1) The diagram below represents the titration curve for the reaction between a particular acid and a particular base.

The equation that best represents the reaction described by the titration curve is
A. HCl(aq) + NH₃(aq)  NH₄Cl(aq)
B. HCl(aq) + NaOH(aq)  NaCl(aq) + H₂O(l)
C. CH₃COOH(aq) + NH₃(aq)  CH₃COONH₄(aq)
D. CH₃COOH(aq) + NaOH(aq)  CH3₃OONa(aq) + H₂O(l)

Solution

2) Oil absorbs a small amount of water from its surroundings. In 1935, Karl Fischer, a German chemist, published a technique for the determination of the water content in oil samples. This technique also involves the conversion of sulfur dioxide to sulfate.
In this analytical technique, the reactants – iodine, sulfur dioxide and a base – are all dissolved in methanol. The base is an organic compound and is represented by B in the balanced equation for this reaction. States are not included in this equation.
H₂O + I₂ + SO₂ + CH₃OH + 3B => [BH]SO₄CH₃ + 2[BH]I

i. What is the mole ratio between iodine and water in this reaction?

Solution

ii. The iodine titrating agent was prepared by dissolving 15.0 g of iodine, I₂, in methanol using a volumetric flask and making up the volume to 500.0 mL. A 10.0 mL sample of oil was analysed using the iodine solution. The mean titre was found to be 4.95 mL.
M(I₂) = 253.8 g mol–1 M(H₂O) = 18.0 g mol^–1
Determine the mass of water present in a 10.0 mL sample of oil.

Solution

iii. Determine the percentage by mass of water present if the density of the oil sample is 0.918 g mL^–1.

Solution