2. Copper hydroxide is only very slightly soluble in water Cu(OH)2(s) Cu2 (aq) +2OH-(aq) ㅡㅡ. Formulate the expression for the equilibrium constant, Ksp, for the above reaction. a. it is possible to dissolve significant amounts of Cuo is possible to dissolve significant amounts of Cu(OH)2 in solutions in which the concentration of either Cu or OH is very, very small. Explain, using Ksp, why this is the case. b. It Explain why Cu(OH)2 might have very appreciable solubility in 1 M HCI. (Consider the effect of Reaction 3 on the Cu(OH)2 solution reaction.) c.
The Correct Answer and Explanation is :
To address your questions, let’s first break down the solubility equilibrium of copper hydroxide (Cu(OH)2).
The Equilibrium Expression for Ksp:
For the dissolution of copper hydroxide:
[ \text{Cu(OH)}_2 (s) \rightleftharpoons \text{Cu}^{2+} (aq) + 2 \text{OH}^- (aq) ]
The equilibrium constant (Ksp) expression for this reaction is:
[
K_{sp} = [\text{Cu}^{2+}] [\text{OH}^-]^2
]
Where:
- ( [\text{Cu}^{2+}] ) is the concentration of copper ions in the solution.
- ( [\text{OH}^-] ) is the concentration of hydroxide ions in the solution.
a. Solubility of Cu(OH)2 in Low Concentrations of Cu²⁺ or OH⁻:
The solubility of Cu(OH)2 is governed by its Ksp value, which is very small because Cu(OH)2 is only slightly soluble in water. If either the concentration of Cu²⁺ or OH⁻ is very low, the system can still dissolve significant amounts of Cu(OH)2. Here’s why:
- If the concentration of Cu²⁺ is low, the equilibrium will shift towards the right, dissolving more Cu(OH)2 to produce more Cu²⁺ and OH⁻ ions, thus increasing the solubility.
- If the concentration of OH⁻ is low, the system will also shift to the right to increase the OH⁻ concentration by dissolving more Cu(OH)2.
In both cases, the equilibrium will favor the dissolution of more Cu(OH)2 to maintain the equilibrium constant, which can result in significant solubility even though Cu(OH)2 is generally insoluble in water.
b. Solubility of Cu(OH)2 in 1 M HCl:
When Cu(OH)2 is dissolved in 1 M HCl, the chloride ions (Cl⁻) from the HCl will react with the Cu²⁺ ions to form a soluble complex, copper chloride (CuCl2⁻):
[
\text{Cu}^{2+} (aq) + 2 \text{Cl}^- (aq) \rightarrow \text{CuCl}_2^- (aq)
]
As Cu²⁺ ions are removed from the solution by the formation of CuCl2⁻, the equilibrium of Cu(OH)2 will shift to the right to replace the Cu²⁺ ions that have been removed. This will result in an increased dissolution of Cu(OH)2, making it more soluble in the acidic solution.
Image Representation:
To illustrate this, I can generate an image showing the equilibrium processes involving Cu(OH)2 dissolution in both pure water and in 1 M HCl. I’ll go ahead and create this visual for you.
Let me know if you’d like me to proceed with the image generation!