Calculate the concentration of potassium iodate solution in g/mol: 500 mL deioni

Question

Calculate the concentration of potassium iodate solution in g/mol:

500 mL deionized H20

1.0838 grams of 9.5 mM KIO3

What is the average moles of iodate ions added to the ascorbic acid solution:

(500mg Ascorbic Acid + binder tablet)

Mass of crushed Ascorbic Acid tablet =

1.4844 g

Volume

Trial 1

Trial 2

Initial Buret Reading

0.14

20.78

Final Buret Reading

18.21

38.86

Volume of KIO3

18.07

18.08

Volume

Trial 1

Trial 2

Initial Buret Reading

0.14

20.78

Final Buret Reading

18.21

38.86

Volume of KIO3

18.07

18.08

Explanation / Answer

mol weight of KIO3 = 214

conc (gram/ml)= 1.0838/500 =0.00216

conc (moles/litre) 1.0838*2 /214 = 0.0101

Ascorbic acid isC6H8O6with MW=176.13. I think you have written 0.400 g and not 400g, right?
So in your sample you have moles C6H8O6 =0.5/176.13 = 2.83*10^-3

The titration reaction is
C6H8O6 + I2 -> C6H6O6 + 2I- +2H+

So 1mole C6H8O6 reacts with 1 mole I2
2.83*10^-3 .. .. .. .. .. .. .. .. . .. . .x
=> x=2.83*10^-3 mole I2

But I2 was formed by the KIO3

IO3- +5I- +6H+ -> 3I2 + 3H2O

1 mole IO3- produces 3 moles I2
x .. .. .. .. .. .. .. .. .. .. 2.83*10^-3

3x=2.83*10^-3 => x =(2.83/3)*10^-3 = 9.43*10^-4 mole KIO3

but you know that M=mole/V => V=mole/M= (9.43*10^-4)/0.0101 = 0.093 ml

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