水电离的氢离子为10-7是否一定显中性

水电离的氢离子为10-7是否一定显中性
2025-03-16 18:33:15
推荐回答(5个)
回答1:

不不不,不一定
ph=-lg(氢离子浓度)
水的离子积=氢离子浓度x氢氧根浓度
水的离子积仅和温度有关,常温时水的离子积等于14
温度升高时水的离子积增大;温度降低时时水的离子积减小
所以,ph=7时在常温(25度)下为中性,温度升高时,ph=7为碱性;温度降低时时,ph=7为酸性

回答2:

对,Ph值的定义就是 -lg(H+)可得Ph=7 溶液显中性

回答3:

不一定,如果常温下,纯水电离出来是10-7则是中性,但是如果有溶质的溶液,里面可能是有一种促进水电离的盐和抑制水电离的酸或碱,导致水电离的为10-7。

回答4:

水电离出的氢等于1乘10负七不一定显中性,试想一个情景,往ph=3的弱酸里加碱,初始时溶液中水电离的氢是1×10的负11在往溶液中加强碱至溶液全为盐时,由于弱酸跟会水解,所以此时水电离的氢大于10的负7,那么再加碱的过程中会存在一个点水电离的氢等于10的负7,此时溶液不呈中性

回答5:

这种说法肯定是错误的!比如常温下醋酸钠溶液中加入NaOH。醋酸根离子会发生水解,水解的本质是在促进水的电离,所以水电离出的氢离子浓度会增大,即大于1*10的-7次方,这时如果向其中逐滴加入NaOH,由于增大了氢氧根的浓度,水的电离又会受到抑制,所以水电离出的氢离子浓度又会减少,在减少的过程中必定会在某个时刻再次等于1*10的-7次方,但是这个时候溶液明显是碱性的!

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