A mixture of 0.02 mole of `KBrO_(3)` and `0.001` mole of `KBr` was treated with excess of KI and acidified. The volume of 0.01M `Na_(2)S_(2)O_(3)` solution required to consume the liberated iodine will be :

Options

A. 1000mL
B. 1200mL
C. 1500mL
D. 800mL

Answer

To solve the problem, we need to determine the volume of 0.01 M Na₂S₂O₃ solution required to consume the liberated iodine when a mixture of KBrO₃ and KBr is treated with excess KI and acidified. Let's break down the steps:

### Step 1: Identify the reactions
1. **Oxidation-Reduction Reaction**: KBrO₃ is a strong oxidizing agent. In acidic medium, it can oxidize Br⁻ ions from KBr to Br₂.
- The balanced reaction is:
\[ 5 \text{Br}^- + \text{KBrO}_3 + 6 \text{H}^+ \rightarrow 3 \text{Br}_2 + 3 \text{H}_2O + \text{K}^+ \]

### Step 2: Calculate the moles of Br₂ produced
2. **Moles of Br⁻ from KBr**:
- We have 0.001 moles of KBr, which gives us 0.001 moles of Br⁻.
3. **Moles of KBrO₃**:
- We have 0.02 moles of KBrO₃.
4. **Determine the limiting reagent**:
- From the reaction, 5 moles of Br⁻ react with 1 mole of KBrO₃ to produce 3 moles of Br₂.
- The moles of Br₂ produced from 0.001 moles of Br⁻:
\[ \text{Moles of Br₂} = \frac{3}{5} \times 0.001 = 0.0006 \text{ moles} \]

### Step 3: Calculate the moles of I₂ produced
5. **Reaction of Br₂ with KI**:
- Br₂ reacts with KI to produce I₂:
\[ \text{Br}_2 + 2 \text{KI} \rightarrow 2 \text{KBr} + \text{I}_2 \]
- From 0.0006 moles of Br₂, the moles of I₂ produced:
\[ \text{Moles of I₂} = 0.0006 \text{ moles} \]

### Step 4: Calculate the moles of Na₂S₂O₃ required
6. **Reaction of I₂ with Na₂S₂O₃**:
- The reaction is:
\[ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \]
- From 1 mole of I₂, 2 moles of Na₂S₂O₃ are required. Therefore, for 0.0006 moles of I₂:
\[ \text{Moles of Na₂S₂O₃} = 2 \times 0.0006 = 0.0012 \text{ moles} \]

### Step 5: Calculate the volume of Na₂S₂O₃ solution needed
7. **Using the concentration of Na₂S₂O₃**:
- The concentration of Na₂S₂O₃ is 0.01 M.
- Using the formula:
\[ \text{Volume (L)} = \frac{\text{Moles}}{\text{Concentration}} = \frac{0.0012}{0.01} = 0.12 \text{ L} = 120 \text{ mL} \]

### Final Answer
The volume of 0.01 M Na₂S₂O₃ solution required to consume the liberated iodine is **120 mL**.

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