Question

A vitamin C (ascorbic acid) tablet was dissolved in approximately 50 mL of distilled water and titrated with the standardized NaOH solution. From the results of this titration, the mg of ascorbic acid in the tablet was calculated.

Molecular formula of ascorbic acid: C6H8O6

Volume of NaOH required to neutralize ascorbic acid in Vitamin C tablet (mL) 14.47

Concentration of NaOH in mol/L,   0.1964

Calculate the amount of ascorbic acid in the Vitamin C tablet in (mg).

Answer 1

Ascorbic acid is monoprotic acid and react with NaOH as,

C₆H₈O₆ (aq) + NaOH (aq) NaC₆H₇O₆ (aq) + H₂O

# of moles of Ascorbic acid C₆H₈O₆ = # of moles of NaOH.

Let us first calculate # of moles of NaOH,

Molarity of NaOH = 0.1964 mol/L,, Volume of NaOH required = 14.47 mL = 0.01447 L.

# of moles of NaOH = Molarity of NaOH * Volume in L

                              = 0.1964 mol/L * 0.01447 L

                             = 2.84*10^-3 mol

Hence,

# of moles of Ascorbic acid = 2.84*10^-3 mol

Molar mass of Ascorbic acid = 176.12 g/mol

Mass of 2.84*10^-3 mol Ascorbic acid = # of moles * Molar mass

                                                     = 2.84*10^-3 mol * 176.12 g/mol

                                                     = 0.501 g

                                                     = 501 mg.

There is 501 mg Ascorbic acid in Vitamin C tablet.

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