Sure, let's break this problem down step-by-step to find the equivalent mass of sodium thiosulfate ([tex]\(Na_2S_2O_3 \cdot 5H_2O\)[/tex]).
### Step 1: Write the Chemical Reaction
The reaction given is:
[tex]\[ 2 S_2O_3^{2-} + I_2 \longrightarrow S_4O_6^{2-} + 2 I^{-} \][/tex]
### Step 2: Identify the Oxidation State Change
The hint suggests:
[tex]\[ 2 S_2O_3^{2-} \longrightarrow S_4O_6^{2-} + 2 e^{-} \][/tex]
From this, we see that two thiosulfate ions ([tex]\(S_2O_3^{2-}\)[/tex]) combine to form one tetrathionate ion ([tex]\(S_4O_6^{2-}\)[/tex]), releasing 2 electrons in the process.
### Step 3: Determine the Molar Mass
We need to determine the molar mass of [tex]\(Na_2S_2O_3 \cdot 5H_2O\)[/tex]. We use the atomic masses of each element:
- Sodium (Na): 23 g/mol
- Sulfur (S): 32 g/mol
- Oxygen (O): 16 g/mol
- Hydrogen (H): 1 g/mol
Calculating the molar mass:
[tex]\[
\text{Molar Mass of } Na_2S_2O_3 \cdot 5H_2O = 2(23) + 2(32) + 3(16) + 5(2(1) + 16) \\
= 2(23) + 2(32) + 3(16) + 5(18) \\
= 46 + 64 + 48 + 90 \\
= 248 \text{ g/mol}
\][/tex]
### Step 4: Number of Electrons Transferred
From the reaction, we see that 2 moles of [tex]\(S_2O_3^{2-}\)[/tex] release 2 electrons (e^-). This means that 1 mole of [tex]\(S_2O_3^{2-}\)[/tex] releases 1 electron.
### Step 5: Calculate the Equivalent Mass
Equivalent mass ([tex]\(E\)[/tex]) is defined as:
[tex]\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{Number of electrons transferred per mole}} \][/tex]
Here, 1 mole of [tex]\( Na_2S_2O_3 \cdot 5H_2O \)[/tex] corresponds to 1 mole of [tex]\( S_2O_3^{2-} \)[/tex] which transfers 1 electron. However, we consider the complete compound itself for the calculation.
Thus, the equivalent mass is:
[tex]\[
E = \frac{\text{Molar Mass of } Na_2S_2O_3 \cdot 5H_2O}{\text{Number of electrons transferred}} = \frac{248 \text{ g/mol}}{2} = 124 \text{ g/equiv}
\][/tex]
### Final Answer:
The equivalent mass of [tex]\(Na_2S_2O_3 \cdot 5H_2O\)[/tex] is [tex]\( 124 \text{ g/equiv} \)[/tex].
