0.001mol/L的硫酸的PH

问题描述:

0.001mol/L的硫酸的PH

PH=3 -lg2

0.001mol/L的硫酸中的H+浓度=0.002mol/L
PH=-lgc(H+)=3-lg2

第二步为弱酸解离,HSO4- H+ + SO42-[HSO4-]=0.001-X [SO42-]=X [H+]=0.001+XK=1.2x10^-2=[SO42-][H+]/[HSO4-]=X(0.001+X)/(0.001-X)X=8.7x10^-4 mol/L.[H+]=0.00187 mol/LpH=2.73