When (g) is dissolved in water, the following equilibrium is established: (aq) + 2(l) ⇌ H3O+(aq) + H — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Express the Equilibrium Constant ($K$) The chemical equation representing the dissolution and subsequent ionization of carbon dioxide in water is: \[\ce{CO2(aq) + 2H2O(l) <=> H3O+(aq) + HCO3-(aq)}\] According to the law of chemical equilibrium, the concentration-based equilibrium constant ($K$) is written as: \[K = \frac{[\ce{H3O+}][\ce{HCO3-}]}{[\ce{CO2}][\ce{H2O}]^2}\] Since liquid water (\ce{H2O(l)}) is a pure liquid solvent, its activity (active mass) is constant and is taken as unity ($1$). Therefore, it is omitted from the equilibrium constant expression: \[K = \frac{[\ce{H3O+}][\ce{HCO3-}]}{[\ce{CO2}]}\] Step 2 - Calculate the Hydronium Ion Concentration ($[\ce{H3O+}]$) from the pH The pH of a solution is defined as the negative logarithm (base 10) of the hydronium ion concentration: \[\text{pH} = -\log_{10}[\ce{H3O+}]\] Rearranging this equation to solve for the concentration of hydronium ions: \[[\ce{H3O+}] = 10^{-\text{pH}}\] Given that the pH of the solution is $6.0$: \[[\ce{H3O+}] = 10^{-6.0}\text{ M} = 1.0 \times 10^{-6}\text{ M}\] Step 3 - Set up the Ratio of Concentrations and Calculate We rearrange our equilibrium constant expression to solve for the ratio of the concentration of bicarbonate ions ($[\ce{HCO3-}]$) to dissolved carbon dioxide ($[\ce{CO2}]$): \[K = [\ce{H3O+}] \times \frac{[\ce{HCO3-}]}{[\ce{CO2}]} \implies \frac{[\ce{HCO3-}]}{[\ce{CO2}]} = \frac{K}{[\ce{H3O+}]}\] Substitute the given equilibrium constant $K = 3.8 \times 10^{-7}$ and the calculated hydronium ion concentration $[\ce{H3O+}] = 1.0 \times 10^{-6}\text{ M}$: \[\frac{[\ce{HCO3-}]}{[\ce{CO2}]} = \frac{3.8 \times 10^{-7}}{1.0 \times 10^{-6}}\] \[\frac{[\ce{HCO3-}]}{[\ce{CO2}]} = 3.8 \times 10^{-7 - (-6)} = 3.8 \times 10^{-1} = 0.38\] Thus, the ratio of the concentration of $\ce{HCO3-(aq)}$ to $\ce{CO2(aq)}$ is: \[\frac{[\ce{HCO3-}]}{[\ce{CO2}]} = \boxed{0.38}\] Step 4 - Evaluate the Options * **Option (A) $3.8 \times 10^{-13}$**: Incorrect. This value is obtained if the equilibrium constant is incorrectly multiplied by the hydronium ion concentration: $K \times [\ce{H3O+}] = 3.8 \times 10^{-7} \times 10^{-6} = 3.8 \times 10^{-13}$. * **Option (B) $6.0$**: Incorrect. This is simply the pH of the solution, which does not represent the required ratio of concentrations. * **Option (C) $0.38$**: Correct. As mathematically demonstrated above, dividing the equilibrium constant by the hydronium ion concentration yields exactly $0.38$. * **Option (D) $13.4$**: Incorrect. This is an arithmetic calculation error.