How many ml of 0.3M `K_2 Cr_2 O_7`(acidic) is required for complete oxidation of 5 ml of 0.2 M `SnC_2 O_4` solution.

Options

A. 3.33 ml
B. 2.22 ml
C. 11 ml
D. 4.44 ml

Answer

To solve the problem of how many ml of 0.3 M \( K_2Cr_2O_7 \) (acidic) is required for the complete oxidation of 5 ml of 0.2 M \( SnC_2O_4 \) solution, we will follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between potassium dichromate (\( K_2Cr_2O_7 \)) and oxalate ion (\( C_2O_4^{2-} \)) in acidic medium can be represented as follows:

\[ K_2Cr_2O_7 + 3 SnC_2O_4 + 8 H_2SO_4 \rightarrow 2 K_2SO_4 + Cr_2(SO_4)_3 + 6 CO_2 + 3 SnO_2 + 8 H_2O \]

From the balanced equation, we can see that 2 moles of \( K_2Cr_2O_7 \) react with 3 moles of \( SnC_2O_4 \).

### Step 2: Calculate moles of \( SnC_2O_4 \)
We need to calculate the moles of \( SnC_2O_4 \) in the given solution:

\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]

Given:
- Molarity of \( SnC_2O_4 = 0.2 \, M \)
- Volume of \( SnC_2O_4 = 5 \, ml = 0.005 \, L \)

Calculating the moles of \( SnC_2O_4 \):

\[ \text{moles of } SnC_2O_4 = 0.2 \times 0.005 = 0.001 \, moles \]

### Step 3: Determine moles of \( K_2Cr_2O_7 \) required
From the balanced equation, we know that:

\[ \frac{2 \, \text{moles of } K_2Cr_2O_7}{3 \, \text{moles of } SnC_2O_4} \]

Using the moles of \( SnC_2O_4 \) calculated:

\[ \text{moles of } K_2Cr_2O_7 = \left( \frac{2}{3} \right) \times 0.001 = \frac{2 \times 10^{-3}}{3} \approx 0.0006667 \, moles \]

### Step 4: Calculate the volume of \( K_2Cr_2O_7 \) needed
Using the molarity of \( K_2Cr_2O_7 \):

\[ \text{Molarity of } K_2Cr_2O_7 = 0.3 \, M \]

Using the formula for molarity:

\[ \text{Molarity} = \frac{\text{moles}}{\text{volume in liters}} \]

Rearranging to find volume:

\[ \text{Volume} = \frac{\text{moles}}{\text{Molarity}} = \frac{0.0006667}{0.3} \approx 0.002222 \, L \]

Converting to ml:

\[ \text{Volume} = 0.002222 \times 1000 = 2.222 \, ml \]

### Final Answer
The volume of 0.3 M \( K_2Cr_2O_7 \) required is approximately **2.22 ml**.

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