A 20 mL sample of 0.40 mol/L bleach solution, NaOCl, is reacted with an excess of KI solution according to the following reaction equation:
2H+ + OCl- + 2I- --> Cl- + H2O + I2
The I2 produced is subsequently reacted according to the equation
2 S2O32- + I2 --> S4O62- + 2I-
What volume of 1.00 mol/L Na2S2O3 is required to titrate the I2?
Thanks! Any help would be appreciated. I dont really care for the answer I just dont know how to go about answering it.
2H+ + OCl- + 2I- --> Cl- + H2O + I2
The I2 produced is subsequently reacted according to the equation
2 S2O32- + I2 --> S4O62- + 2I-
What volume of 1.00 mol/L Na2S2O3 is required to titrate the I2?
Thanks! Any help would be appreciated. I dont really care for the answer I just dont know how to go about answering it.
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Since all of your equations are balanced, it is just a matter of a couple stoichiometry calculations:
0.020 L X 0.40 mol/L OCl- = 8.0 X 10^-3 mol OCl-
8.0 X 10^-3 mol OCl- X (1 mol I2 / 1 mol OCl-) X (2 mol S2O32-/ 1 mol I2) = 0.016 mol S2O32-
0.016 mol S2O32- X (1 L/1 mol) = 0.016 L = 16 mL
Hope that helps...
0.020 L X 0.40 mol/L OCl- = 8.0 X 10^-3 mol OCl-
8.0 X 10^-3 mol OCl- X (1 mol I2 / 1 mol OCl-) X (2 mol S2O32-/ 1 mol I2) = 0.016 mol S2O32-
0.016 mol S2O32- X (1 L/1 mol) = 0.016 L = 16 mL
Hope that helps...