in the experiment, a vitamin C tablet was dissolved in water and diluted to 100
ID: 816009 • Letter: I
Question
in the experiment, a vitamin C tablet was dissolved in water and diluted to 100 ml. A student took 50 ml of solution to titrate. The titration volume of 0.025 M KIO3 (aq) was 15.8 ml. Calculate the ascorbic acid content per tablet. The molar mass of vitamin C is 176 g/mol
IO3- + 5I- + 6H+ = 3I2 + H20 C6H8O6 + I2 = C6H8O6 + 2I- + 2H+
According to the experiment, whihc of the following is correct
a) vitamin c is also called ascorbic acid
b) Iodide ion I- react with stgarch indicator to form the complex
c) Vitamin C is an ionic compound and soluble in water
d) vitamin C is used as a reducing agent
Explanation / Answer
part 1:
IO3- + 5I- + 6H+ = 3I2 + H20
here KIO3 (IO3-) reacts to form I2
number of moles of KIO3 = Molarity of KIO3 * Volume of KIO3
= 0.025 M * 15.8 mL
= 0.395 mmol
From above reaction:
number of moles of I2 formed = 3* number of moles of KIO3 = 3 * 0.395 = 1.185 mmol
Now look at 2nd reaction:
C6H8O6 + I2 = C6H6O6 + 2I- + 2H+
Here 1 mol ascorbic acid reacts with 1 mol of I2
number of moles of ascorbic acid reacted = number of moles of I2 = 1.185 mmol = 0.001185 mol
mass of ascorbic acid = number of moles of ascorbic acid * molar mass of ascorbic acid
= 0.001185 mol* 176 g/mol
=0.20856 gm
=0.21 gm (approx)
Answer: ascorbic acid content per tablet = 0.21 gm
Part 2:
In 2nd reaction iodine is reduced; it gains two electrons and we have the REDUCTION of iodine.
So, Vitamin C is oxidized by losing two electrons. So we have the OXIDATION of vitamin C and hence it will act as reducing agent.
Answer : d) vitamin C is used as a reducing agent
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