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?
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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