`'a' gm` `KHC_(2)O_(4)` is used to neutralize 100 mL of 0.02 M `KMnO_(4)` in acid medium, where as `'b' gm` `KHC_(2)O_(4)` is used to neutralize 100 mL of 0.02 M `Ca(OH)_(2)` then,

Options

A. `a=b`
B. `4a=5b`
C. `5a=4b`
D. `2a=b`

Answer

To solve the problem, we need to find the relationship between the masses 'a' and 'b' of potassium hydrogen tartrate (KHC₂O₄) used to neutralize potassium permanganate (KMnO₄) and calcium hydroxide (Ca(OH)₂) respectively.

### Step-by-Step Solution:

1. **Calculate the number of equivalents of KMnO₄:**
- The molarity of KMnO₄ is given as 0.02 M and the volume is 100 mL (0.1 L).
- The number of moles of KMnO₄ can be calculated using the formula:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (L)} = 0.02 \, \text{mol/L} \times 0.1 \, \text{L} = 0.002 \, \text{mol} \]
- In acidic medium, KMnO₄ undergoes reduction where Mn changes from +7 to +2, which involves a change of 5 electrons.
- Therefore, the number of equivalents of KMnO₄ is:
\[ \text{Number of equivalents} = \text{Number of moles} \times \text{n} = 0.002 \, \text{mol} \times 5 = 0.01 \, \text{equivalents} \]

2. **Calculate the number of equivalents of KHC₂O₄:**
- Let 'a' grams of KHC₂O₄ be used to neutralize KMnO₄.
- The equivalent weight of KHC₂O₄ is given by:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{\text{x-factor}} \]
- KHC₂O₄ dissociates to give one H⁺ ion, hence its x-factor is 1.
- The molar mass of KHC₂O₄ is approximately 188 g/mol.
- Therefore, the equivalent weight of KHC₂O₄ is:
\[ \text{Equivalent weight} = \frac{188 \, \text{g/mol}}{1} = 188 \, \text{g/equiv} \]
- The number of equivalents of KHC₂O₄ can be expressed as:
\[ \text{Number of equivalents} = \frac{a}{188} \]
- Setting the equivalents equal gives:
\[ \frac{a}{188} = 0.01 \implies a = 0.01 \times 188 = 1.88 \, \text{g} \]

3. **Calculate the number of equivalents of Ca(OH)₂:**
- The molarity of Ca(OH)₂ is also given as 0.02 M and the volume is 100 mL (0.1 L).
- The number of moles of Ca(OH)₂ is:
\[ \text{Number of moles} = 0.02 \, \text{mol/L} \times 0.1 \, \text{L} = 0.002 \, \text{mol} \]
- Each mole of Ca(OH)₂ can donate 2 OH⁻ ions, thus:
\[ \text{Number of equivalents} = \text{Number of moles} \times 2 = 0.002 \, \text{mol} \times 2 = 0.004 \, \text{equivalents} \]

4. **Calculate the number of equivalents of KHC₂O₄ used to neutralize Ca(OH)₂:**
- Let 'b' grams of KHC₂O₄ be used to neutralize Ca(OH)₂.
- Using the equivalent weight calculated earlier:
\[ \frac{b}{188} = 0.004 \implies b = 0.004 \times 188 = 0.752 \, \text{g} \]

5. **Establish the relationship between A and B:**
- From the calculations, we have:
\[ a = 1.88 \, \text{g} \quad \text{and} \quad b = 0.752 \, \text{g} \]
- To find the relationship, we can express it as:
\[ \frac{4a}{5} = b \]
- Rearranging gives:
\[ 4a = 5b \]

### Final Result:
The relationship between 'a' and 'b' is:
\[ 4a = 5b \]