Math Problem Statement

(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  ---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

Solution

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

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

Given:

  • Solubility of CuCO3\text{CuCO}_3 in water at 25°C is 1.46×1031.46 \times 10^{-3} g/L.
  • Molar mass of CuCO3=123.55g/mol\text{CuCO}_3 = 123.55 \, \text{g/mol}.

The solubility-product constant, KspK_{sp}, is calculated using the concentration of ions at equilibrium: CuCO3(s)Cu2+(aq)+CO32(aq)\text{CuCO}_3(s) \rightleftharpoons \text{Cu}^{2+}(aq) + \text{CO}_3^{2-}(aq)

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

    Molarity of CuCO3=1.46×103 g/L123.55 g/mol=1.18×105 mol/L\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 Cu2+\text{Cu}^{2+} and CO32\text{CO}_3^{2-} ions at equilibrium.

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

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

  3. Calculate the value of KspK_{sp}:

    Since CuCO3\text{CuCO}_3 dissociates into one Cu2+\text{Cu}^{2+} ion and one CO32\text{CO}_3^{2-} ion, their concentrations are both equal to the molarity of CuCO3\text{CuCO}_3: Ksp=(1.18×105)(1.18×105)=1.39×1010K_{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 H+\text{H}^+ or H3O+\text{H}_3\text{O}^+ ions in the solution. Since CO32\text{CO}_3^{2-} is a basic ion, it will react with H+\text{H}^+: CO32+H+HCO3\text{CO}_3^{2-} + \text{H}^+ \rightarrow \text{HCO}_3^-

This reaction reduces the concentration of CO32\text{CO}_3^{2-} in the solution, which causes more CuCO3\text{CuCO}_3 to dissolve to re-establish equilibrium. Therefore, the solubility of CuCO3\text{CuCO}_3 increases when the pH is lowered.

  • The solubility of CuCO3\text{CuCO}_3 increases when the pH is lowered.
  • The dissociation of CuCO3\text{CuCO}_3 also increases.

Part (c): Effect of MgCO3\text{MgCO}_3 on Solubility

In the presence of MgCO3\text{MgCO}_3, the solution already contains the common ion CO32\text{CO}_3^{2-}. According to the common ion effect, adding more CO32\text{CO}_3^{2-} decreases the solubility of CuCO3\text{CuCO}_3, because the equilibrium shifts to the left to favor the undissolved form of the salt.

  • In the presence of MgCO3\text{MgCO}_3, the solubility of CuCO3\text{CuCO}_3 decreases.
  • The dissociation of CuCO3\text{CuCO}_3 decreases due to the common ion effect.

Final Summary of Answers

  1. (a) The solubility-product constant KspK_{sp} of CuCO3\text{CuCO}_3 is 1.39×10101.39 \times 10^{-10}.
  2. (b) The solubility of CuCO3\text{CuCO}_3 increases when the pH is lowered, as CO32\text{CO}_3^{2-} reacts with H+\text{H}^+.
  3. (c) The solubility of CuCO3\text{CuCO}_3 decreases in the presence of MgCO3\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 CuCO3\text{CuCO}_3?
  2. How does the presence of other cations (like Ca2+\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 KspK_{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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Chemistry Equilibrium
Solubility Product
Common Ion Effect
pH and Solubility

Formulas

Ksp = [Cu²⁺][CO₃²⁻]
Solubility (mol/L) = (given solubility in g/L) / (molar mass of CuCO₃)

Theorems

Solubility Product Constant (Ksp)
Le Chatelier's Principle
Common Ion Effect

Suitable Grade Level

Grades 11-12