A solution was prepared such that the initial concentrations of Cu²⁺(aq) and CN^-(aq) were 0.0120 M and 0.0400 M, respectively. These ions react according to the following chemical equation:
Cu²⁺(aq) + 4CN^-(aq)
Kc= 1.0 x 10²⁵ ⇌ Cu(CN) ₄^2-(aq)
What will be the concentration of Cu²⁺(aq) at equilibrium

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the reaction is:Kc = [Cu(CN)42-] / [Cu2+][CN-]4Substituting the initial concentrations and the equilibrium constant into the expression, we get:1.0 x 1025 = [Cu(CN)42-] / (0.0120 M)(0.0400 M)4Solving for [Cu(CN)42-], we get:[Cu(CN)42-] = 2.40 x 10-19 MThe concentration of Cu2+ at equilibrium is then:[Cu2+] = [Cu(CN)42-] / (4[CN-]4)Substituting the equilibrium concentration of Cu(CN)42- and the initial concentration of CN-, we get:[Cu2+] = (2.40 x 10-19 M) / (4(0.0400 M)4)= 2.00 x 10-6 MTherefore, the concentration of Cu2+ at equilibrium is 2.00 x 10-6 M, which is closest to choice A, 0.0020 M.

A Solution Was Prepared Such That the Initial Concentrations of Cu2+(aq)

A Solution Was Prepared Such That the Initial Concentrations of Cu²⁺(aq)