Given the following values of equilibrium constants: What is the value of the equilibrium constant for the following reaction? Cu(OH)₂(s) + 4NH₃(aq) Cu(NH₃)₄²⁺(aq) + 2OH^-(aq) A) 1.6 × 10^-19 B) 1.6 × 10^-6 C) 6.2 × 10³¹ D) 1.0 × 10¹³ E) 1.6 × 10^-32

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is:

K = (Cu(NH$_3$)$_4$(OH)$_2^{2+}$)(OH$^-$)$_2$ / (Cu(OH)$_2$)(NH$_3$)$_4$(OH)$_2^{2+}$)(NH$_3$)$_2$

The given equilibrium constants can be used to form a series of equations that relate K for the given reaction to the given equilibrium constants:

K$_1$ = (Cu(OH)$_2$)/(Cu$^{2+}$)(OH$^-$)$^2$
K$_2$ = (Cu(NH$_3$)$_4$(OH)$_2^{2+}$)/(Cu$^{2+}$)(OH$^-$)$^2$(NH$_3$)$_2$
K$_3$ = (Cu(NH$_3$)$_4^{2+}$)/(Cu$^{2+}$)(NH$_3$)$_4$
K$_4$ = (OH$^-$)$_2$/ (H$_2$O)

By multiplying these equations together and simplifying, we get:

K = K$_2$ / K$_1$ = K$_3$(OH$^-$)$_2$/ (Cu(OH)$_2$)(NH$_3$)$_2$

Plugging in the given values for the equilibrium constants, we get:

K = (1.6 × 10$^{-6}$)(10$^{-4}$)$^2$ / (1.1 × 10$^{-18}$)(10$^{-4}$)$^2$ = 1.6 × 10$^{-6}$

Therefore, the correct answer is choice B, 1.6 × 10$^{-6}$.

Given the Following Values of Equilibrium Constants: What Is