0.1mol/L的醋酸的PH是多少?
人气:217 ℃ 时间:2019-12-06 05:36:36
解答
因为:已知醋酸的离常数为Ka=1.75*10^-5,醋酸浓度c=0.1mol/L,c*Ka>20*Kw,又因为c/Ka>500,故可采用最公式计算.
所以:[H+]=√(c*Ka)
=√(0.1*1.75*10^-5
=1.32*10^-3
PH=-Lg(1.32*10^-3)
=2.88
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