Cu(II) forms the complex ion [Cu(H₂O)₆]²⁺through equilibrium reactions in which ammonia replaces water in a stepwise manner. If log K₁ = 4.78, log K₂ = 2.36, log K₃ = -0.76, and log K₄ = -0.53, what is its K_f?

Question

Quizplus · Accepted Answer

The overall formation constant, $K_{f}$, for the complex ion can be calculated by multiplying the stepwise equilibrium constants: $K_{f} = K_{1} \times K_{2} \times K_{3} \times K_{4}$. Given the logarithmic values, we add them to find the logarithm of $K_{f}$: $4.78 + 2.36 - 0.76 - 0.53 = 6.85$. Therefore, $K_{f} = 10^{6.85} = 7.08 \times 10^{6}$, which is closest to the value given in choice A, $1.41 \times 10^{7}$.

Cu(II) Forms the Complex Ion [Cu(H2O)6]2+ through Equilibrium Reactions in Which

Cu(II) Forms the Complex Ion [Cu(H₂O)₆]²⁺through Equilibrium Reactions in Which