This is the procedure: http://www3.chem21labs.com/labfiles/UofC_GL22_Lab.pdf?rf=

Question

This is the procedure: http://www3.chem21labs.com/labfiles/UofC_GL22_Lab.pdf?rf=3899

At the end point of a titration in the standardization of the Na2S2O3 solution (using KIO3), you have a colorless (and clear) solution. At the end point in the determination of Cu(IO3)2 solubility, you have a colorless (but cloudy) heterogeneous mixture. Explain this difference. (Hint: look at the reaction equations on the first page of the handout.)

At the end point of a titration in the standardization of the Na2S2O3 solution (using KIO3), you have a colorless (and clear) solution. At the end point in the determination of Cu(IO3)2 solubility, you have a colorless (but cloudy) heterogeneous mixture. Explain this difference. (Hint: look at the reaction equations on the first page of the handout.)

Explanation / Answer

1) The amount of I3- produced in the following two reactions :

2 Cu2+ + 5 I- -------> 2 CuI(s) + I3-

IO3- + 8 I- + 6 H3O+ --------> 3 I3-.9 H2O

is determined by titration with a standardized Na2S2O3 solution using a starch indicator. This liberated iodine forms a deep-blue complex with starch indicator. The deep blue color will disappear when all the brown I3- has been converted to colorless I- after the reaction with S2O32- .

2 S2O32- + I3- --------> S4O62- + 3 I- (colourless)

2) Cu(IO3)2(s) <------> Cu2+(aq) + 2 IO3-

If the solution is cloudy, it means it contains undissolved solid (Cu(IO3)2), remove it from the solution by filtration and then titrate with Na2S2O3.

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