(a) If the solubility of CuCO3 in water at 25°C is 1.46 ✕ 10−3 g/L, calculate the solubility-product constant (*K*sp) of CuCO3, assuming complete dissociation of the CuCO3 that has dissolved.

(b) How will the solubility of CuCO3 be affected by pH? Explain your answer.

The solubility of CuCO3  ---Select--- increases stays the same decreases when the pH is lowered. The CO32− anions are basic and react with H3O+ ions, and the dissociation of CuCO3  ---Select--- increases stays the same decreases .

(c) How will the solubility of CuCO3 be affected if the solution includes MgCO3? Explain your answer.

In the presence of MgCO3, the solubility of CuCO3  ---Select--- increases stays the same decreases . MgCO3 provides the common ion CO32− and the dissociation of CuCO3

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.

(a) If the solubility of CuCO3 in water at 25°C is 1.46 ✕ 10−3 g/L, calculate the solubility-product constant (Ksp) of CuCO3, assuming complete dissociation of the CuCO3 that has dissolved. (b) How will the solubility of CuCO3 be affected by pH? Explain your answer. The solubility of CuCO3 increases when the pH is lowered. The CO32− anions are basic and react with H3O+ ions, and the dissociation of CuCO3 increases. (c) How will the solubility of CuCO3 be affected if the solution includes MgCO3? Explain your answer. In the presence of MgCO3, the solubility of CuCO3 decreases.

Learn how to calculate the solubility-product constant (Ksp) of CuCO3 and understand how pH and the presence of MgCO3 affect its solubility. This detailed solution covers chemistry equilibrium concepts like Ksp, the common ion effect, and Le Chatelier's principle. Suitable for high school students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation