浓度均为0.1mol/L的下列溶液中由水电离出的c(H+)H2O大小关系为:①盐酸②醋酸溶液③硫酸溶液④氢氧化钠溶
② > ① =④ > ③
为什么?
人气:147 ℃ 时间:2019-11-15 00:49:32
解答
(1)盐酸电离出[H+]=0.1mol/L水电离出[H+]=[OH-]=10^-14/0.1=10^-13mol/L(2)醋酸电离出[H+]大约是10^-5mol/L水电离出[H+]=[OH-]=10^-14/10^-5=10^-9mol/L(3)硫酸电离出[H+]=0.2mol/L水电离出[H+]=[OH-]=10^-14/0.2=0....
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