`'a' g KHC_(2)O_(4)` are required to reduce `100mL` of `0.02M KMnO_(4)` in acid medium and `'b' g KHC_(2)O_(4)` neutralises `100mL` of `0.05M Ca(OH)_(2)` then `:`

Options

A. `a=b`
B. `2a=b`
C. `a=2b`
D. None of these

Explanation

To solve the problem step by step, we need to analyze the two reactions involving potassium hydrogen oxalate (KHC₂O₄) and find the relationship between the grams of KHC₂O₄ required in each case.

### Step 1: Calculate the amount of KHC₂O₄ required to reduce KMnO₄

1. **Identify the reaction**: KHC₂O₄ reduces KMnO₄ in an acidic medium.
2. **Determine oxidation states**:
- In KMnO₄, Mn is in the +7 oxidation state and is reduced to +2 in Mn²⁺.
- The change in oxidation state for Mn is \( +7 \to +2 \), which is a change of 5 units.
3. **Determine the oxidation state of carbon in KHC₂O₄**:
- For KHC₂O₄, the oxidation states of carbon can be calculated as follows:
- Let the oxidation state of carbon be \( x \).
- The equation is \( 2x + 1 - 4 = 0 \) (considering the charge balance).
- Solving gives \( x = +3 \) for KHC₂O₄, and when it is oxidized to CO₂, the oxidation state becomes +4.
- Therefore, the change in oxidation state per carbon atom is \( +1 \) and for 2 carbon atoms, it is \( 2 \times +1 = +2 \).
4. **Calculate the n-factor for KHC₂O₄**:
- The n-factor for KHC₂O₄ in this reaction is 2 (since it has 2 carbon atoms contributing to the oxidation).
5. **Calculate the equivalents of KMnO₄**:
- Molarity of KMnO₄ = 0.02 M, Volume = 100 mL = 0.1 L.
- Number of moles of KMnO₄ = Molarity × Volume = \( 0.02 \times 0.1 = 0.002 \) moles.
- Equivalents of KMnO₄ = Moles × n-factor = \( 0.002 \times 5 = 0.01 \) equivalents.
6. **Calculate the equivalents of KHC₂O₄**:
- Let \( A \) be the grams of KHC₂O₄ required.
- Equivalents of KHC₂O₄ = \( \frac{A}{Molar\ mass\ of\ KHC₂O₄} \).
- Setting the equivalents equal: \( 0.01 = \frac{A}{Molar\ mass\ of\ KHC₂O₄} \).

### Step 2: Calculate the amount of KHC₂O₄ required to neutralize Ca(OH)₂

1. **Identify the reaction**: KHC₂O₄ neutralizes Ca(OH)₂.
2. **Determine the n-factor**:
- For Ca(OH)₂, the n-factor is 2 (it can donate 2 OH⁻ ions).
- For KHC₂O₄, the n-factor is 1 (it can accept 1 H⁺ ion).
3. **Calculate the equivalents of Ca(OH)₂**:
- Molarity of Ca(OH)₂ = 0.05 M, Volume = 100 mL = 0.1 L.
- Number of moles of Ca(OH)₂ = \( 0.05 \times 0.1 = 0.005 \) moles.
- Equivalents of Ca(OH)₂ = Moles × n-factor = \( 0.005 \times 2 = 0.01 \) equivalents.
4. **Calculate the equivalents of KHC₂O₄**:
- Let \( B \) be the grams of KHC₂O₄ required.
- Equivalents of KHC₂O₄ = \( \frac{B}{Molar\ mass\ of\ KHC₂O₄} \).
- Setting the equivalents equal: \( 0.01 = \frac{B}{Molar\ mass\ of\ KHC₂O₄} \).

### Step 3: Relate A and B

1. From the first reaction, we have:
\[ A = 0.01 \times Molar\ mass\ of\ KHC₂O₄ \]
2. From the second reaction, we have:
\[ B = 0.01 \times Molar\ mass\ of\ KHC₂O₄ \]
3. Therefore, we can write the relationship:
\[ A = \frac{1}{2}B \quad \text{or} \quad B = 2A \]

### Conclusion

The relationship between the grams of KHC₂O₄ required in the two reactions is:
\[ B = 2A \]