Asked by anonymous
Calculate [Cu^2+] in a 0.15 M CuSO4 solution that also 6.5 M in free NH3.
Cu^2+ + 4NH3 >>>>>> Cu(NH3)4
kf = 1.1 * 10^11.
thx
Cu^2+ + 4NH3 >>>>>> Cu(NH3)4
kf = 1.1 * 10^11.
thx