In permanganate titrations, potassium pemanganats is used an an oxidizing agent in acidic medium.
The medium is maintained acidic by the use of dilute sulphuric acid. Potassium permanganate acts as self indicator . The potential equation, when potassium permanganate acts as an oxidizing agent is
`2KMnO_(4) + 3H_(2)SO_(4) to K_(2)SO_(4) + 2MnSO_(4) + 3H_(2)O + 5[O]`
or `MnO_(4)^(-) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_(2)O`
Before the end point, the solution remains colourless but after teh equivalence point only one extra drop of `KMnO_(4)` solution imparts pink colour , i.e. appearance of pink colour indicates end point. These titrations are used for estimation of ferrous salts, oxalic acid, oxalates, hydrogen peroxide, `As_(2)O_(3)` etc.
Mass of `KHC_(2)O_(4)` (potassium acid oxalate ) required to reduce 100 ml of `0.02 KMnO_(4)` in acidic medium `(MnO_(4)^(-)" changes to " Mn^(2+))` is x gm and to neutralize 100 ml of `0.05M Ca(OH)_(2)` is y gm ,then

Options

A. `x = y`
B. `2x = y`
C. `x = 2y`
D. none of above

Answer

To solve the problem, we need to determine the mass of potassium acid oxalate (KHC₂O₄) required to reduce 100 ml of 0.02 M KMnO₄ and the mass required to neutralize 100 ml of 0.05 M Ca(OH)₂. We will denote the mass required for the KMnO₄ reaction as \( x \) grams and for the Ca(OH)₂ neutralization as \( y \) grams.

### Step 1: Calculate moles of KMnO₄

Given:
- Volume of KMnO₄ solution = 100 ml = 0.1 L
- Molarity of KMnO₄ = 0.02 M

Using the formula for moles:
\[ \text{Moles of KMnO₄} = \text{Molarity} \times \text{Volume} = 0.02 \, \text{mol/L} \times 0.1 \, \text{L} = 0.002 \, \text{mol} = 2 \, \text{mmol} \]

### Step 2: Determine moles of KHC₂O₄ required

From the balanced reaction:
\[ 5 \, \text{HC}_2\text{O}_4^- + 2 \, \text{MnO}_4^- \rightarrow 2 \, \text{Mn}^{2+} + 10 \, \text{CO}_2 + 6 \, \text{H}_2\text{O} \]

From the stoichiometry, 2 moles of KMnO₄ react with 5 moles of HC₂O₄⁻. Therefore, for 2 mmol of KMnO₄:
\[ \text{Moles of HC}_2\text{O}_4^- = \frac{5}{2} \times 2 \, \text{mmol} = 5 \, \text{mmol} \]

### Step 3: Calculate mass of KHC₂O₄

The molar mass of KHC₂O₄ (Potassium acid oxalate) is approximately:
- K = 39.1 g/mol
- H = 1.0 g/mol
- C = 12.0 g/mol × 2 = 24.0 g/mol
- O = 16.0 g/mol × 4 = 64.0 g/mol

Total molar mass = 39.1 + 1.0 + 24.0 + 64.0 = 128.1 g/mol

Now, calculate the mass:
\[ \text{Mass of KHC}_2\text{O}_4 = \text{Moles} \times \text{Molar Mass} = 5 \, \text{mmol} \times \frac{128.1 \, \text{g/mol}}{1000} = 0.6405 \, \text{g} \]
Thus, \( x = 0.6405 \, \text{g} \).

### Step 4: Calculate moles of Ca(OH)₂

Given:
- Volume of Ca(OH)₂ solution = 100 ml = 0.1 L
- Molarity of Ca(OH)₂ = 0.05 M

Using the formula for moles:
\[ \text{Moles of Ca(OH)}_2 = \text{Molarity} \times \text{Volume} = 0.05 \, \text{mol/L} \times 0.1 \, \text{L} = 0.005 \, \text{mol} = 5 \, \text{mmol} \]

### Step 5: Determine moles of KHC₂O₄ required for neutralization

Since Ca(OH)₂ is a dibasic base, it will require twice the amount of KHC₂O₄ to neutralize:
\[ \text{Moles of HC}_2\text{O}_4^- = 2 \times \text{Moles of Ca(OH)}_2 = 2 \times 5 \, \text{mmol} = 10 \, \text{mmol} \]

### Step 6: Calculate mass of KHC₂O₄ for neutralization

Using the same molar mass:
\[ \text{Mass of KHC}_2\text{O}_4 = 10 \, \text{mmol} \times \frac{128.1 \, \text{g/mol}}{1000} = 1.281 \, \text{g} \]
Thus, \( y = 1.281 \, \text{g} \).

### Step 7: Calculate the ratio \( \frac{x}{y} \)

Now we have:
- \( x = 0.6405 \, \text{g} \)
- \( y = 1.281 \, \text{g} \)

The ratio:
\[ \frac{x}{y} = \frac{0.6405}{1.281} = \frac{1}{2} \]

### Conclusion

Thus, the relationship between \( x \) and \( y \) is:
\[ x : y = 1 : 2 \]