Let's solve the problem step-by-step: --- ## (a) Calculate \([CO_2(aq)]\) at 25°C We use **Henry's Law**: \[ [CO_2(aq)] = K_H \times P_{CO_2} \] Given: - \( K_{H,CO_2} = 10^{-1.46} \) M/atm - \( P_{CO_2} = 2 \) atm Calculate \( K_H \): \[ K_H = 10^{-1.46} \approx .0347 \text{ M/atm} \] Now calculate \([CO_2(aq)]\): \[ [CO_2(aq)] = .0347 \times 2 = .0694 \text{ M} \] **Final answer for (a):** \[ \boxed{[CO_2(aq)] = .0694 \text{ M}} \] --- ## (b) Calculate the pH of the soda The relevant equilibrium is: \[ CO_2(aq) + H_2O \leftrightarrow HCO_3^- + H^+ \quad K_{a,1} = 10^{-6.33} \] Let \( x = [H^+] = [HCO_3^-] \) produced. The initial concentration of \( CO_2(aq) \) is \( .0694 \) M, and at equilibrium, it will be \( .0694 - x \). The equilibrium expression: \[ K_{a,1} = \frac{[HCO_3^-][H^+]}{[CO_2(aq)]} \] \[ 10^{-6.33} = \frac{x^2}{.0694 - x} \] Since \( K_{a,1} \) is small, \( x \ll .0694 \), so \( .0694 - x \approx .0694 \): \[ 10^{-6.33} = \frac{x^2}{.0694} \] \[ x^2 = 10^{-6.33} \times .0694 \] First, calculate \( 10^{-6.33} \): \[ 10^{-6.33} \approx 4.68 \times 10^{-7} \] Now, \[ x^2 = (4.68 \times 10^{-7}) \times .0694 = 3.247 \times 10^{-8} \] \[ x = \sqrt{3.247 \times 10^{-8}} \approx 1.8 \times 10^{-4} \text{ M} \] Now, pH is: \[ pH = -\log_{10}(x) = -\log_{10}(1.8 \times 10^{-4}) \approx 3.74 \] **Final answer for (b):** \[ \boxed{pH = 3.74} \] --- ### **Summary of Answers** (a) \([CO_2(aq)] = .0694\) M (b) \(pH = 3.74\)