1. & 2 Oxidation of ethanol in wine
Ethanol is a primary alcohol. When this comes into contact with atmospheric oxygen, it becomes oxidized to an aldehyde. Since wine contains ethanol oxidation can this happen if, for example, recommends his wine without a cork. Oxidation reaction is shown below (the reaction equation is shown in particular for the substances).
2CH 3 CH 2 OH + O 2 ® 2CH 3 CHO + 2H 2 O
Methanol + oxygen ® ethanal (acetataldehyd) + water (not aligned)
Oxideringsprocessen does not stop when the aldehydes are formed, and a further oxidation to a carboxylic acid will therefore be:
2CH 3 CHO + O 2 ® 2CH 3 COOH + 2H 2 O
Ethanal ethane + oxygen ® acid (eddiekesyre) + water (not aligned)
3. Reconciled reaction equation for the "balloon test"
Ethanol is oxidized by the strong oxidant potassium dichromate under acidic conditions.
The orange potassium dichromate is reduced to the green crom (III) ion. Color shift is used as an indicator of alcohol content by a so called "balloon test" is used if it is suspected a person has too much alcohol in the blood. The reaction of the process are:
3CH 3 CH 2 OH + 2Cr 2 O 7 2 - + 16H + ® 3CH 3 COOH + 4Cr 3 + + 11H 2 O
4. Ionligninger of the chemical processes in steps 1, 2 and 3
1.
Yeast proteins contain sulfur, and this may go beyond the wine. Therefore, we have found out how much free sulfur, the total is in the wine. Sulfur dioxide reacts with the wine's other components and only a small fraction is free as dissolved sulfur dioxide or sulfite. So first we have transformed sulfite to sulfur dioxide. This is done by adding H 3 PO 4, which dissolves and gives free H + ions and these ions can react with sulfite:
SO 3 2 -- (aq) + 2 H + (aq) → H 2 SO 3 (aq)
And dihydrogensulfit can decompose to sulfur dioxide and water:
H 2 SO 3 (aq) → H 2 O (l) + SO 2 (aq)
2.
We now have transferred the escaping sulfur dioxide to a solution of hydrogen peroxide. This is done
during heating as well as N 2 bubbling of the acid liquid:
SO 2 (g) + H 2 O 2 (aq) -> H 2 SO 4 (aq)
3.
To determine the total sulfur dioxide in wine, the sulfuric acid formed is titrated with NaOH:
H 2 SO 4 (aq) + 2 NaOH (aq) → Na 2 SO 4 (aq) + 2 H 2 O (l)
5. Total molar concentration of SO 2 in the studied wine
Based on the approach described in the previous section, add 40 ml of white wine and 10 ml of 25% orthophosphersyre. This mixture is boiled with N 2 bubbling and the liberated sulfur dioxide is transferred to 10 ml 0.3% hydrogen peroxide. This solution with 15 ml 0.01 M NaOH to cover.
When the cover has been devoted following drug volume of NaOH:
n NaOH = 0.01 M × (15 × 10 -- 3 L ) = 1.5 × 10 -4 mol
Since sulfuric acid and sodium hydroxide react in the ratio of 1 to 2, the quantity of drug consumed sulfuric acid 7.5 × 10 -5 mol.
This drug quantity must now be counted on to what the total molar concentration of sulfur dioxide in wine. If we look at the responses shows that SO 2 forming H 2 SO 4 at a 1 to 1 and H 2 SO 4 decompose to SO 4 2 - in the ratio 1 to 1 Then the following statements must be true:
n SO42-= n SO2
We have been informed by volume of white wine and therefore its concentration should be:
c SO2 = 1.875 * 10 ^ -3 M
6. Number ppm SO 2 in the studied wine
They calculated the mass of the studied white wines. Since volume is 40 ml and the density is 1 kg / l, must:
m = white 40 g
Then calculated the mass of sulfur dioxide. Molar mass is 64.06 g / mol, so the mass is:
m SO2 = 64.06 g / mol × (7.5 × 10 -5 mol) = 0.0048 g
The concentration is then: = 120.11 ppm
