If 20.06 mL of Na2S2O3 is required to react with 10.00 mL of 0.0104 M KIO3, what

Question

If 20.06 mL of Na2S2O3 is required to react with 10.00 mL of 0.0104 M KIO3, what is the concentration of the Na2S2O3

Explanation / Answer

The reaction involved is KIO3 + 5KI + 3H2SO4 ---> 3K2SO4 + 3H2O + 3I2 1 mole of KIO3 gives 3 moles of iodine Further 3I2 + 6Na2S2O3 ----> 6NaI + 3Na2S4O6 3 moles of I2 react with 6 moles of Na2S2O3 So 1 mole of KIO3 reacts with 6 moles of Na2S2O3 One important formula is Molarity = moles of solute/liter of solution Moles of KIO3 = 0.010 l of KIO3 x 0.0104 M = 0.000104 moles of KIO3 we know 1 mole of KIO3 reacts with 6 moles of Na2S2O3 0.000104 moles of KIO3 eacts with 0.000624 moles of Na2S2O3 Molarity of Na2S2O3 = 0.000624 moles of Na2S2O3 / 0.02006 l = 0.0311 M

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