1. The Keq is a huge number which means that the complex (the product) is favored (big time) over the reactants. So adding ammonia obviously decreases Cu^+2 (big time since Keq is so large).
2. Look at Keq for #2. The way it is written is very small BUT if you reverse it, the number is gain, a very large number. You add Na2S, that ppts CuS (big time) and that reduces the concn of Cu^+2 quickly because Keq is so large.
Qualitively explain the effect of changing concentration of Cu^2+ using your experimental observations and the Keq values below (no calculations required)
Cu^2+(aq) + 4 NH3(aq) --> [Cu(NH3)4]^2+(aq) , Keq = 1.2x10^12
CuS(s) --> Cu^2+(aq) + S^2-(aq) , Keq = 8.5 x 10^-45
Cu/Zn = 0.525 V
after 1 drop NH3: 0.505 V
after 2 drops NH3: 0.435 V
Cu/Zn = 0.510 V
after 1 drop Na2S: 0.452 V
after 2 drops Na2S: 0.393 V
So I know Voltage obviously decreases with added concentration, but how does the Keq account for that?
1 answer
1 answer