Ascorbic acid (C6H8O6) is a fairly good reducing agent (Eº = + 0.39 V). Due to this property, it is — Organic Chemistry Chemistry Question
PREPARATORY PRACTICAL PROBLEM P6: Determination of the silver content in an unknown sample
Ascorbic acid (C6H8O6) is a fairly good reducing agent (Eº = + 0.39 V). Due to this property, it is widely used in volumetric analysis. It can be used for the determination of various cations (e.g., Au3+, Ag+, Hg2+) and anions (e.g., ClO3–, BrO3–, IO3–, VO43–, Fe(CN)63–). During the titration it is oxidized via loss of 2 electrons forming dehydroascorbic acid (C6H6O6) – as shown in the previous problem.
Chemicals and reagents
* Unknown sample (containing Ag+ ions)
* Ascorbic acid, aqueous solution, c = 0.05 mol dm–3
* Potassium iodate, solid
* Potassium iodide, solid
* Hydrochloric acid, c = 2 mol dm–3
* Sodium acetate, 20 % aqueous solution,
* Variamine Blue, 0.2 % aqueous solution
Preparation of aqueous ascorbic acid solution (c = 0.05 mol dm–3)
Weigh approximately 8.9 g ascorbic acid and dissolve it in a small amount of water. Do not use metal vessels or spoons. Transfer the solution to a standard volumetric flask and dilute it to 1.000 dm3 with freshly prepared cold distilled water.
Preparation of potassium hydrogen iodate solution (c = 0.00833 mol dm–3)
Weigh accurately 3.2492 g KH(IO3)2 in a 50 cm3 beaker and dissolve it in 50 cm3 distilled water. Transfer the solution in a standard volumetric flask and dilute it to 1000 cm3 with freshly prepared distilled water.
Standardization of the ascorbic acid solution (its c is near to 0.05 mol dm–3)
Pipette 20.00 cm3 of KH(IO3)2 solution (0.00833 mol dm–3) into a clean conical flask. Add approx. 1 g KI and 5 cm3 of HCl solution (c = 2 mol dm–3). Titrate the liberated iodine with 0.05 mol dm–3 ascorbic acid solution. When the colour has faded to pale yellow, add 10 drops of Variamine Blue indicator (hydrogen sulfate, 0.2 % (by mass) solution in water). Slowly add 20 % sodium acetate solution until the deep violet colour of the indicator appears, then add 2 cm3 more. Titrate the solution slowly until the deep violet colour disappears. Repeat the procedure as necessary.
Determination of the silver content of the unknown sample
Dilute the given sample solution to 100 cm3 in a volumetric flask using distilled water. Pipette 10.00 cm3 of the unknown solution into a clean conical flask. Dilute the sample to 50 - 70 cm3 with distilled water. Heat the solution to 60 ºC. Add 1 cm3 of Variamine Blue indicator and titrate rapidly with 0.05 mol dm–3 ascorbic acid solution. The solution must be thoroughly swirled during the titration. If the temperature falls under 40 ºC, reheat to 60 ºC. When the blue or violet colour of the indicator has disappeared and the greyish-white colour of the precipitated silver metal has become apparent, add 20% sodium acetate solution to re-establish the colour of the indicator. Then titrate slowly adding titrant dropwise, until the colour of the indicator fades away. Repeat the procedure as necessary.
Variamine Blue has the structure shown below:
[VISUAL]
H NH3CO NH2
Write a balanced equation for the formation of iodine and the titration of iodine with ascorbic acid solution in Procedure 2.
Model Answer
H(IO3)2– + 10 I– + 11 H+ → 6 I2 + 6 H2O
C6H8O6 + I2 → C6H6O6 + 2 HI
Calculate the concentration of the ascorbic acid solution prepared.
Model Answer
The amount of iodine liberated is 1 mmol. If V1 cm3 is the volume of the ascorbic acid solution consumed, its concentration is 1/V1 mol/dm3.
Variamine Blue is shown in its reduced form. Draw the structure of the oxidized form given that it is oxidized with the loss of 2 electrons. Which form is responsible for the blue-violet colour?
Model Answer
The oxidized form is shown below:
[VISUAL]
H3CO N NH
The oxidized form is responsible for the blue-violet colour because of its non-aromatic (chinoidal) structure.
Write a balanced equation of the reaction between ascorbic acid and silver ions.
Model Answer
C6H8O6 + 2 Ag+ → C6H6O6 + 2 Ag + 2 H+
Determine the silver content of the unknown sample.
Model Answer
If the consumption is V2, the amount of silver in the unknown is 20V2/V1 mmol. If consumption between 10 and 20 cm3 is required, the AgNO3 content in the unknown sample must be between 1.7 – 3.4 g.