Final answer:
The molar solubility of calcium carbonate in a 0.285 M potassium carbonate solution will be lower than in pure water due to the common ion effect, which is not explicitly calculable without additional information or assumptions.
Step-by-step explanation:
The student asked about the molar solubility of calcium carbonate (CaCO₃) in a 0.285 M potassium carbonate solution. The Ksp (solubility product constant) of calcium carbonate is 4.5 × 10-9. However, when calcium carbonate is placed in a solution of potassium carbonate, there is a common ion effect due to the presence of carbonate ions, which will affect the solubility of calcium carbonate.
To understand this effect, we must apply Le Chatelier's principle, which predicts that the addition of a common ion will decrease the solubility of the compound. In general, the increased concentration of carbonate ions in solution will shift the equilibrium towards the solid (undissolved) calcium carbonate, thereby decreasing its molar solubility.
In a pure water system, the dissociation of calcium carbonate can be represented by the following equilibrium reaction, and the solubility can be calculated through an ICE table:
- CaCO₃ (s) → Ca²⁺ (aq) + CO₃²⁺ (aq)
We would normally set up an ICE table and calculate the solubility 's'. But in this case, because of the common ion effect with the given 0.285 M of carbonate ions already in solution, the solubility will be even lower than the solubility in pure water and it will require a more complex calculation that considers the initial concentration of carbonate ions.