Question

常温下,将1mL pH=1的H2SO4溶液加水稀释到100mL,由水电离出的c(H+)=( )

Answer 1

解,PH=1 可求得C(H+)=10^-1molL

1ml 稀释到100ml 有C1*V1=C2*V2 ,稀释后C(H+)=1.0X10^-3molL

根据Kw=C(H+)XC(OH-) 而C(H+)总=C(H+)a+C(H+)w

C(H+)a＞＞C(H+)w 则C(H+)w=C(OH-)w=KwC(H+)=1.0X10^-3molL

C(H+)w=10^-14÷10^-3=1.0×10^-11molL

即由水电离的氢离子浓度1.0×10^-11molL