溶液中含有0.10mol/L游离NH3,0.10mol/LNH4Cl==> 缓冲溶液
pH＝pKa＋log(0.10/0.10)=9.24 ==>pOH=4.76==>[OH-]=10^-4.76=1.74x10^-5 mol/L
又,0.15mol/L[Cu(NH3)4]2+
Cu2+ + 4NH3 [Cu(NH3)4]2+ 设[Cu2+]=X mol/L
Kf=10^12=[ [Cu(NH3)4]2+]/[Cu2+][NH3]^4=(0.15-X)/{X(0.1)^4}=(0.15-X)/X10^-4
X10^8=0.15-X
X=1.5x10^-9 mol/L
所以,[Cu2+][OH-]^2=1.5x10^-9x(1.74x10^-5)^2=4.5x10^-19>Ksp ===>有Cu(OH)2生成!