Let's tackle each part of the question in detail.

Part (a): Solubility-Product Constant ($K_{sp}$) Calculation

Given:

• Solubility of $\text{CuCO}_3$ in water at 25°C is $1.46 \times 10^{-3}$ g/L.
• Molar mass of $\text{CuCO}_3 = 123.55 \, \text{g/mol}$.

The solubility-product constant, $K_{sp}$, is calculated using the concentration of ions at equilibrium: $\text{CuCO}_3(s) \rightleftharpoons \text{Cu}^{2+}(aq) + \text{CO}_3^{2-}(aq)$

1. Convert solubility from g/L to mol/L (molarity):

   $\text{Molarity of CuCO}_3 = \frac{1.46 \times 10^{-3} \text{ g/L}}{123.55 \text{ g/mol}} = 1.18 \times 10^{-5} \text{ mol/L}$

   This concentration represents the concentrations of $\text{Cu}^{2+}$ and $\text{CO}_3^{2-}$ ions at equilibrium.

2. Write the expression for the solubility product constant $K_{sp}$:

   [ K_{sp} = [\text{Cu}^{2+}][\text{CO}_3^{2-}] ]

3. Calculate the value of $K_{sp}$:

   Since $\text{CuCO}_3$ dissociates into one $\text{Cu}^{2+}$ ion and one $\text{CO}_3^{2-}$ ion, their concentrations are both equal to the molarity of $\text{CuCO}_3$: $K_{sp} = (1.18 \times 10^{-5})(1.18 \times 10^{-5}) = 1.39 \times 10^{-10}$

Part (b): Effect of pH on Solubility

Lowering the pH increases the concentration of $\text{H}^+$ or $\text{H}_3\text{O}^+$ ions in the solution. Since $\text{CO}_3^{2-}$ is a basic ion, it will react with $\text{H}^+$: $\text{CO}_3^{2-} + \text{H}^+ \rightarrow \text{HCO}_3^-$

This reaction reduces the concentration of $\text{CO}_3^{2-}$ in the solution, which causes more $\text{CuCO}_3$ to dissolve to re-establish equilibrium. Therefore, the solubility of $\text{CuCO}_3$ increases when the pH is lowered.

• The solubility of $\text{CuCO}_3$ increases when the pH is lowered.
• The dissociation of $\text{CuCO}_3$ also increases.

Part (c): Effect of $\text{MgCO}_3$ on Solubility

In the presence of $\text{MgCO}_3$, the solution already contains the common ion $\text{CO}_3^{2-}$. According to the common ion effect, adding more $\text{CO}_3^{2-}$ decreases the solubility of $\text{CuCO}_3$, because the equilibrium shifts to the left to favor the undissolved form of the salt.

• In the presence of $\text{MgCO}_3$, the solubility of $\text{CuCO}_3$ decreases.
• The dissociation of $\text{CuCO}_3$ decreases due to the common ion effect.

Final Summary of Answers

1. (a) The solubility-product constant $K_{sp}$ of $\text{CuCO}_3$ is $1.39 \times 10^{-10}$.
2. (b) The solubility of $\text{CuCO}_3$ increases when the pH is lowered, as $\text{CO}_3^{2-}$ reacts with $\text{H}^+$.
3. (c) The solubility of $\text{CuCO}_3$ decreases in the presence of $\text{MgCO}_3$ due to the common ion effect.

Would you like more details or have any questions about this explanation?

Related Questions

1. What is the impact of temperature on the solubility of $\text{CuCO}_3$?
2. How does the presence of other cations (like $\text{Ca}^{2+}$) affect the solubility of carbonates?
3. What are some real-world applications of the solubility product concept?
4. Can you describe how the common ion effect influences chemical equilibria in general?
5. What is the role of the $K_{sp}$ value in predicting precipitation reactions?

Tip:

When dealing with solubility equilibria, always consider how changes in ion concentration, temperature, or pH can shift the equilibrium according to Le Chatelier's principle.