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*, K_f, of Cu(NH₃)₄²⁺ is 1.70 x 10¹³

 Suppose you have a solution that contains 0.0440 M Ca²⁺ and 0.0980 M Ag⁺. If solid Na₃PO₄ is added to this mixture, which of the following phosphate species would precipitate out of solution first?

 When the second cation just starts to precipitate, what percentage of the first cation remains in solution?

Answer 1

Cu(NH3)4]2+↔Cu2+ +4NH3

ICE table

		[Cu2+]			[NH3]			Cu(NH3)4]2+
initial		0.0300M			0.400M			0
Amount reacted		0.03-0.0300			0.4-4*0.0300			+0.0300
		0			0.28			0.0300
	change			+x		+4x		-x
equilibrium			x		0.28+4x		0.03-x

Kf=[ Cu(NH3)42+]/[Cu2+][NH3]^4=1.7*10^13

1/kf=[Cu2+][NH3]^4/[ Cu(NH3)42+]

Or, 1/1.7*10^13=x*(0.28+4x)^4/(0.03-x)[ignore x as x<<<0.028,0.03)

Or,0.59*10^-13=x*(0.28)^4/0.03

Or, 0.59*10^-13=x*(0.204)

Or,x=2.89*10^-13M

X=[Cu2+]=2.89*10^-13M

[NH3]=0.28+4x=0.12+4(2.89*10^-13M)=0.28M

Cu(NH3)4]2+=0.03-x=0.03- 2.89*10^-13M=0.03M

Answer 2