Practice Quiz

SAT Chemistry

Category - Multiple Choice

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In the following titration reaction:

I2 + 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 I2 (indicated by the solution turning clear). What was the original concentration of iodine?
  1. 0.10M
  2. 0.25M
  3. 0.5M
  4. 0.6M
  5. 1.0M
Explanation
Answer: C. Solution is as follows:

VS2O3 = 100mL = 0.1 L
[S2O3] = 0.500M
VI2 = 50mL = 0.05L
[I2] = ??

moles of S2O3 = 0.500M * 0.1L = 0.05mol S2O3
from the mole ratio in the balanced chemical equation, 1mol I2 = 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
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