In the titration of Na2S2O3 against KIO3 the average titre value was 23.1cm3,the volume of iodine used in the titration was 25cm3 and the concentration of Na2S2O3 was 0.1M.calculate the molarity of KIO3
2.34 moles
And why should Sarah believe the answer that andrew threw out there? He explained nothing.
Na2S2O3 + 2 KIO3 = K2S2O3 + 2 NaIO3
So,
molesKIO3=1/2 molesNa2S2O3
M*.025=.5*.1*.0231
M=??
and if course, Sarah knows what Andrew posted cannot possibly be correct.
To calculate the molarity of KIO3 in the given titration, we will use the equation balanced for the reaction between Na2S2O3 and KIO3:
2Na2S2O3 + KIO3 → Na2SO4 + 2NaI + KIO3
In this equation, the stoichiometric ratio between Na2S2O3 and KIO3 is 2:1. This means that 2 moles of Na2S2O3 react with 1 mole of KIO3.
Given that the average titre value is 23.1 cm3, and the volume of iodine used in the titration is 25 cm3, we can assume that the reaction is complete in 23.1 cm3 of Na2S2O3 solution.
To find the number of moles of Na2S2O3 used in the reaction, we need to calculate the moles of iodine (I2) produced using the volume of iodine (25 cm3) and the concentration of Na2S2O3 (0.1 M):
moles of I2 = concentration × volume
moles of I2 = 0.1 M × 25 cm3 = 0.0025 moles
Since the stoichiometric ratio between Na2S2O3 and I2 is 2:1, we can determine the moles of Na2S2O3:
moles of Na2S2O3 = 0.0025 moles/2 = 0.00125 moles
Now, we can calculate the molarity of KIO3:
Molarity of KIO3 = moles of KIO3 / volume of KIO3 solution
Given that the moles of KIO3 is equal to the moles of Na2S2O3 (0.00125 moles), and the volume of KIO3 used in the titration is 25 cm3, we can substitute these values into the formula:
Molarity of KIO3 = 0.00125 moles / 0.025 L
Molarity of KIO3 ≈ 0.05 M
Therefore, the molarity of KIO3 in the given titration is approximately 0.05 M.