OCu2+ is a Lewis acid in the reaction, and CN" is a Lewis base O Removing some of the Cu2+ (e.g., by precipitation) will cause some of the complex ion to dissociate The complex ion is a Lewis acid-base adduct O If enough CN" is added, the equilibrium mixture would contain zero Cu2+

Chemistry for Engineering Students
4th Edition
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Lawrence S. Brown, Tom Holme
Chapter12: Chemical Equilibrium
Section: Chapter Questions
Problem 12.37PAE: Again the experiment in Exercise 12.33 was redesigned. This time, 0.15 mol each of N, and O2 was...
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Please answer 4,5,6 statement only

Choose the best answer below. Consider the following complex ion formation
equilibrium and it's equilibrium constant, Kf. Which of the following statements is
FALSE?
Cu2+ (aq) + 4CN" (aq) Û Cu(CN)4²- (aq)
Kf = 1.0x1025
O Adding more CN¯ will cause more complex ion to form
The equilibrium lies in favor of the complex ion
Cu2+ is a Lewis acid in the reaction, and CN- is a Lewis base
Removing some of the Cu2* (e.g., by precipitation) will cause some of the
complex ion to dissociate
The complex ion is a Lewis acid-base adduct
O If enough CN" is added, the equilibrium mixture would contain zero Cu2+
Transcribed Image Text:Choose the best answer below. Consider the following complex ion formation equilibrium and it's equilibrium constant, Kf. Which of the following statements is FALSE? Cu2+ (aq) + 4CN" (aq) Û Cu(CN)4²- (aq) Kf = 1.0x1025 O Adding more CN¯ will cause more complex ion to form The equilibrium lies in favor of the complex ion Cu2+ is a Lewis acid in the reaction, and CN- is a Lewis base Removing some of the Cu2* (e.g., by precipitation) will cause some of the complex ion to dissociate The complex ion is a Lewis acid-base adduct O If enough CN" is added, the equilibrium mixture would contain zero Cu2+
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