A mixture of `NaHC_(2)O_(4) and KHC_(2)O_(4) .H_(2)C_(2)O_(4)` required equal volumess of `0.2 N KMnO_(4) and 0.12N NaOH ` separtely. What is the molar ration `NaHC_(2)O_(4) and KHC_(2)O_(4) .H_(2)O_(4)` in the mixture?

Options

A. `6:1`
B. `1:6`
C. `1:3`
D. `3:1`

Answer

To find the molar ratio of NaHC₂O₄ and KHC₂O₄·H₂C₂O₄ in the mixture, we can follow these steps:

### Step 1: Define Variables
Let:
- \( X \) = moles of NaHC₂O₄
- \( Y \) = moles of KHC₂O₄·H₂C₂O₄

### Step 2: Analyze the Reactions
1. **KMnO₄ Titration**:
- KMnO₄ reacts with both NaHC₂O₄ and KHC₂O₄·H₂C₂O₄.
- The reaction with NaHC₂O₄ can be represented as:
\[ \text{C}_2\text{O}_4^{2-} + \text{MnO}_4^{-} \rightarrow \text{CO}_2 + \text{Mn}^{2+} \]
- The equivalent factor (nF) for NaHC₂O₄ is 2 (since 2 electrons are involved).
- For KHC₂O₄·H₂C₂O₄, it provides 2 moles of C₂O₄²⁻, thus its nF is 4.

### Step 3: Set Up the First Equation
Using the normality and volume, we can express the equivalents:
\[ \text{Equivalents of KMnO₄} = \text{Equivalents of NaHC₂O₄} + \text{Equivalents of KHC₂O₄·H₂C₂O₄} \]
Given that both solutions require equal volumes \( V \):
\[ 0.2V = 2X + 4Y \quad \text{(1)} \]

### Step 4: Analyze the NaOH Reaction
2. **NaOH Neutralization**:
- NaOH reacts with NaHC₂O₄ (nF = 1) and KHC₂O₄·H₂C₂O₄ (nF = 3).
- The equation for NaOH can be set up as:
\[ \text{Equivalents of NaOH} = \text{Equivalents of NaHC₂O₄} + \text{Equivalents of KHC₂O₄·H₂C₂O₄} \]
Using the normality and volume:
\[ 0.12V = X + 3Y \quad \text{(2)} \]

### Step 5: Solve the System of Equations
Now we have two equations:
1. \( 0.2V = 2X + 4Y \)
2. \( 0.12V = X + 3Y \)

We can divide equation (1) by \( V \) and equation (2) by \( V \):
\[ 0.2 = 2X/V + 4Y/V \quad \text{(1')} \]
\[ 0.12 = X/V + 3Y/V \quad \text{(2')} \]

Let \( a = X/V \) and \( b = Y/V \):
1. \( 0.2 = 2a + 4b \)
2. \( 0.12 = a + 3b \)

### Step 6: Rearranging and Solving
From equation (2'):
\[ a = 0.12 - 3b \quad \text{(3)} \]
Substituting (3) into (1'):
\[ 0.2 = 2(0.12 - 3b) + 4b \]
\[ 0.2 = 0.24 - 6b + 4b \]
\[ 0.2 = 0.24 - 2b \]
\[ 2b = 0.24 - 0.2 = 0.04 \implies b = 0.02 \]

Substituting \( b \) back into (3):
\[ a = 0.12 - 3(0.02) = 0.12 - 0.06 = 0.06 \]

### Step 7: Finding the Molar Ratio
Now we have:
- \( a = X/V = 0.06 \)
- \( b = Y/V = 0.02 \)

The molar ratio \( \frac{X}{Y} \) is:
\[ \frac{X}{Y} = \frac{0.06}{0.02} = 3 \]

Thus, the molar ratio of NaHC₂O₄ to KHC₂O₄·H₂C₂O₄ is **3:1**.