Question

In: Chemistry

A vitamin C tablet containing 250 mg of ascorbic acid (C6H8O6; Ka=8.0

A vitamin C tablet containing 250 mg of ascorbic acid (C6H8O6; Ka=8.0

Solutions

Expert Solution

                           We have , 250 mg of ascorabicacid (C6H8O6) dissolved in 250ml water . Ka = 8.0*10-5

            We know that , M = wt / mwt * 1000/v

                                       = 0.250/176.12 * 1000/250

         Intial concentration of ascorbic acid is , = 5.68 * 10-3 moles/lit

Ascorbic acid dissolved in water , C6H8O6 +H2O ------------.> C6H7O6- + H3O+

   intial concentration ---            5.68*10-3                      0                0

        concentration change ----              -x                              +x              +x

       equilibrium concentration ----     5.68*10-3 - x                   x                 x

                               Ka    =   [C6H7O6-][H3O+] / [C6H8O6]

                     SO , 8 * 10-5 = (X)(X) / ( 5.68*10-3 - x )

                  8 * 10-5 ( 5.68*10-3 - x ) - X2 = 0

                X2 + 8 * 10-5 X - 4.54*10-7 =0

               X = -b b2-4ac / 2a

             X    = -(8*10-5) (8*10-5)2 - 4(1)(- 4.54*10-7) / 2(1)

       [H3O+]    = X   = 6.3 * 10-4 mol/lit

Since , [H3O+] = [C6H7O6-] = 6.3 * 10-4 mol/lit

pH = -log[H3O+] = -log(6.3*10-4)

pH = 3.2


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