Question

A 0.2387 g sample was analyzed for ascorbic acid. it required 24.75 ml of .009687 M KBrO₃ to cause the solution to turn yellow, signaling complete consumption of the ascorbic acid and formation of Br₂. The solution was then treated with KI and starch and back titrated with Na₂S₂O₃. It required 2.98mL of 0.05118 M Na₂S₂O₃ to reach the final endpoint, signaled by the disappearance of the blue color. Calculate the percentage of ascorbic acid C₆H₈O₆ in the sample.

Answer 1

Required KBrO3 are calculated:

n KBrO3 = M * V = 0.009687 * 0.02475 = 2.40x10 ^ -4 mol

The required moles of Na2S2O3 are calculated:

n Na2S2O3 = 0.05118 * 0.00298 = 1.50x10 ^ -4 mol

The difference in moles is calculated:

n CA = (2.4x10 ^ -4) - (1.5x10 ^ -4) = 9x10 ^ -5 mol

The acid mass is calculated:

g CA = n * MM = 9x10 ^ -5 * 192.12 = 0.0173 g

The percentage of acid is calculated:

% CA = g Acid * 100 / g sample = 0.0173 * 100 / 0.2387 = 7.25%

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