In the following titration reaction:
I ₂+ 2 s₂O₃- --> S₄O₆-2 + 2I^-
Starch is used as an indicator. When starch is exposed to iodine (I₂) it is a deep blue color. When the iodine is reduced to iodide (I^-) , the solution will turn clear. Assume that it takes 100mL of 0.500M s₂O₃^- to neutralize 50mL of I ₂(indicated by the solution turning clear) . What was the original concentration of iodine?

Question

Quizplus · Accepted Answer

Solution is as follows: V_s2O3 = 100mL = 0.1 L [s₂O₃] = 0.500M V_{I 2}= 50mL = 0.05L [I₂] = ?? moles of s₂O₃ = 0.500M * 0.1L = 0.05mol s₂O₃ from the mole ratio in the balanced chemical equation, 1mol I ₂= 2mol s₂O₃ 0.05mol s₂O₃ * 1mol I₂/2mol s₂O₃ = 0.025mol I₂ M=n/V which means the concentration of iodine is 0.025mol/0.05L = .5M

In the Following Titration Reaction