Question

A soft drink was bottled with a partial pressure of CO$_2$ of 3 bar over the liquid at room temperature. The partial pressure of CO$_2$ over the solution approaches a value of 30 bar when 44 g of CO$_2$ is dissolved in 1 kg of water at room temperature. The approximate pH of the soft drink is ______ $\times 10^{-1}$.

(First dissociation constant of H$_2$CO$_3$ = 4.0 $\times 10^{-7}$; log 2 = 0.3; density of the soft drink = 1 g mL$^{-1}$)

Answer 1

37

Answer 2

Here's how to solve this problem:

1. Finding Dissolved Concentration:

   For 30 bar, dissolving 44 g CO₂ in 1 kg water gives 1 mole CO₂ dissolved.

   By Henry’s law, at 3 bar the dissolved CO₂ moles per kg water are:

   n_CO₂ = (3/30) × 1 = 0.1 mole

   Since 1 kg water ≈ 1 L, the concentration, C ≈ 0.1 M.

2. Acid Dissociation of Carbonic Acid:

   We assume that the dissolved CO₂ behaves effectively as carbonic acid, H₂CO₃.

   For the weak acid dissociation:

   H₂CO₃ ⇌ H⁺ + HCO₃⁻

   with Ka = 4.0 × 10⁻⁷.

   Let x = [H⁺] = [HCO₃⁻]. Since x is small, [H₂CO₃] ≈ 0.1 M.

   Then,

   x² = Ka × (0.1) = 4.0 × 10⁻⁷ × 0.1 = 4.0 × 10⁻⁸

   ⇒ x = √(4.0 × 10⁻⁸) = 2.0 × 10⁻⁴ M

3. Calculating pH:

   pH = –log[H⁺] = –log(2.0 × 10⁻⁴)

   = –(log 2.0 + log 10⁻⁴)

   = –(0.3 – 4)

   = 3.7

   Expressed as a multiple of 10⁻¹, pH = 37 × 10⁻¹.

Therefore, the answer is 37.