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Solution: 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, and 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 for vitamin C.

Problem

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, and 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 for vitamin C.

Solution

We’re being asked to calculate the Molar mass of Ascorbic acid if 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, and the equivalence point occurred at 28.42 mL

Recall that at the equivalence point of a titration:


moles acid = moles base


Since Ascorbic acid is a monoprotic acid, there will be just one equivalence point for the titration. 


This means:

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