A soft drink was bottled with a partial pressure of CO of 3 bar over the liquid at room temperature. The partial pressure of CO over the solution approaches a value of 30 bar when 44g of CO is dissolv

Question

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

Prashant Kotian · Accepted Answer

Correct answer: 37.00. **Step 1: Apply Henry's Law to find CO₂ concentration** Henry's Law: C = kH × P Using the equilibrium data: kH = C/P = 44g/44g·mol⁻¹ ÷ 30 bar = 1 mol/30 bar = 0.033 mol/(L·bar) At bottling pressure (3 bar): C = 0.033 × 3 = 0.1 mol/L **Step 2: Write the dissociation equation for carbonic acid** CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻ For the first dissociation: Ka1 = [H⁺][HCO₃⁻]/[H₂CO₃] = 4.0 × 10⁻⁷ **Step 3: Set up the equilibrium expression** Assuming [H₂CO₃] ≈ 0.1 mol/L and [H⁺] = [HCO₃⁻] = x: Ka1 = x²/0.1 = 4.0 × 10⁻⁷ **Step 4: Solve for [H⁺]** x² = 4.0 × 10⁻⁸ x = 2.0 × 10⁻⁴ mol/L = [H⁺] **Step 5: Calculate pH** pH = -log[H⁺] = -log(2.0 × 10⁻⁴) pH = -log(2) - log(10⁻⁴) = -0.3 + 4 = 3.7 **Step 6: Express as requested** pH ≈ 3.7 = 37 × 10⁻¹ Therefore, the answer is 37.00.

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