第1个回答 2014-04-08
Ag(NH3)2+ + 2 H+ =Ag+ +2NH4+
K=[NH4+]^2 /{ [Ag(NH3)2+]*[H+]^2}
={[Ag+][NH4+]^2*[OH-]^2}/{ [Ag(NH3)2+]*[H+]^2[OH-]^2}
={[NH4+]^2*[OH-]^2}/{ [Ag(NH3)2+]*[H+]^2[OH-]^2/[Ag+]}
={[NH4+]^2*[OH-]^2}/{ [Ag(NH3)2+]*Kw^2 /[Ag+]}
={[NH4+]^2*[OH-]^2)/ [NH3]^2}/{ [Ag(NH3)2+]*Kw^2 /([Ag+]*[NH3]^2)}
=Kb^2 /( K稳*Kw^2)
=(1.77*10^-5)^2 / [1.1*10^7 * (1.0*10^-14)^2]
=2.85*10^11
可见K很大,完全反应
2) Ag(NH3)2+ + 2 H+ = Ag+ + 2NH4+
0.2V 0.4 V 0.2V 0.4V
溶液[Ag+]=0.2V /2V=0.1mol/L
[H+]=(2V-0.4V)/2V=0.8mol/L
由于H+大量存在,NH4+基本上没有水解
[NH4+]=0.4V/2V=0.2mol/L
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上面求K太繁杂
其实可以分三步来求:
1) Ag(NH3)2+ <=>Ag+ +2NH3 K=1/K稳
2)2NH3 +2 H2O<==>2NH4+ +2 OH- K=Kb^2
3)2H+ + 2OH-=2H2O K=(1/Kw)^2
三式相加,K相乘即行