Question

Sulfur dioxide, SO2, is used in wine as an antioxidant, slowing the oxidation of ethanol
to ethanoic acid. To determine the SO2 content of a wine sample, sodium hydroxide
and sulfuric acid were added to ensure that all sulfur was present as SO2. A 20.00 mL
aliquot was then titrated against standardised iodine solution, using starch indicator.
Reaction occurred according to the equation:
SO2(aq) + 2H2O(l) + 12(aq) → 4H+ (aq) + 21- (aq) + SO42- (aq)
The end point was reached with 15.62 mL of 0.00501 M iodine solution. The
concentration (in M) of SO2 in the wine was ...

A)6.2 x 10-3
B)7.8 x 10-3
C)3.9 x 10-3
D)5.0 x 10-3

Answer 1

The  concentration (in M) of SO₂ in the wine : c. 3.9 x 10⁻³

Further explanation

Given

Reaction

SO2(aq) + 2H2O(l) + I2(aq) → 4H+ (aq) + 21- (aq) + SO42- (aq)

Required

The  concentration (in M) of SO2

Solution

Titration ⇒ mol SO₂ = mol iodine(I)

mol iodine :

=15.62 ml x 0.00501 M

=0.078256 mlmol

From equation, mol ratio of of SO₂ : I₂ = 1 : 1, so mol SO₂=0.078256

The concentration of SO₂(in 20 ml sample) :

M = n/V

M = 0.078256 mlmol/20 ml

M = 0.0039128

M = 3.9 x 10⁻³