A mixture of `K_(2)C_(2)O_(4)` and `KHC_(2)O_(4)` required equal volumes of `0.1 M K_(2)Cr_(2)O_(7)` for oxidation and 0.1 M NOH for neutralisation is separate titration. The molar ratio of `K_(2)CrO_(4)` and `KHC_(2)O_(4)` in the mixture is

Options

A. `1:1`
B. `2:1`
C. `1:2`
D. `3:1`

Answer

To solve the problem, we need to determine the molar ratio of \( K_2Cr_2O_7 \) (potassium dichromate) reacting with the two compounds in the mixture: \( K_2C_2O_4 \) (potassium oxalate) and \( KHC_2O_4 \) (potassium hydrogen oxalate).

### Step-by-Step Solution:

1. **Identify the Compounds and Their Reactions**:
- The two compounds in the mixture are \( K_2C_2O_4 \) and \( KHC_2O_4 \).
- Both compounds will react with \( K_2Cr_2O_7 \) during oxidation and with \( NaOH \) during neutralization.

2. **Write the Oxidation Reactions**:
- The oxidation of \( K_2C_2O_4 \) can be represented as:
\[ K_2C_2O_4 + K_2Cr_2O_7 \rightarrow 2 K_2CO_3 + Cr_2O_7^{2-} \]
- The oxidation of \( KHC_2O_4 \) can be represented as:
\[ KHC_2O_4 + K_2Cr_2O_7 \rightarrow K_2CO_3 + Cr_2O_7^{2-} + CO_2 + H_2O \]

3. **Determine Moles of Reactants**:
- Let \( x \) be the moles of \( K_2C_2O_4 \) and \( y \) be the moles of \( KHC_2O_4 \).
- The volume of \( K_2Cr_2O_7 \) used for oxidation is equal for both compounds, hence:
\[ \text{Moles of } K_2C_2O_4 \text{ oxidized} = \frac{x}{2} \]
\[ \text{Moles of } KHC_2O_4 \text{ oxidized} = y \]

4. **Equate the Moles**:
- Since equal volumes of \( K_2Cr_2O_7 \) are used, we can set the moles equal:
\[ \frac{x}{2} = y \]
- Rearranging gives:
\[ x = 2y \]

5. **Neutralization Reaction**:
- The neutralization of \( K_2C_2O_4 \) and \( KHC_2O_4 \) with \( NaOH \) also occurs in a similar manner:
- \( K_2C_2O_4 \) requires 2 moles of \( NaOH \) for neutralization.
- \( KHC_2O_4 \) requires 1 mole of \( NaOH \) for neutralization.

6. **Set Up the Neutralization Equation**:
- Let the moles of \( NaOH \) used for \( K_2C_2O_4 \) be \( 2x \) and for \( KHC_2O_4 \) be \( y \).
- Since equal volumes are used for neutralization:
\[ 2x = y \]

7. **Combine the Equations**:
- From \( x = 2y \) and \( 2x = y \), substituting \( x \) from the first equation into the second gives:
\[ 2(2y) = y \implies 4y = y \implies 3y = 0 \]
- This indicates a relationship between \( x \) and \( y \).

8. **Final Molar Ratio**:
- From \( x = 2y \), we can express the molar ratio of \( K_2C_2O_4 \) to \( KHC_2O_4 \):
\[ \text{Molar ratio of } K_2C_2O_4 : KHC_2O_4 = 2 : 1 \]

### Conclusion:
The molar ratio of \( K_2C_2O_4 \) to \( KHC_2O_4 \) in the mixture is \( 2:1 \).

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