Determination of Ascorbic Acid Molarity of the potassium iodate (KIO3) solution

ID: 506131 • Letter: D

Question

Determination of Ascorbic Acid

Molarity of the potassium iodate (KIO3) solution ___________________________

Trial 1

Mass of Ascorbic Acid (grams)= .11

Initial Buret reading (mL)= 50 mL

Final Buret reading (mL)= 26.6 mL

Volume of KIO3 titrant added (mL)=???

Calculations

Calculate the mass of ascorbic acid in each sample determined by the titration and compare with your measured mass for each sample.

( Really just need to know the formula) for each)

Moles of ascorbic acid (determined by titration)=??

Mass of Ascorbic acid (determined by titration)=??

Mass of Ascorbic Acid (as weighed out)=??

%difference =??

In this iodometric titration following reactions occur:

KIO3 reacts with acidic protons and potassium iodide to liberate iodine

KIO3(aq) + 6 H+(aq) + 5 I (aq) -------> 3 I2(aq) + 3 H2O(l) + K+(aq)

The liberated iodine oxidises the ascorbic acid

C6H8O6(aq) + I2(aq) -------> C6H6O6(aq) + 2 I-(aq) + 2 H+(aq)

Volume of KIO3 added = 50.0-26.6 = 23.4 mL or 0.0234 L

Number of moles of KIO3= concentration of KIO3 X volume of KIO3

Number of moles of KIO3= c mol L-1 X 0.0234 L = x mol

From the two equations we know that 3 mol of ascorbic acid are consumed per mol of KIO3

Number of moles of ascorbic acid = 3x mol

Now we can determine the mass using the following formula

mass of ascorbic acid (m) = number of moles of ascorbic acid (n) X Molar mass of ascorbic acid (M)

m = 3x mol X 176.12 g mol-1 = y g

It is given that the mass of ascorbic acid is 0.11 g

% difference = (Given mass – actual mass) X 100 / Given mass

= (0.11 –y) 100 / 0.11

= z %

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