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

Options

A. `200 mL`
B. `400 mL`
C. `20 mL`
D. `40 mL`

Explanation

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

### Step 1: Determine the reaction and the n-factor
`H2SO3` (sulfurous acid) dissociates in water to produce `H+` ions and `HSO3-` ions. The first ionization can be represented as:
\[ H2SO3 \rightarrow H^+ + HSO3^- \]

For this reaction, the n-factor (number of H+ ions that can react) is 1 because one mole of `H2SO3` produces one mole of `H+`.

### Step 2: Calculate the equivalents of `H2SO3`
To find the equivalents of `H2SO3`, we use the formula:
\[ \text{Equivalents} = \text{Molarity} \times \text{Volume (in L)} \]
Given:
- Molarity of `H2SO3` = 0.2 M
- Volume of `H2SO3` = 200 mL = 0.2 L

Calculating the equivalents:
\[ \text{Equivalents of } H2SO3 = 0.2 \, \text{M} \times 0.2 \, \text{L} = 0.04 \, \text{equivalents} \]

### Step 3: Determine the n-factor for `Ca(OH)2`
`Ca(OH)2` dissociates in water to produce:
\[ Ca(OH)2 \rightarrow Ca^{2+} + 2OH^- \]
The n-factor for `Ca(OH)2` is 2 because it can donate 2 moles of `OH-` ions.

### Step 4: Set up the equation for neutralization
Using the equivalence concept:
\[ n_1 \times V_1 = n_2 \times V_2 \]
Where:
- \( n_1 \) = n-factor of `Ca(OH)2` = 2
- \( V_1 \) = volume of `Ca(OH)2` required (in L)
- \( n_2 \) = n-factor of `H2SO3` = 1
- \( V_2 \) = volume of `H2SO3` = 0.2 L

Substituting the values:
\[ 2 \times V_1 = 1 \times 0.04 \]
\[ 2 \times V_1 = 0.04 \]

### Step 5: Solve for \( V_1 \)
\[ V_1 = \frac{0.04}{2} = 0.02 \, \text{L} \]

### Step 6: Convert to mL
To convert liters to milliliters:
\[ V_1 = 0.02 \, \text{L} \times 1000 \, \text{mL/L} = 20 \, \text{mL} \]

### 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 **20 mL**.

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