Problem 26 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

When \(\mathrm{CO}_{2}(\mathrm{~g})\) is dissolved in water, the following equilibrium is established: \(\mathrm{CO}_{2}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\) \(+\mathrm{HCO}_{3}^{-}(\mathrm{aq})\) for which the equilibrium constant is \(3.8 \times 10^{-7}\). If the pH of solution is \(6.0\), what would be the ratio of concentration of \(\mathrm{HCO}_{3}^{-}(\mathrm{aq})\) to \(\mathrm{CO}_{2}(\mathrm{aq})\) ? (a) \(3.8 \times 10^{-13}\) (b) \(6.0\) (c) \(0.38\) (d) \(13.4\)

Step-by-Step Solution

Answer

The ratio of concentration of \(\mathrm{HCO}_3^-\) to \(\mathrm{CO}_2\) is approximately 0.38.

Step 1: Understanding the Equilibrium Constant

The equilibrium constant expression for the reaction is: \[ K = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{HCO}_3^-]}{[\mathrm{CO}_2]} \]Given that the equilibrium constant \( K \) is \( 3.8 \times 10^{-7} \), we can use this value in our calculations.

Step 2: Calculate the concentration of hydronium ion

pH is given as 6.0, which is the negative logarithm of the hydronium ion concentration: \[ pH = -\log [\mathrm{H}_3\mathrm{O}^+] \]Therefore, \[ [\mathrm{H}_3\mathrm{O}^+] = 10^{-pH} = 10^{-6.0} \]\[ [\mathrm{H}_3\mathrm{O}^+] = 1.0 \times 10^{-6} \,M \]

Step 3: Establish the ratio of concentrations

Now we can express the ratio of \( [\mathrm{HCO}_3^-] \) to \( [\mathrm{CO}_2] \) using the equilibrium constant expression: \[ 3.8 \times 10^{-7} = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{HCO}_3^-]}{[\mathrm{CO}_2]} \]We know that \( [\mathrm{H}_3\mathrm{O}^+] = 1.0 \times 10^{-6} \), so we can rearrange to find the ratio: \[ [\mathrm{HCO}_3^-] = 3.8 \times 10^{-7} \times \frac{[\mathrm{CO}_2]}{[\mathrm{H}_3\mathrm{O}^+]} \]\[ [\mathrm{HCO}_3^-] = 3.8 \times 10^{-7} \times \frac{[\mathrm{CO}_2]}{1.0 \times 10^{-6}} \]\[ [\mathrm{HCO}_3^-] = \frac{3.8 \times 10^{-7}}{1.0 \times 10^{-6}} \times [\mathrm{CO}_2] \]\[ [\mathrm{HCO}_3^-] \approx 0.38 \times [\mathrm{CO}_2] \]

Step 4: Identify the correct ratio

The ratio of the concentration of \( [\mathrm{HCO}_3^-] \) to \( [\mathrm{CO}_2] \) is approximately \( 0.38 \). Therefore, the correct answer is (c) \( 0.38 \).

Key Concepts

Equilibrium Constant

The equilibrium constant (K) is a fundamental concept in chemistry that represents the ratio of the concentrations of the products to the reactants at equilibrium in a reversible chemical reaction. It is given by the law of mass action, which states that at a constant temperature, the equilibrium constant is constant for a particular reaction.

For the reaction involving the dissolution of carbon dioxide in water, the equilibrium constant expression is formulated as: \[ K = \frac{[\mathrm{H}_3\mathrm{O}^+][\mathrm{HCO}_3^-]}{[\mathrm{CO}_2]} \]
This means that the product of the hydronium ion and bicarbonate ion concentrations, divided by the carbon dioxide concentration, remains constant at equilibrium. A large equilibrium constant indicates a reaction with a high tendency to form products, while a small value denotes a reaction that favors reactants.

pH Calculation

pH is a scale used to specify the acidity or basicity of an aqueous solution. It is a measure of the concentration of hydronium ions (\([\mathrm{H}_3\mathrm{O}^+]\)) in the solution. The pH is calculated using the following relationship: \[ pH = -\log [\mathrm{H}_3\mathrm{O}^+] \]

This logarithmic scale means that as the hydronium ion concentration decreases, the pH value increases and the solution becomes less acidic. Conversely, an increase in the hydronium ion concentration leads to a lower pH, indicating a more acidic solution. In the given exercise, the pH value of 6.0 indicates a hydronium ion concentration of \(1.0 \times 10^{-6}\) M.

Hydronium Ion Concentration

Hydronium ions (\([\mathrm{H}_3\mathrm{O}^+]\)) are formed when an acid dissolves in water and contributes to the solution's acidity. The concentration of these ions is directly related to the pH and can be calculated from it. As seen in the exercise, the hydronium ion concentration can be determined using the pH value: \[ [\mathrm{H}_3\mathrm{O}^+] = 10^{-pH} \]
With a pH of 6.0, the hydronium ion concentration is \(1.0 \times 10^{-6}\) M. The concentration of hydronium ions is crucial for calculating the equilibrium constant and for understanding the chemical behavior of the system.

Concentration Ratio

In chemical equilibrium, the concentration ratio of reactants and products provides insight into the extent of the reaction. It's especially important when considering reactions that do not go to completion. In the context of the given problem, the concentration ratio of bicarbonate ions to carbon dioxide is derived from the equilibrium constant and the known hydronium ion concentration.

By rearranging the equilibrium expression such that \( [\mathrm{HCO}_3^-] \) is isolated, it becomes possible to solve for the ratio using the known values. Therefore, the concentration ratio reflects the relative amounts of different species in the solution at equilibrium. It is essential for making predictions about the behavior of the system under various conditions and understanding the dynamic nature of chemical equilibria.