A soft drink was bottled with a partial pressure of CO₂ of 3 bar over the liquid at room temperature. The partial pressure of CO₂ over the solution approaches a value of 30 bar when 44 g of CO₂ is dissolved in 1 kg of water at room temperature. The approximate pH of the soft drink is ______ × 10 –1 .
(First dissociation constant of
H₂CO₃ = 4.0 × 10 –7 ; log 2 = 0.3; density
of the soft drink = 1 g mL –1 ) .

Solution

Chemistry ma’am · 4 steps

First use Henry's law to find how much CO2 dissolves at 3 bar, then treat it as a weak acid.

1. Henry's law. Dissolved gas is proportional to its partial pressure. At 30 bar, 44 g of CO₂ (that is, 1 mol) dissolves in 1 kg of water. At 3 bar, one tenth of the pressure, only 0.1 mol dissolves.
2. Concentration. The density is 1 g mL⁻¹, so 1 kg of drink is about 1 L. That gives C ≈ 0.1 mol L⁻¹ of H₂CO₃.

   [H⁺] = √(Kₐ₁ C) = √(4.0 × 10⁻⁷ × 0.1) = √(4 × 10⁻⁸) = 2 × 10⁻⁴ M

3. Take the log. pH = -log(2 × 10⁻⁴) = 4 - log 2 = 4 - 0.3 = 3.7. Only the first dissociation matters because Kₐ₂ is far smaller.
4. So. pH = 3.7 = 37 × 10⁻¹, so the answer is 37.

The common slip: 32. Using the full 1 mol of CO₂ from the 30 bar data, without scaling down to 3 bar, gives pH 3.2, which is 32.

The idea to keep. Gas dissolved in a drink? Henry's law turns pressure into concentration first. Then a weak acid gives [H⁺] = √(Kₐ C).

Answer from the NTA final answer key. Solution by PrepWiser.