Part A. Calculate the concentration of KIO3 solution in moles/litres 9.5 mM / gr

Question

Part A.

Calculate the concentration of KIO3 solution in moles/litres

9.5 mM / gram (KIO3)

mass of KIO3 = 1.0838

added to 500mL deionized water.

Part B.

Use your closest 2 titrations to calculate the average moles of iodate ions added to the ascorbic acid solution:

(ASCORBIC ACID SOLUTION = 500mg Ascorbic Acid + binder in every tablet ~ crushed tablet weighed 1.4844 grams added to 250 mL deionized water)

Trial 1 -

Volume KIO3 added = 18.07

Trial 2 -

Volume KIO3 added = 18.08

Part C:

Calculate the average moles of Ascorbic acid in the pipetted sample used for tirtration:

(ASCORBIC ACID SOLUTION = 500mg Ascorbic Acid + binder in every tablet ~ crushed tablet weighed 1.4844 grams added to 250 mL deionized water)

Explanation / Answer

Part A

1gm of KIO3 contain =9.5 mM

1.0838 gm will contain = 9.5*1.0838=10.296 mM

Concentration : no. of moles of KIO3/ Volume of solvent= 10.2961 mM/0.5 L =2.05922 * 10^-2 moles/Litres (ans)

Part B

1st trail

no.of moles of KIO3 =2.05922*10^-2 * 0.01807=3.721 *10^-4 moles

2nd trail

no.of moles of KIO3 =2.05922*10^-2 * 0.01808=3.723 *10^-4moles

Average No. of moles of Iodate ion added =3.722 *10^-4 moles

Part C

Average No. of moles of Iodate ion added =3.722 *10^-4 moles= average no. of moles of ascorbic acid in the pipetted sample (As at equivalence point no. of moles ofIodate ion and ascorbic acid will be same)

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