Question

A soft drink was bottled with a partial pressure of \(\mathrm{CO_2}\) of 3 bar over the liquid at room temperature. The partial pressure of \(\mathrm{CO_2}\) over the solution approaches a value of 30 bar when 44 g of \(\mathrm{CO_2}\) 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 \(\mathrm{H_2CO_3}\) = \(4.0\times 10^{-7}\); \(\log 2 = 0.3\); density of the soft drink = 1 g \(\mathrm{mL^{-1}}\)) .

Hint:

Use Henry's law to find the dissolved \(\mathrm{CO_2}\) at 3 bar (about 0.1 M), then \([\mathrm{H^+}] = \sqrt{K_{a1}C}\).

Answer 1

Correct Answer: 37

44 g \(\mathrm{CO_2}\) = 1 mol dissolved in 1 kg water at 30 bar. By Henry's law (solubility proportional to pressure), at 3 bar the amount dissolved = 1 \(\times\) (3/30) = 0.1 mol per kg water.
With density 1 g \(\mathrm{mL^{-1}}\), the concentration is about 0.1 M of \(\mathrm{H_2CO_3}\).
\([\mathrm{H^+}] = \sqrt{K_{a1}C} = \sqrt{4\times 10^{-7}\times 0.1} = \sqrt{4\times 10^{-8}} = 2\times 10^{-4}\) M.
\(\mathrm{pH} = -\log(2\times 10^{-4}) = 4 - 0.3 = 3.7 = 37\times 10^{-1}\).
Answer: 37