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85问答库 > 计算0.1mol⼀lNH4Cl和0.1mol⼀lNH3H2O等体积相混的PH值

计算0.1mol⼀lNH4Cl和0.1mol⼀lNH3H2O等体积相混的PH值

2025-04-07 10:50:11

推荐回答（1个）

回答1：

0.1mol/lNH4Cl和0.1mol/lNH3H2O等体积相混
显然是缓冲溶液,NH4+
的水解和NH3·H2O
的电离都受到抑制
则：[NH4+]≈c(NH4+)=0.05mol/L
[NH3·H2O]≈c(NH3·H2O)=0.05mol/L
NH3·H2O<==>NH4+
+
OH-
Kb=
[OH-]
*
[NH4+]
/[NH3·H2O]=[OH-]*0.05mol/L
/0.05mol/L
=[OH-]
[OH-]=Kb=1.77*10^-5
[H+]=Kw/[OH-]=10^-14
/
1.77*10^-5
=5.65*10^-10
pH=-lg[H+]=-lg5.65*10^-10=9.25

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