A 0.552-g sample of ascorbic acid (vitamin C) was dissolved in water to a total volume of 20.0 mL and titrated with 0.1103 M KOH. The equivalence point occurred at 28.42 mL. The pH of the solution at 10.0 mL of added base was 3.72. From this data, determine the molar mass and K, for vitamin C.

176.36g/mol

Explanation:

It was given that:

Mass of ascorbic acid=0.552 g

Volume of water=20.0 mL

Concentration of KOH=0.1103 M

Volume of KOH=28.42 mL. = 0.02842l

pH of solution at 10.0 mL= 3.72

At equivalence point, number of moles of acid is equal to the number of moles of base.

Number of moles of base= (KOH) x Volume

                        =0.1103 x 0.02842 L

                                =0.00313 moles.

Therefore number of miles of acid= 0.00313moles

No of moles= Mass/molar mass

Molar Mass= Mass of Acid/ No of moles of acid

= 0.552g/0.00312moles

= 176.36g/mol

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