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85问答库 > 0.4摩尔每升氨水50毫升与0.2摩尔每升盐酸50毫升混合,溶液PH是多少?

0.4摩尔每升氨水50毫升与0.2摩尔每升盐酸50毫升混合,溶液PH是多少?

2025-04-04 05:53:47

推荐回答（3个）

回答1：

. NH3 + H+ = NH4+
起始浓度 / mol·L-1 0.2 0.1 0
平衡浓度 / mol·L-1 0.1 0 0.1
溶液中存在共轭缓冲对NH3-NH4+，故溶液的pH值由缓冲溶液pH值计算公式求得：
. pH = 14 - pKb + lgc(NH3)/c(NH4+)
. = 14 - 4.75 + lg1 = 9.25

回答2：

混合后氨水的浓度比氯化铵的浓度为一。那么
PH=14-PKb=14-4.75=9.25

回答3：

生成氯化铵与氨水摩尔数一样，所以是缓冲溶液，用公式一算就可以了，希望能帮到你。

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