Question

Vitamin C is ascorbic acid, HC₆H₇O_6, which can be titrated with a strong base. A certain 25 mL sample of ascorbic acid solution is titrated to endpoint with 23.4 ml of 0.09081 M NaOH solution.

A. What is the molarity of the ascorbic acid solution?

B. How many grams of Vitamin C were in this 25 mL sample?, HC₆H₇O₆(aq) + NaOH(aq) = NaC₆H₇O₆(aq) + H₂O(l)

Answer 1

a) The balanced equation is

HC6H7O6 (aq) + NaOH (aq) --> NaC6H7O6(aq) + H2O (l)

NaOH moles used = M x V ( in L)             where V = 23.4 ml = 0.0234 L

NaOH moles = 0.09081 x 0.0234 = 0.002125

ascorbic acid moles reacted = NaOH moles = 0.002125

Ascorbic acid volume used = 25 ml = 0.025 L

Molarity of ascorbic acid = moles / volume = 0.002125 /0.025 = 0.085 M

b) Number of grams of vitamin C = moles x molar mass of vitamin C

              = 0.002125 mol x 176.12 g/mol = 0.374 g