要配制PH=7的缓冲溶液100ml,需要0.05mol/l的KH2PO4和0.05mol/l的K2HPO4溶液各多少毫升

题目

要配制PH=7的缓冲溶液100ml,需要0.05mol/l的KH2PO4和0.05mol/l的K2HPO4溶液各多少毫升
已知磷酸的PK2=7.21

答案

pK2=7.21=-lg([HPO4 2-][H+]/[H2PO4-])=-lg([H+])-lg([HPO4 2-]/[H2PO4-])=pH-lg([HPO4 2-]/[H2PO4-])=7-lg([HPO4 2-]/[H2PO4-])得到：0.21=-lg([HPO4 2-]/[H2PO4-])10^(-0.21)=0.62=[HPO4 2-]/[H2PO4-]得到V(HPO4 2...