80℃ 时,0.01mol/l的盐酸pH=,
问题描述：
80℃ 时,0.01mol/l的盐酸pH=,
答
0.01mol/l的盐酸pH=-lg0.01=2
答
溶液的PH值,只与溶液中氢离子浓度有关
所以80℃ 时,0.01mol/l的盐酸pH=-lgC(H+)=-lg0.01=2