Florida Drinking Water Operator "B" Practice Test

How much chlorine is required to completely oxidize 1 mg/l of hydrogen sulfide to the sulfate form?

1. 4.16 mg/l

2. 8.32 mg/l

3. 2.12 mg/l

4. 16.64 mg/l

The correct answer is: 8.32 mg/l

The correct answer, which indicates that 8.32 mg/l of chlorine is required to completely oxidize 1 mg/l of hydrogen sulfide (H₂S) to the sulfate form (SO₄²⁻), is based on the stoichiometry of the oxidation reaction. To understand this, it helps to know that each molecule of hydrogen sulfide requires a specific amount of chlorine to facilitate the oxidation process. The typical reaction for the oxidation of hydrogen sulfide with chlorine can be represented as follows: \[ \text{H}_2\text{S} + 4 \text{Cl}_2 + 2 \text{H}_2\text{O} \rightarrow \text{SO}_4^{2-} + 8 \text{HCl} \] From this reaction, it is clear that one molecule of hydrogen sulfide reacts with four molecules of chlorine. To convert this into workable concentrations, we need to consider the molecular weights: the molecular weight of H₂S is approximately 34 g/mol, and that of Cl₂ is about 71 g/mol. For each 1 mg/l of hydrogen sulfide: - The stoichiometry indicates we need four times the equivalent of chlorine (