Question

The complex ion Cu(NH3)42 is formed in a solution made of 0.0300 M Cu(NO3)2 and 0.400 M NH3. What are the concentrations of Cu2 , NH3, and Cu(NH3)42 at equilibrium? The formation constant*, Kf, of Cu(NH3)42 is 1.70 × 1013.

This is what the hint gave me,

Since the formation constant, Kf, is very large, it can be assumed that the reaction goes nearly to completion to form Cu(NH3)42 . Cu2 is the limiting reagent and will be used up first. Additionally, some NH3 will also be consumed to form the complex ion. Initially, there are 0.400 M of NH3 and 0.0300 M of Cu2 . Consider the stoichiometric relationship between Cu2 and the other species in the balanced reaction in ordered to determine how much NH3 is consumed and how much Cu(NH3)42 is formed when Cu2 and NH3 are mixed. Every 1 mole of Cu2 will react with 4 moles of NH3; therefore, NH3 will consume 4 times the initial concentration of Cu2 . Additionally, 1 mole of Cu2 reacts to form 1 mole of Cu(NH3)42 ; therefore, Cu(NH3)42 will form the same amount as the initial concentration of Cu2 . Use these relationships to calculate the concentrations of all the species after Cu2 mixes with NH3.

Concentration (M)	Cu2+ (aq)	+	4NH3 (aq)	<---->	Cu(NH3)42+ (aq)
Initial	[Cu2+]o		[NH3]0		0
Change	-[Cu2+]o		-4[NH3]0		+[Cu2+]0
Final	~0		[NH3]0-4[Cu2+]0		[Cu2+]0

[Cu2 ]0 and [NH3]0 represent the initial concentrations of Cu2 and NH3, respectively. Note above that the final concentration of Cu2 is approximately 0 M. We initially assumed that the reaction went to completion, however, though this assumption was made, Cu(NH3)42 will still dissociate, or back-react, to produce Cu2 and NH3 in order to reach equilibrium. Therefore, set up an ICE table to calculate the equilibrium values for each species where the final concentrations in the above table are now the initial values used in the ICE table.

Please help I need to finish this today!

Answer 1