Consider the reaction If the K_sp for Cu(OH)₂ is 2.2 $\times$ 10^-20,what is the value of the equilibrium constant,K,for the reaction below? A) 1.0 $\times$ 10^-33 B) 4.6 $\times$ 10^-7 C) 2.1 $\times$ 10¹³ D) 2.2 $\times$ 10⁶ E) 9.5 $\times$ 10³²

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is: K = [Cu2+][OH-]^2/[Cu(OH)2] Using the given value of K_sp, we can calculate the solubility product of Cu(OH)2 as: K_sp = [Cu2+][OH-]^2 = 2.2 × 10^-16 Therefore, the equilibrium constant K can be written as: K = [Cu2+][OH-]^2/[Cu(OH)2] = [Cu2+][OH-]^2/K_sp Substituting the values: K = 2.2 × 10^-16/2.2 × 10^-12 = 10^-4 Therefore, the answer is option D: 2.2 × 10^(-6) or 0.0000022.

Consider the Reaction If the Ksp for Cu(OH)2 Is ×\times×

Consider the Reaction

If the K_sp for Cu(OH)₂ Is $\times$