25℃时,0.01mol/l的NaOH溶液和0.0005mol/l的Ba(OH)2溶液中H+之比为

问题描述:

25℃时,0.01mol/l的NaOH溶液和0.0005mol/l的Ba(OH)2溶液中H+之比为

0.01mol/l的NaOH溶液中,c(OH-)=0.01mol/L,c(H+)=10^-14/0.01=10^-12mol/L
0.0005mol/l的Ba(OH)2中,c(OH-)=0.0005*2=0.001mol/L,c(H+)=10^-14/0.001=10^-11mol/L,溶液中H+之比=10^-12/10^-11=10:1很好,不过是1比10你把顺序搞反了不信你再算