硫化氢发生电离但是其电离程度较小,S2-第一步水解程度大于第二步,导致相同浓度的KHS溶液、K2S

2025-03-16 04:23:22
推荐回答(5个)
回答1:

前部分,你的理解没有错,硫离子的水解程度比硫氢离子水解程度大,是指硫离子水解成硫氢离子大于硫氢离子水解成硫化氢分子。硫离子第一步水解得到硫氢离子的浓度很小,远小于硫氢氧化钾浓度。俩者水解,当然后者产生的硫化氢浓度更大。而且硫离子第一步水解产生的氢氧根离子会抑制硫氢离子的水解。

回答2:

1、这两种物质无论是水解还是电离,程度都是很小的,因此,假设KHS溶液、K2S溶液物质的量浓度都是1mol/L,c(HS-)约等于c(S2-)约等于1mol/L。
2、两种溶液中的H2S都是靠HS-的水解得到的。
按之前假设KHS溶液的c(HS-)约等于1mol/L;
K2S溶液的c(HS-)是水解产生的,相对于1mol/L而言小到可以忽略。
即:前者溶液中的c(HS-)远远大于后者,显然,前者溶液中的c (H2S)更大。

回答3:

S2- 水解生成硫化氢,得经过两步,第一步先水解成HS-离子,然后第二步水解才是硫化氢,在KHS和K2S浓度一样大的情况下,前者一步水解形成硫化氢

回答4:

(1)[H2S]反比于溶液的PH!
(2)PH : KHS溶液< K2S溶液
(3)[H2S] : KHS溶液> K2S溶液

回答5:

硫化氢第一电离常数Ka1
HS-水解常数Kh=Kw/Ka1
查表计算

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