The Henry's law constant for \(\mathrm{CO}_{2}\) in water at $25^{\circ} \mathrm{C}\( is \)3.4 \times 10^{-4} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}(\mathbf{a})$ What is the solubility of \(\mathrm{CO}_{2}\) in water at this temperature if the solution is in contact with air at normal atmospheric pressure? (b) Assume that all of this \(\mathrm{CO}_{2}\) is in the form of \(\mathrm{H}_{2} \mathrm{CO}_{3}\) produced by the reaction between \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}:\) $$ \mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(a q) $$ What is the pH of this solution?

First, we use the Henry's law formula to calculate the solubility of \(\mathrm{CO}_{2}\) in water at \(25^{\circ}\mathrm{C}\) and normal atmospheric pressure. The formula for Henry's law is: $$ C = k_H \times P $$ Where \(C\) is the solubility of the gas in the solution, \(k_H\) is the Henry's law constant for the gas, and \(P\) is the partial pressure of the gas. For this exercise, we have the Henry's law constant for \(\mathrm{CO}_{2}\) (\(k_H\)) as \(3.4 \times 10^{-4} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}\), and the partial pressure of \(\mathrm{CO}_{2}\) in the atmosphere can be approximated as normal atmospheric pressure, which is 101325 Pa. Now, we can calculate the solubility of \(\mathrm{CO}_{2}\) in water: $$ C = (3.4 \times 10^{-4} \mathrm{~mol} / \mathrm{m}^{3}-\mathrm{Pa}) \times 101325\,\text{Pa} $$

As we assumed that all of the \(\mathrm{CO}_{2}\) is in the form of \(\mathrm{H}_{2}\mathrm{CO}_{3}\), the concentration of \(\mathrm{H}_{2}\mathrm{CO}_{3}\) in the solution is the same as the solubility of \(\mathrm{CO}_{2}\) in water (\(C\)). So: $$ [\mathrm{H}_{2}\mathrm{CO}_{3}] = C $$

To calculate the pH of the solution, we first need to know the dissociation constant (\(K_a\)) for \(\mathrm{H}_{2}\mathrm{CO}_{3}\). The \(K_a\) value for \(\mathrm{H}_{2}\mathrm{CO}_{3}\) is approximately \(4.3 \times 10^{-7}\). The \(K_a\) expression for \(\mathrm{H}_{2}\mathrm{CO}_{3}\) is: $$ K_a = \frac{[\mathrm{H}^+][\mathrm{HCO}_3^-]}{[\mathrm{H}_{2}\mathrm{CO}_{3}]} $$ Since the solution is quite dilute, we can assume that \([\mathrm{H}^+] \approx [\mathrm{HCO}_3^-]\). Thus, the equation simplifies to: $$ K_a = \frac{[\mathrm{H}^+]^2}{[\mathrm{H}_{2}\mathrm{CO}_{3}]} $$ We can solve for \([\mathrm{H}^+]\): $$ [\mathrm{H}^+] = \sqrt{K_a \times [\mathrm{H}_{2}\mathrm{CO}_{3}]} $$ Finally, we can calculate the pH of the solution using the formula: $$ \text{pH} = -\log_{10}([\mathrm{H}^+]) $$