The complex ion Cu(NH3)4 2+ is formed in a solution made of .0200M Cu(NO3)2 and

Question

The complex ion Cu(NH3)4 2+ is formed in a solution made of .0200M Cu(NO3)2 and .500M NH3. What are the concentrations below.

Im not sure how to reward extra points but i definitely can for whoever can help me answer this one, and also explain where I got to add more point

The complex ion Cu(NH3)42 is formed in a solution made of O.0200 M Cu(NO3)2 and 0.500 M NH3 What are the concentrations of Cu2+, NHs, and Cu(NH3) at equilibrium? The formation constant*, K, of Cu(NH3)42 is 1.70 x1013. Number 2+ Number Number Cu(NH,

Explanation / Answer

                          Cu(NO3)2 +4NH3 ? Cu(NO3)2(NH3)4          Kf = 1.70*10^13

Initial                     0.02        0.5                   0

At equilibrium         0.02-x      0.5-4x              x

Since Kf is very very high. So limiting with finish almost.

Limiting reagent is Cu(NO3)2 because limiting reagent is the one which finishes first. Cu(NO3)2 has very less initial concentration as compared to NH3.

So, x= 0.02

[NH3] = 0.5-4x=0.42

[Cu+2]?0M

[NH3]?0.42M

[Cu(NO3)2(NH3)4 ]?0.02M

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