2 H⁺ ions:

- n-factor of `H_(2)SO_(3)` = 2

### Step 3: Calculate the Equivalents of `H_(2)SO_(3)`
To find the equivalents of `H_(2)SO_(3)`, we use the formula:

\[ \text{Equivalents} = \text{Molarity} \times \text{Volume (L)} \]

Convert 200 mL to liters:

\[ 200 \text{ mL} = 0.2 \text{ L} \]

Now, calculate the equivalents:

\[ \text{Equivalents of } H_2SO_3 = 0.2 \, \text{M} \times 0.2 \, \text{L} \times 2 = 0.08 \, \text{equivalents} \]

### Step 4: Determine the n-factor for `Ca(OH)_(2)`
The n-factor for `Ca(OH)_(2)` is 2 because it can provide 2 OH⁻ ions:

- n-factor of `Ca(OH)_(2)` = 2

### Step 5: Set Up the Equation for Neutralization
Using the equivalence concept, we set up the equation:

\[ \text{Equivalents of } H_2SO_3 = \text{Equivalents of } Ca(OH)_2 \]

Let \( V_1 \) be the volume of `Ca(OH)_(2)` required in liters. The equation becomes:

\[ 0.08 = 0.1 \, \text{M} \times V_1 \times 2 \]

### Step 6: Solve for \( V_1 \)
Rearranging the equation gives:

\[ V_1 = \frac{0.08}{0.1 \times 2} = \frac{0.08}{0.2} = 0.4 \, \text{L} \]

Convert liters back to milliliters:

\[ V_1 = 0.4 \, \text{L} \times 1000 = 400 \, \text{mL} \]

### Final Answer
The volume of 0.1 M `Ca(OH)_(2)` required to neutralize 200 mL of 0.2 M `H_(2)SO_(3)` is **400 mL**.

---