Answer:
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
