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 () Calculation
Given:
- Solubility of in water at 25°C is g/L.
- Molar mass of .
The solubility-product constant, , is calculated using the concentration of ions at equilibrium:
-
Convert solubility from g/L to mol/L (molarity):
This concentration represents the concentrations of and ions at equilibrium.
-
Write the expression for the solubility product constant :
[ K_{sp} = [\text{Cu}^{2+}][\text{CO}_3^{2-}] ]
-
Calculate the value of :
Since dissociates into one ion and one ion, their concentrations are both equal to the molarity of :
Part (b): Effect of pH on Solubility
Lowering the pH increases the concentration of or ions in the solution. Since is a basic ion, it will react with :
This reaction reduces the concentration of in the solution, which causes more to dissolve to re-establish equilibrium. Therefore, the solubility of increases when the pH is lowered.
- The solubility of increases when the pH is lowered.
- The dissociation of also increases.
Part (c): Effect of on Solubility
In the presence of , the solution already contains the common ion . According to the common ion effect, adding more decreases the solubility of , because the equilibrium shifts to the left to favor the undissolved form of the salt.
- In the presence of , the solubility of decreases.
- The dissociation of decreases due to the common ion effect.
Final Summary of Answers
- (a) The solubility-product constant of is .
- (b) The solubility of increases when the pH is lowered, as reacts with .
- (c) The solubility of decreases in the presence of due to the common ion effect.
Would you like more details or have any questions about this explanation?
Related Questions
- What is the impact of temperature on the solubility of ?
- How does the presence of other cations (like ) affect the solubility of carbonates?
- What are some real-world applications of the solubility product concept?
- Can you describe how the common ion effect influences chemical equilibria in general?
- What is the role of the 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.
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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