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.
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