Question

# A 10.0 mL sample of aqueous NaOCL is treated with excess KI in an acidic solution....

A 10.0 mL sample of aqueous NaOCL is treated with excess KI in an acidic solution. The quantity of iodine that is liberated is such that 28.02 mL of 0.025M Na2S2O3 solution must be added to cause the disappearance of the dark blue color due to the starch indicator. What is the molarity of the solution of NaOCL?

#### Homework Answers

Answer #1

NaOCl+ 2KI + H2O --> I2 + NaCl +2KOH

the above liberated I2 reacts with Na2S2O3

2 Na2S2O3 + I2 ........> Na2S4O6 + 2 NaI

from the abovev reaction we can say that

2 moles of Na2S2O3 reacts with 1 mole of I2

given no.of moles = molarity*vol. in lt. = 0.025*0.02802= 7.005*10^-4 moles

2 moles of Na2S2O3 reacts with 1 mole of I2

for 7.005*10^-4 moles of Na2S2O3 .............?

= 7.005*10^-4*1/2 = 3.5*10^-4 moles of I2

For 1 mole of NaOCl ......................1mole of I2 releases.

?................................................. 3.5*10^-4 moles of I2

= 3.5*10^-4 moles of NaOCl

now

Molarity of NaOCl = (no.of moles/vol. in lt.) = (3.5*10^-4/0.01) = 3.5*10^-2 M

molarity of the solution of NaOCL is 0.035M

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