(1)纯水中水电离出的氢离子和氢氧根离子浓度是相等的,所以纯水中c(H+ )=5.0×10-7 mol/L=c(OH-),此温度下Kw=2.5×10-13mol2/L2,溶液中,Kw=c(H+)?c(OH-),所以0.01mol/L稀硫酸溶液中,c(OH-)=
=Kw c(H+)
=1.25×10-11mol/L,故答案为:5.0×10-7 mol/L;1.25×10-11mol/L;2.5×10?13
0.02
(2)将pH=12NaOH溶液与pH=2的H2SO4溶液混合后pH=11(忽略混合后体积的变化),设NaOH溶液与H2SO4溶液的体积分别为X、Y,则
=10-3,解得X:Y=11:9,故答案为:11:9; 10?2X?10?2Y X+Y
(3)设两溶液体积是V,0.1mol/L NaOH溶液中氢氧根离子物质的量是0.1V,0.06mol/L的H2SO4溶液的物质的量是0.12V,混合后酸剩余,所以剩余氢离子浓度c=
=0.01mol/L,即pH=2,故答案为:2.0.12V?0.1V 2V