The concentration of ozone in ground-level air can be determined by allowing the gas to react with an aqueous solution of potassium iodide, KI, in a redox reaction that produces molecular iodine, molecular oxygen, and potassium hydroxide. Assume 1 atm and 25.0 C

(a) Deduce the balanced reaction for the overall process
(b) Determine the ozone concentration, in ppb, in a 10.0 L sample of outdoor air if it required 17.0 ug of KI to react with it.

a) 2KI + O₃ + H₂O > I₂ + O₂ + 2KOH

b) Ozone concentration = 0.246 ppb

Explanation:

a) The balanced equation for the reaction is

2KI + O₃ + H₂O > I₂ + O₂ + 2KOH

b) We first convert 17 μg of KI to number of moles

Number of moles = (mass)/(molar mass)

Molar mass of KI = 166 g/mol

Mass of KI that reacted = 17 μg = (17 × 10⁻⁶) g

Number of moles = (17 × 10⁻⁶)/166

Number of moles of KI that reacted = (1.0241 × 10⁻⁷) moles

From the stoichiometric balance of the reaction,

2 moles of KI reacts with 1 mole of O₃

Then, (1.0241 × 10⁻⁷) moles of KI will react with (1.0241 × 10⁻⁷ × 1/2) moles of O₃

Number of moles of O₃ that reacted = (5.12 × 10⁻⁸) moles.

To express the amount of O₃ in 10.0 L of air in ppb, we need to convert the amount of O₃ that reacted.

Mass = (number of moles) × (molar mass)

Molar mass of O₃ = 48 g/mol

Mass of O₃ that reacted = (5.12 × 10⁻⁸) × 48 = 0.0000024578 g = (2.46 × 10⁻⁶) g

Concentration in ppb = (Mass of solute in μg)/(volume of solution in L)

Mass of solute = Mass of O₃ = (2.46 × 10⁻⁶) g = 2.46 μg

Volume of solution = 10.0 L

Concentration of O₃ in air in ppb = 2.46/10 = 0.246 ppb

a. a clone will be physically identical to the original, but won’t be genetically identical

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