Moles of `KHC_(2)O_(4)` (potassium acid oxalate) required to reduce 100ml of 0.02M `KMnO_(4)` in acidic medium (to `Mn^(2+)`) is :

Explanation

To solve the problem of determining the moles of potassium acid oxalate (KHC₂O₄) required to reduce 100 mL of 0.02 M KMnO₄ in acidic medium to Mn²⁺, we can follow these steps:

### Step 1: Calculate the number of moles of KMnO₄
First, we need to calculate the number of moles of KMnO₄ present in the solution. The formula to calculate moles is:

\[ \text{Moles} = \text{Molarity} \times \text{Volume (in liters)} \]

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

Calculating the moles:

\[ \text{Moles of KMnO₄} = 0.02 \, \text{mol/L} \times 0.1 \, \text{L} = 0.002 \, \text{mol} \]

### Step 2: Determine the n-factor of KMnO₄
In acidic medium, KMnO₄ is reduced from Mn(VII) to Mn(II). The change in oxidation state is:

\[ \text{Change in oxidation state} = 7 - 2 = 5 \]

Thus, the n-factor for KMnO₄ is 5.

### Step 3: Calculate the equivalents of KMnO₄
The number of equivalents of KMnO₄ can be calculated using the formula:

\[ \text{Equivalents} = \text{Moles} \times \text{n-factor} \]

Calculating the equivalents:

\[ \text{Equivalents of KMnO₄} = 0.002 \, \text{mol} \times 5 = 0.01 \, \text{equivalents} \]

### Step 4: Determine the n-factor of KHC₂O₄
Potassium acid oxalate (KHC₂O₄) is oxidized to CO₂. The oxidation state of carbon in KHC₂O₄ is +3, and it changes to +4 in CO₂. Therefore, the change in oxidation state is:

\[ \text{Change in oxidation state} = 4 - 3 = 1 \]

Since there are two carbon atoms in KHC₂O₄, the n-factor for KHC₂O₄ is:

\[ \text{n-factor} = 2 \times 1 = 2 \]

### Step 5: Calculate the equivalents of KHC₂O₄
Let the number of moles of KHC₂O₄ be \( n \). The equivalents of KHC₂O₄ can be expressed as:

\[ \text{Equivalents of KHC₂O₄} = n \times 2 \]

### Step 6: Set up the equivalence equation
Since the equivalents of the reducing agent (KHC₂O₄) must equal the equivalents of the oxidizing agent (KMnO₄), we can set up the equation:

\[ n \times 2 = 0.01 \]

### Step 7: Solve for n (moles of KHC₂O₄)
Now, we can solve for \( n \):

\[ n = \frac{0.01}{2} = 0.005 \, \text{mol} \]

### Final Answer
The moles of KHC₂O₄ required to reduce 100 mL of 0.02 M KMnO₄ in acidic medium is **0.005 moles**.

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