A mixture contaning 0.05 moleof `K_(2)Cr_(2)O_(7) and 0.02 "mole of" KMnO_(4)` was treated eoith excess of KI in acidic medium. The liberated iodine required `1.0L "of" Na_(2)S_(2)O_(3)` solution for titration. Concentration of `Na_(2)S_(2)O_(3)` solution was:

Options

A. `0.4 molL^(-1)`
B. `0.20 molL^(-1)`
C. `0.25 molL^(-1)`
D. `0.30 molL^(-1)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the number of equivalents of K2Cr2O7 and KMnO4
1. **K2Cr2O7**:
- Moles = 0.05
- n-factor (from +6 to +3) = 6 (since 2 Cr atoms each gain 3 electrons)
- Equivalents = Moles × n-factor = 0.05 × 6 = 0.30

2. **KMnO4**:
- Moles = 0.02
- n-factor (from +7 to +2) = 5 (since 1 Mn atom gains 5 electrons)
- Equivalents = Moles × n-factor = 0.02 × 5 = 0.10

### Step 2: Calculate total equivalents of iodine liberated
- Total equivalents of iodine = Equivalents from K2Cr2O7 + Equivalents from KMnO4
- Total equivalents of iodine = 0.30 + 0.10 = 0.40

### Step 3: Relate the equivalents of iodine to Na2S2O3
- The reaction between iodine (I2) and sodium thiosulfate (Na2S2O3) is:
\[ I_2 + 2Na_2S_2O_3 \rightarrow Na_2S_4O_6 + 2NaI \]
- From the reaction, 1 equivalent of I2 reacts with 2 equivalents of Na2S2O3. Therefore, the equivalents of Na2S2O3 needed will be equal to the equivalents of iodine.

### Step 4: Calculate the equivalents of Na2S2O3
- Since we have 0.40 equivalents of iodine, we also need 0.40 equivalents of Na2S2O3.

### Step 5: Use the volume of Na2S2O3 solution to find its concentration
- Volume of Na2S2O3 solution = 1.0 L
- Using the formula:
\[ \text{Equivalents} = \text{Concentration (M)} \times \text{Volume (L)} \]
\[ 0.40 = \text{Concentration} \times 1.0 \]
- Therefore, Concentration = 0.40 M

### Final Answer
The concentration of the Na2S2O3 solution is **0.40 M**.

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