Answer:
The balanced net ionic equation to show why the solubility of CaCO₃(s) increases in the presence of a strong acid is:
CaCO₃(s) + 2 H⁺(aq) → Ca²⁺(aq) + CO₂(g) + H₂O(l)
In this reaction, the strong acid (H⁺) reacts with the carbonate ion (CO₃²⁻) in CaCO₃(s) to form carbonic acid (H₂CO₃), which is unstable and decomposes into water and carbon dioxide gas (CO₂). As CO₂(g) escapes into the air, the equilibrium shifts towards the products, resulting in increased solubility of CaCO₃(s) in the acid solution.
To calculate the equilibrium constant (K) for this reaction, we can use the following equation:
K = [Ca²⁺][CO₂]/[H⁺]²
The concentrations of Ca²⁺ and CO₂ are determined by the solubility product constant for CaCO₃, which is 4.8 × 10⁻⁹ at 25°C. Since CaCO₃(s) is a sparingly soluble salt, we can assume that the concentration of Ca²⁺ in the solution is negligible compared to the initial concentration of H⁺, and that the concentration of CO₂ can be approximated as the partial pressure of CO₂ in the gas phase. Therefore, we can simplify the expression for K as:
K ≈ Ksp[CO₂]/[H⁺]²
Substituting the values, we get:
K ≈ (4.8 × 10⁻⁹) × (Pco₂/[H⁺]²)
At standard conditions (1 atm and 25°C), the partial pressure of CO₂ is 0.03 atm. If we assume that the initial concentration of H⁺ is 0.1 M, we can calculate the equilibrium constant:
K ≈ (4.8 × 10⁻⁹) × (0.03/0.1)² ≈ 1.44 × 10⁻⁶
Therefore, the equilibrium constant for the reaction between CaCO₃(s) and a strong acid is approximately 1.44 × 10⁻⁶ at 25°C.
Step-by-step explanation:
please follow me for more if you need any help