$20 \;mL$ iodine solution of unknown molarity are titrated with a $0.02\;M $ sodium thiosulfate solution. The equivalence point is reached after adding $12.1\;mL$ of the thiosulfate solution. Remember → this reaction and give its ionic and "molecular" equation!
$4Na^+$ $+$ $2S_2O_3^{2-}$ $+$ $I_2$ $\longrightarrow$ $2Na^+$ $+$ $S_4O_6^{2-}$ $+$ $2Na^+$ $+$ $2I^-$ $2Na_2S_2O_3$ $+$ $I_2$ $\longrightarrow$ $Na_2S_4O_6$ $+$ $2NaI$ Sodium thiosulfate and iodine produce sodium tetrathionate and sodium iodide.
What is the ratio by which sodium thiosulfate and iodine react ?
Moles of thiosulfate/ moles of iodine = $\frac{2}{1}$
Calculate the number of moles of thiosulfate used!
$n_{Na_2S_2O_3}$ $=$ $12.1\cdot 10^{-3}\cdot 0.02$ $=$ $2.42\cdot10^{-4}\; mol$
Calculate the number of moles of iodine initially present!
$n_{I_2}$ $=$ $\frac{1}{2}\cdot 2.42\cdot10^{-4}= 1.21 \cdot10^{-4}\; mol$
Calculate the molarity of the initial iodine solution!
$c_{I_2}$ $=$ $\frac{1.21 \cdot10^{-4}}{20\cdot10^{-3}}$ $=$ $0.00605\; M$