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write a balanced net ionic equation to show why the solubility of coco3(s) increases in the presence of a strong acid and calculate the equilibrium constant for the reaction of this sparingly soluble salt with acid. consider only the first step in the reaction with strong acid. use the pull-down boxes to specify states such as (aq) or (s).

User Liontass
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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:

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User GustyWind
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