Question

Calculate the weight percent of ascorbic acid in a tablet of Vitamin C from the following data:A 80 mg sample of a crushed Vitamin C tablet was dissolved in 40 mL of H2SO4 and 20 mL of water. Two grams of KI and 40. mL of 0.00653 M KIO3 solution was added, and the mixture titrated to a starch endpoint. The titration required 15 mL of 0.0510 M thiosulfate solution. I know the answer is 88%, but how did they get it?

Answer 1

Answer:

The Answer is 88%

Explanation:

The balanced ionic equation of the reaction

   $IO^{3- }+ 8 I^- + 6 H^+ => 3 I^{3- }+ 3 H_2O$

  $I^{3-} + 2 S_2O_3^{2-} => 3 I^- + S_4O_6^{2-}$

 

  $C_6H_8O_6 + I^{3- }+ H_2O => C_6H_6O_6 + 3 I^- + 2 H^+$

Looking at the above reactions

   The original number of moles of $I^{3-}$ = 3 × number of moles of $IO^{3-}$

                                                                = 3 × volume × concentration of $KIO_3$

Note: The formula for number of moles is volume × concentration

                                                                $=3 * \frac{40}{1000} * 0.00653 =0.000784$ mol

The number of moles of $I^{3-}$ left after its reaction with ascorbic  acid

                         = $\frac{1}{2}$ x moles of $S_2O_3^{2-}$

                          = $\frac{1}{2}$ x volume x concentration of    $S_2O_3^{2-}$

                          $= \frac{1}{2} * \frac{15}{1000} * 0.0510 =0.000383 \ mol$

Note: The division by 1000 is to convert mill liter to liter

                             Moles of ascorbic acid = moles of $I^{3-}$ reacted

                           $= initial \ moles \ of \ I^{3-} \ - remaining \ moles \ of \ I^{3-}$

                           $=0.000784 - 0.000383$

                          $=0.000401 \ mols$

              $Mass = moles \ * molar \ mass$

Hence

              Mass of ascorbic acid = moles of ascorbic acid × molar mass of ascorbic acid

                                                      $= 0.000401* 176.12 = 0.07055\ g$

                                                      $= 70.55\ mg$

 Weight% of ascorbic acid = mass of ascorbic acid/mass of sample x 100%

                                             = 70.55/80  × 100%

                                              = 88.1%

                       

Answer 2

The weight percent of ascorbic acid in a tablet of Vitamin C  is 88%.

Calculation of the weight percentage:

Since

Initial moles of I3- = 3 x moles of IO3-

= 3 x volume x concentration of KIO3

= 3 x 40/1000 x 0.00653

= 0.0007836 mol

Now

Moles of I3- remaining after reaction with ascorbic acid

= 1/2 x moles of S2O32-

= 1/2 x volume x concentration of S2O32-

= 1/2 x 15/1000 x 0.0510

= 0.0003825 mol

Now

Moles of ascorbic acid = moles of I3- reacted

= initial moles of I3- - remaining moles of I3-

= 0.0007836 - 0.0003825

= 0.0004011 mol

Now

Mass of ascorbic acid = moles x molar mass of ascorbic acid

= 0.0004011 x 176.12

= 0.07064 g = 70.64 mg

Now finally

Weight% of ascorbic acid = mass of ascorbic acid/mass of sample x 100%

= 70.64/80 x 100%

= 88.3%

= 88%

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