To solve the problem step by step, we will analyze the reactions involved and use the concept of equivalents to find the value of \( x \).
### Step 1: Understand the Problem
We have a \( 100 \, \text{mL} \) solution of \( \text{Na}_2\text{S}_2\text{O}_3 \) that is divided into two equal parts (A and B). Part A reacts with \( 12.5 \, \text{mL} \) of \( 0.2 \, \text{M} \, \text{I}_2 \) in an acidic medium, and part B is diluted \( x \) times, with \( 50 \, \text{mL} \) of the diluted solution requiring \( 5 \, \text{mL} \) of \( 0.8 \, \text{M} \, \text{I}_2 \) in a basic medium. We need to find the value of \( x \).
### Step 2: Calculate the Moles of \( \text{I}_2 \) in Part A
The moles of \( \text{I}_2 \) used in part A can be calculated using the formula:
\[
\text{Moles of } I_2 = \text{Volume (L)} \times \text{Molarity}
\]
\[
\text{Moles of } I_2 = 0.0125 \, \text{L} \times 0.2 \, \text{mol/L} = 0.0025 \, \text{mol}
\]
### Step 3: Determine the Equivalent of \( \text{Na}_2\text{S}_2\text{O}_3 \) in Part A
The reaction for part A is:
\[
2 \, \text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2 \, \text{NaI}
\]
From the reaction, we see that 2 moles of \( \text{Na}_2\text{S}_2\text{O}_3 \) react with 1 mole of \( \text{I}_2 \). Therefore, the moles of \( \text{Na}_2\text{S}_2\text{O}_3 \) that reacted is:
\[
\text{Moles of } \text{Na}_2\text{S}_2\text{O}_3 = 2 \times \text{Moles of } I_2 = 2 \times 0.0025 = 0.005 \, \text{mol}
\]
### Step 4: Calculate the Millimoles of \( \text{Na}_2\text{S}_2\text{O}_3 \)
Since we have \( 100 \, \text{mL} \) of the solution, the total millimoles of \( \text{Na}_2\text{S}_2\text{O}_3 \) in the original solution is:
\[
\text{Millimoles} = 0.005 \, \text{mol} \times 1000 = 5 \, \text{mmol}
\]
### Step 5: Analyze Part B
In part B, the reaction is:
\[
\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow 2 \, \text{SO}_4^{2-} + 2 \, \text{I}^-
\]
Here, the moles of \( \text{I}_2 \) used in part B is:
\[
\text{Moles of } I_2 = 5 \, \text{mL} \times 0.8 \, \text{mol/L} = 0.004 \, \text{mol}
\]
### Step 6: Calculate the Equivalent of \( \text{Na}_2\text{S}_2\text{O}_3 \) in Part B
From the reaction stoichiometry, 1 mole of \( \text{S}_2\text{O}_3^{2-} \) reacts with 1 mole of \( \text{I}_2 \). Therefore, the moles of \( \text{S}_2\text{O}_3^{2-} \) that reacted is also \( 0.004 \, \text{mol} \).
### Step 7: Calculate the Total Millimoles of \( \text{S}_2\text{O}_3^{2-} \) in Part B
Since we are using \( 50 \, \text{mL} \) of the diluted solution:
\[
\text{Millimoles of } S_2O_3^{2-} = 0.004 \, \text{mol} \times 1000 = 4 \, \text{mmol}
\]
### Step 8: Set Up the Dilution Equation
Let \( M_1 \) be the concentration of \( \text{Na}_2\text{S}_2\text{O}_3 \) in the original solution, and \( M_2 \) be the concentration after dilution. The relationship between the original and diluted concentrations is given by:
\[
M_1 \times 50 \, \text{mL} = M_2 \times 50 \, \text{mL} \times x
\]
Substituting the values:
\[
5 \, \text{mmol} = 4 \, \text{mmol} \times x
\]
Solving for \( x \):
\[
x = \frac{5}{4} = 1.25
\]
### Final Answer
The value of \( x \) is \( 1.25 \).
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