What is the solubility of copper hydroxide Cu(OH)2 in mL? Mwt of Cu(OH)2 = 97.57; Ksp = 5.5 x 10^21, Cu^2+ (aq) + 2 OH^- → Cu(OH)2 (s) (aq)
The Correct Answer and Explanation is:
To calculate the solubility of copper hydroxide (Cu(OH)₂), we need to use its solubility product constant (Ksp) and apply stoichiometry. Here’s the step-by-step approach:
Given Data:
- Molar mass of Cu(OH)₂: 97.57 g/mol
- Ksp = 5.5 × 10⁻²¹
- The dissolution reaction for Cu(OH)₂ is: Cu²⁺(aq) + 2OH⁻(aq) ⇌ Cu(OH)₂(s)
Let the solubility of Cu(OH)₂ be “S” mol/L. This represents the amount of Cu(OH)₂ that dissolves in water. When Cu(OH)₂ dissolves, it dissociates into Cu²⁺ and OH⁻ ions according to the reaction:
- Cu(OH)₂ (s) ⇌ Cu²⁺ (aq) + 2 OH⁻ (aq)
From this, we can see that for every 1 mole of Cu(OH)₂ that dissolves, 1 mole of Cu²⁺ and 2 moles of OH⁻ are produced. So, if the solubility of Cu(OH)₂ is “S” mol/L:
- [Cu²⁺] = S mol/L
- [OH⁻] = 2S mol/L
Now, use the Ksp expression for the dissociation equilibrium: Ksp=[Cu2+][OH−]2Ksp = [Cu²⁺][OH⁻]^2Ksp=[Cu2+][OH−]2
Substituting the concentrations into the Ksp expression: Ksp=(S)(2S)2=4S3Ksp = (S)(2S)² = 4S³Ksp=(S)(2S)2=4S3
Given that Ksp = 5.5 × 10⁻²¹, we can solve for S: 5.5×10−21=4S35.5 × 10⁻²¹ = 4S³5.5×10−21=4S3 S3=5.5×10−214S³ = \frac{5.5 × 10⁻²¹}{4}S3=45.5×10−21 S3=1.375×10−21S³ = 1.375 × 10⁻²¹S3=1.375×10−21 S=1.375×10−213≈5.13×10−8mol/LS = \sqrt[3]{1.375 × 10⁻²¹} \approx 5.13 × 10⁻⁸ mol/LS=31.375×10−21≈5.13×10−8mol/L
So the solubility of Cu(OH)₂ in moles per liter is approximately 5.13 × 10⁻⁸ mol/L.
To convert the solubility to grams per liter:
Now, to find the solubility in grams per liter, we multiply the molar solubility by the molar mass of Cu(OH)₂: Solubility (g/L)=S×Molar mass of Cu(OH)₂\text{Solubility (g/L)} = S \times \text{Molar mass of Cu(OH)₂}Solubility (g/L)=S×Molar mass of Cu(OH)₂ Solubility (g/L)=(5.13×10−8mol/L)×97.57g/mol\text{Solubility (g/L)} = (5.13 × 10⁻⁸ mol/L) × 97.57 g/molSolubility (g/L)=(5.13×10−8mol/L)×97.57g/mol Solubility (g/L)≈5.0×10−6g/L\text{Solubility (g/L)} \approx 5.0 × 10⁻⁶ g/LSolubility (g/L)≈5.0×10−6g/L
This means the solubility of Cu(OH)₂ in water is approximately 5.0 × 10⁻⁶ g/L, or 5.0 μg/L.
Conclusion:
Copper hydroxide (Cu(OH)₂) is extremely insoluble in water, as indicated by its very low solubility value of approximately 5.0 μg/L. This is consistent with the very small Ksp value, which reflects the limited dissociation of Cu(OH)₂ in aqueous solution.
