题目

将0.2mol/L的［Ag(NH3)2]＋溶液与0．6mol/L的HNO3溶液等体积混合计算[Ag(NH3)2]＋的浓度

提问时间：2020-12-15

答案

[Ag(NH3)2]+ +2H+ =Ag+ + 2NH4+
K=[Ag+]× [NH4+]^2 ÷ { [Ag(NH3)2+ ]× [H+]^2}
=[Ag+]×[NH4+]^2× [OH]^2 / { [Ag(NH3)2+ ]×[H+]^2× [OH-]^2}
=[Ag+]×[NH4+]^2× [OH]^2 /{ [Ag(NH3)2+ ] × Kw^2 }
=[NH4+]^2× [OH]^2 /{ [Ag(NH3)2+ ] ÷[Ag+] × Kw^2}
=[NH4+]^2× [OH]^2÷ [NH3]^2 /{ [Ag(NH3)2+ ] ÷[Ag+] ÷ [NH3]^2 × Kw^2}
=KbNH3^2 /{K稳× Kw^2}
=(1.77*10^-5）)^2 /{10^7.40*10^-28}
=1.25*10^11
K很大,说明基本完全反应
[Ag(NH3)2]+ +2H+ =Ag+ + 2NH4+
0.2V 0.6V 0 0
0 0.2V 0.2V 0.4V
因此溶液中[H+]=0.2V/2V=0.1mol/L
[Ag+]=0.2V/2V=0.1mol/L
[NH4+]=0.4V/2V=0.2mol/L
K很大,说明基本完全反应,溶液中[Ag(NH3)2]＋浓度很小,可以认为[H+]、[Ag+]、[NH4+]基本不变
设 [Ag(NH3)2]＋为x
[Ag(NH3)2]+ +2H+ =Ag+ + 2NH4+
x 0.1 0.1 0.2
K =[Ag+]× [NH4+]^2 ÷ { [Ag(NH3)2+ ]× [H+]^2}
= 0.1× ( 0.2)^2 ÷ { x ×(0.1)^2}= 1.25*10^11
x= 3.2*10^-12mol/L
即 [Ag(NH3)2]＋的浓度= 3.2*10^-12mol/L