Asked by David

Calculate the concentrations of Cd+2
, Cd(CN)4^-2 and CN at equilibrium when dissolvs 0.42 moles of Cd(NO3)2 (2.50 mol/L)
Kf for Cd(NO3)2 = 7,1x10^16.
Answered by DrBob222
I must be missing something. With no CN^- there how does any [Cd(CN)4]^2- get formed?