Consider the following Redox Reaction:
I3- + 2S2O32- --> 3I- + S4O62-
Given that it requires 36.40 mL of 0.3300 M S2O32- solution to titrate I3- in a 15.00 mL sample. Calculate the molarity of I3- in the aliquot.
Here:
M(S2O32-)=0.3300 M
V(S2O32-)=36.4 mL
V(I3-)=15.0 mL
According to balanced reaction:
1*number of mol of S2O32- =2*number of mol of I3-
1*M(S2O32-)*V(S2O32-) =2*M(I3-)*V(I3-)
1*0.33*36.4 = 2*M(I3-)*15.0
M(I3-) = 0.4004 M
Answer: 0.4004 M
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