Henry's law constant for \(\mathrm{CO}_{2}\) at \(38^{\circ} \mathrm{C}\) is \(2.28 \times 10^{-3}\) \(\mathrm{mol} / \mathrm{L} \cdot \mathrm{atm} .\) Calculate the \(\mathrm{pH}\) of a solution of \(\mathrm{CO}_{2}\) at \(38^{\circ} \mathrm{C}\) in equilibrium with the gas at a partial pressure of 3.20 atm.

Determine the molar concentration of \( \mathrm{CO}_{2}\) in the solution using Henry's law, which says that the concentration of a gas in a solution is proportional to the partial pressure of that gas above the solution. The constant of proportionality is Henry's Law constant. Here, Henry's law constant for \( \mathrm{CO}_{2}\) is given to be \(2.28 \times 10^{-3}\, \mathrm{mol/L}\cdot \mathrm{atm}\) and the partial pressure of \( \mathrm{CO}_{2}\) is \(3.20 \, \mathrm{atm}\). Therefore, the concentration of \( \mathrm{CO}_{2}\) can be calculated as \( \mathrm{CO}_{2}\) concentration = Henry's law constant \( \times \) Partial pressure of \(\mathrm{CO}_{2}\) = \(2.28 \times 10^{-3}\, \mathrm{mol/L}\cdot\mathrm{atm} \times 3.20 \, \mathrm{atm} = 0.007296 \, \mathrm{mol/L}\).

Further, consider that each \( \mathrm{CO}_{2}\) molecule produces a hydrogen ion, \( \mathrm{H}^{+}\), in solution. This is because \( \mathrm{CO}_{2}\) can form carbonic acid, \( \mathrm{H}_{2}\mathrm{CO}_{3}\), in water, which then disassociates into \( \mathrm{H}^{+}\) and \( \mathrm{HCO}_{3}^{-}\). Thus, the concentration of \( \mathrm{H}^{+}\) ions is the same as the concentration of the \( \mathrm{CO}_{2}\), calculated in step 1, which is \(0.007296 \, \mathrm{mol/L}\).

Finally, calculate the pH of the solution using the formula \( \mathrm{pH} = -\log[\mathrm{H}^{+}]\) where \([\mathrm{H}^{+}]\) is the hydrogen concentration in \( \mathrm{mol/L}\). Substituting the calculated value of \( \mathrm{H}^{+}\) concentration into this equation gives \(\mathrm{pH} = -\log(0.007296)= 2.14\).