Asked by Bam
                In an excess of NH3(aq), Cu2+ ion forms a deep blue complex ion, Cu(NH3)4^2+ (Kf=5.6x10^11). Calculate the concentration of Cu2+ in a solution adding 5.0x10^-3 mol of CuSO4 to .5 L of .4 M NH3
            
            
        Answers
                    Answered by
            DrBob222
            
    Cu^+2 + 4NH3 ==> Cu(NH3)4^+2
Kf = [Cu(NH3)4^+2]/([Cu^+2][NH3]^4
Substitute into the above (in M units) and solve for )Cu^+)
You can assume, with such a large constant, that [Cu(NH3)4^+2] = 0.01 and NH3 is about 0.4-(4*0.01) = about 0.36
So (Cu^+2) is about 1 x 10^-12M but you should confirm that.
    
Kf = [Cu(NH3)4^+2]/([Cu^+2][NH3]^4
Substitute into the above (in M units) and solve for )Cu^+)
You can assume, with such a large constant, that [Cu(NH3)4^+2] = 0.01 and NH3 is about 0.4-(4*0.01) = about 0.36
So (Cu^+2) is about 1 x 10^-12M but you should confirm that.
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.