To solve the problem of determining the amount of `Na2S2O3.5H2O` (sodium thiosulfate pentahydrate) required to completely reduce 100 mL of a 0.25 N iodine solution, we can follow these steps:
### Step 1: Understand the Reaction
The reaction between iodine (I2) and sodium thiosulfate can be represented as:
\[
I_2 + 2 Na_2S_2O_3 \rightarrow 2 NaI + Na_2S_4O_6
\]
From this balanced equation, we can see that 1 mole of I2 reacts with 2 moles of Na2S2O3.
### Step 2: Calculate Millimoles of Iodine
We need to calculate the millimoles of iodine in the solution. The normality (N) of the iodine solution is given as 0.25 N, and the volume is 100 mL.
\[
\text{Millimoles of I2} = \text{Normality} \times \text{Volume (in L)}
\]
\[
\text{Millimoles of I2} = 0.25 \, \text{N} \times 0.1 \, \text{L} = 0.025 \, \text{equivalents} = 25 \, \text{millimoles}
\]
### Step 3: Calculate Millimoles of Sodium Thiosulfate Required
From the balanced equation, we know that 1 mole of I2 reacts with 2 moles of Na2S2O3. Therefore, the millimoles of Na2S2O3 required will be:
\[
\text{Millimoles of Na2S2O3} = 2 \times \text{Millimoles of I2} = 2 \times 25 = 50 \, \text{millimoles}
\]
### Step 4: Calculate the Equivalent Weight of Sodium Thiosulfate
The formula for sodium thiosulfate pentahydrate is `Na2S2O3.5H2O`. We need to calculate its molecular weight:
- Sodium (Na): 23 g/mol × 2 = 46 g/mol
- Sulfur (S): 32 g/mol × 2 = 64 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol
- Water (H2O): 18 g/mol × 5 = 90 g/mol
Adding these together:
\[
\text{Molecular weight of Na2S2O3.5H2O} = 46 + 64 + 48 + 90 = 248 \, \text{g/mol}
\]
### Step 5: Calculate the Weight of Sodium Thiosulfate Required
Now, we can calculate the weight of sodium thiosulfate required using the formula:
\[
\text{Weight} = \text{Millimoles} \times \text{Molecular weight} \times \frac{1}{1000}
\]
\[
\text{Weight} = 50 \, \text{millimoles} \times 248 \, \text{g/mol} \times \frac{1}{1000} = 12.4 \, \text{g}
\]
### Step 6: Final Calculation
The weight of sodium thiosulfate required is:
\[
\text{Weight} = 0.050 \, \text{mol} \times 248 \, \text{g/mol} = 12.4 \, \text{g}
\]
### Conclusion
Thus, the amount of `Na2S2O3.5H2O` required to completely reduce 100 mL of 0.25 N iodine solution is **12.4 grams**.
