The atmosphere will be 412 ppm (volume) of CO2. What will be the pH of rain without other pollutants at 25 °C and 1 atm?

pH of rain with only CO₂ as the acidic pollutant in the atmosphere = 5.58

Explanation:

If other sources of acidity like CO₂, and acidic particles are minimal,

P꜀ₒ₂ = 1 atm × 412ppm = 1 × 412 × 10⁻⁶ = 4.12 × 10⁻⁴ atm

With the CO₂ conc of 412ppm,

Henry's law constant for CO₂, Kₕ = 0.038 M/atm

[CO₂(aq)] = KₕP꜀ₒ₂ = (0.038)(4.12 x 10⁻⁴) = 1.57 x 10⁻⁵ M

Chemical Reaction

CO₂ + H₂0 <> H⁺ + HCO₃⁻ Kₐ = 4.45 x 10⁻⁷ (from literature)

Note: although we can not really use an initial change equilibrium table because

CO₂ will not be depleted from the gas phase and CO₂ will he constant, we can expect that [H⁺] = [HCO₃⁻] (assuming no other sources of H⁺

Kₐ = 4.45 x 10⁻⁷ = [H⁺] [HCO₃⁻]/[CO₂] = [H⁺]²/(1.57 x 10⁻⁵)

[H⁺] = √[(4.45 x 10⁻⁷)(1.57 x 10⁻⁵)] = 2.64 x 10⁻⁶ M

pH = -log [H⁺] = - log (2.64 x 10⁻⁶) = 5.58

Hope this Helps!!!

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