10 mL solution of `H_(2)SO_(4)` and `H_(2)C_(2)O_(4)` (oxalic acid), on titration with with 0.1 M KOH, required 20 " mL of " the base. 10 " mL of " the same solution on titration with `(M)/(300)K_(2)Cr_(2)O_(7)` required 50 " mL of " `K_(2)Cr_(2)O_(7)`.
Q. What should be the volume strength of `H_(2)Cr_(2)O_(7)`, if `H_(2)O_(2)` react with the same volume of `(M)/(300)K_(2)Cr_(2)O_(7)` solution.

Options

A. 5.6 V
B. 0.56 V
C. 11.2 V
D. 1.12 V

Explanation

To solve the problem step by step, we will analyze the titration reactions and calculate the required values.

### Step 1: Determine the equivalents of the acids in the solution
We have a mixture of `H₂SO₄` and `H₂C₂O₄` titrated with `0.1 M KOH`.

- The reaction of `H₂SO₄` with `KOH` is:
\[ H₂SO₄ + 2 KOH \rightarrow K₂SO₄ + 2 H₂O \]
Here, 1 mole of `H₂SO₄` reacts with 2 moles of `KOH`.

- The reaction of `H₂C₂O₄` (oxalic acid) with `KOH` is:
\[ H₂C₂O₄ + 2 KOH \rightarrow K₂C₂O₄ + 2 H₂O \]
Similarly, 1 mole of `H₂C₂O₄` reacts with 2 moles of `KOH`.

### Step 2: Calculate the total equivalents of the acids
Given that 10 mL of the acid solution requires 20 mL of `0.1 M KOH`:
- Moles of `KOH` used:
\[ \text{Moles of KOH} = \text{Volume (L)} \times \text{Molarity} = 0.020 \, \text{L} \times 0.1 \, \text{mol/L} = 0.002 \, \text{mol} \]
- Since `H₂SO₄` and `H₂C₂O₄` both contribute 2 equivalents per mole, the total equivalents of the acids in the solution can be calculated as:
\[ \text{Total equivalents} = 0.002 \, \text{mol} \times 1 \, \text{eq/mol} = 0.002 \, \text{eq} \]

### Step 3: Determine the equivalents of `K₂Cr₂O₇`
Now, we titrate the same 10 mL solution with `K₂Cr₂O₇`:
- Given that 10 mL of the solution requires 50 mL of `K₂Cr₂O₇`:
- Moles of `K₂Cr₂O₇` used:
\[ \text{Moles of K₂Cr₂O₇} = \text{Volume (L)} \times \text{Molarity} = 0.050 \, \text{L} \times \frac{1}{300} \, \text{mol/L} = \frac{0.050}{300} \, \text{mol} = \frac{1}{6000} \, \text{mol} \]

### Step 4: Calculate the equivalents of `K₂Cr₂O₇`
The reaction of `K₂Cr₂O₇` in acidic medium is:
\[ K₂Cr₂O₇ + 6 H^+ + 3 e^- \rightarrow 2 Cr^{3+} + 3 H₂O \]
This means 1 mole of `K₂Cr₂O₇` gives 6 equivalents. Therefore, the equivalents of `K₂Cr₂O₇` used are:
\[ \text{Equivalents of K₂Cr₂O₇} = \frac{1}{6000} \, \text{mol} \times 6 \, \text{eq/mol} = \frac{6}{6000} = \frac{1}{1000} \, \text{eq} \]

### Step 5: Set up the equivalence for `H₂O₂`
Now, we know that the equivalents of `H₂O₂` will equal the equivalents of `K₂Cr₂O₇`:
\[ \text{mEq of H₂O₂} = \text{mEq of K₂Cr₂O₇} \]
Given that `H₂O₂` has a 1:1 stoichiometry with `K₂Cr₂O₇`, we have:
\[ \text{mEq of H₂O₂} = \frac{1}{1000} \, \text{eq} \]

### Step 6: Calculate the normality of `H₂O₂`
Since we have 10 mL of `H₂O₂` solution reacting:
\[ \text{Normality of H₂O₂} = \frac{\text{Equivalents}}{\text{Volume (L)}} = \frac{\frac{1}{1000}}{0.010} = 0.1 \, \text{N} \]

### Step 7: Calculate the volume strength of `H₂Cr₂O₇`
The volume strength of `H₂Cr₂O₇` is given by:
\[ \text{Volume strength} = \text{Normality} \times 5.6 \]
Thus,
\[ \text{Volume strength of H₂Cr₂O₇} = 0.1 \times 5.6 = 0.56 \]

### Final Answer
The volume strength of `H₂Cr₂O₇` is **0.56**.

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