Calculate the molar solubility of Cu(OH)2, Ksp = 2.2 × 10–20, in 0.91 M NH3. Don't forget to use the complexation reaction Cu2+ + 4 NH3 ⇌ Cu(NH3)42+, K = 5.0 × 1013. Round your answer to two significant figures and do NOT use scientific notation.
Cu(OH)2, Ksp =1.6×10^-19
To calculate the molar solubility of Cu(OH)2 in 0.91 M NH3, we can use the concept of the common ion effect and the solubility product constant (Ksp).
First, let's write the balanced equation for the dissolution of Cu(OH)2:
Cu(OH)2 (s) ⇌ Cu2+ (aq) + 2 OH- (aq)
The solubility product constant, Ksp, is given as 2.2 × 10^–20, which represents the equilibrium expression for the dissolution of Cu(OH)2:
Ksp = [Cu2+][OH-]^2
Now, let's consider the complexation reaction between Cu2+ and NH3:
Cu2+ (aq) + 4 NH3 (aq) ⇌ Cu(NH3)4^2+ (aq)
The equilibrium constant for this reaction, K, is given as 5.0 × 10^13.
The complex Cu(NH3)4^2+ can also be considered as a source of Cu2+ ions. So let's define 'x' as the molar solubility of Cu(OH)2 in moles per liter (mol/L). Then, the concentration of Cu2+ ions can be calculated as 'x' (mol/L) from the dissolution of Cu(OH)2, and '4x' (mol/L) from the complex Cu(NH3)4^2+.
The concentration of OH- ions can be given as '2x' (mol/L) from the dissolution of Cu(OH)2.
NH3 is a weak base, and its concentration in the solution is given as 0.91 M.
Now, let's set up an expression for Ksp using the concentrations we just defined:
Ksp = [Cu2+][OH-]^2
2.2 × 10^–20 = (x + 4x)(2x)^2
2.2 × 10^–20 = 8x^3
Rearranging this equation:
8x^3 = 2.2 × 10^–20
Solving for x:
x^3 = (2.2 × 10^–20) / 8
x^3 = 2.75 × 10^–21
x ≈ 0.033
The molar solubility of Cu(OH)2 in 0.91 M NH3 is approximately 0.033 mol/L, rounded to two significant figures.