x moles of sulfur get disproportionated in a basic medium and form $N a_{2} S_{2} O_{3}$ and $N a_{2} S$ as:
$S + N a O H \to N a_{2} S_{2} O_{3} + N a_{2} S$
Sodium thiosulfate reacts with iodine in a basic medium to form $N a I$ and $N a_{2} S O_{4}$ by the reaction:
$N a_{2} S_{2} O_{3} + I_{2} + N a O H \to N a I + N a_{2} S O_{4} + H_{2} O$
If $N a I$ is completely precipitated by 2.8 N 100 ml $A g N O_{3}$ solution then what is the value of x?

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Solution

We can use balanced reactions to find the value of x as:
$4 S + 6 N a O H \to N a_{2} S_{2} O_{3} + 2 N a_{2} S + 3 H_{2} O$
$N a_{2} S_{2} O_{3} + 4 I_{2} + 10 N a O H \to 8 N a I + 2 N a_{2} S O_{4} + 5 H_{2} O$
$A g N O_{3} + N a I \to A g I ↓ + N a N O_{3}$
we have the number of moles of sodium iodide and silver nitrate to be equal to 0.28 moles
since x mole sulfur gives x/4 mole of $N a_{2} S_{2} O_{3}$
These produce (x/4)*8 moles of NaI in the second reaction:
2x =0.28 moles of NaI, x = 0.14

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