5H3PO4+7NaOH=3NaH2PO4+2Na2HPO4[H2PO4-]/[HPO42-]=3/2根据PH=PKa2-lg[H2PO4-]/[HPO42-]=7.21-lg3/2=7.03
n(NaH2PO4)=0.3moln(Na2HPO4)=0.2moln[H+]=Ka2*n[NaH2PO4]/[Na2PO4]pH=-lg{n[H+]/V}