A 200 ml sample of a citrus fruit drink containing ascorbic acid (vitamin C) was acidified with H 2 S O 4 and 20 ml of 0.025 M − I 2 was added. Some of the I 2 was reduced by the ascorbic acid

Question

A 200 ml sample of a citrus fruit drink containing ascorbic acid (vitamin C) was acidified with $H_{2} S O_{4}$ and 20 ml of 0.025 $M - I_{2}$ was added. Some of the $I_{2}$ was reduced by the ascorbic acid to $I^{-}$ . The excess of $I_{2}$ required 5.0 ml of 0.01 $M - N a_{2} S_{2} O_{3}$ for reduction. What was the vitamin C content of the drink in microgram vitamin per ml drink? $C_{6} H_{8} O_{6}$ + $I_{2}$ $\to$ $C_{6} H_{6} O_{6}$ + 2HI; $5 H_{2} O$ + $S_{2} O_{3}^{2 -}$ + $4 I_{2}$ $\to$ $2S O_{4}^{2 -}$ + $8 I^{-}$ + $10 H^{+}$

Answer:

264.00

Solution:

$I_{2}$ added = $20 \times 0.025$ = 0.5 mmol. Excess $I_{2}$ = 4×(5×0.01) = 0.2 mmol. $I_{2}$ on vitamin C = 0.3 mmol = mmol vitamin C = $0.3 \times 176$ = 52.8 mg in 200 ml = 52800/200 = 264\ \mug/ml. Ans 0264.

TestHub · Accepted Answer

I 2 ​ added = 20 × 0.025 = 0.5 mmol. Excess I 2 ​ = 4×(5×0.01) = 0.2 mmol. I 2 ​ on vitamin C = 0.3 mmol = mmol vitamin C = 0.3 × 176 = 52.8 mg in 200 ml = 52800/200 = 264\ \mug/ml. Ans 0264.

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