需要知道Ka值.Na3PO4+HCl--->Na2HPO4+NaCl；Na2HPO4+HCl--->NaH2PO4+NaCl；NaH2PO4+HCl--->H3PO4+NaCl.[H+][H2PO4-]/[H3PO4]=Ka1[H+][HPO42-]/[H2PO4-]=Ka2[H+][PO43-]/[HPO42-]=Ka36．今欲用Na3PO4与HCl来配制pH = 7.20的缓冲溶液,则Na3PO4与HCl物质的量之比n(Na3PO4)∶n(HCl)应当是()(H3PO4的pKa1~pKa3分别是2.12,7.20,12.36)(A) 1:1 (B) 1:2(C) 2:3(D) 3:2原题(*_*)谢谢老师了，帮忙解一下吧0.0pKa2=7,20，与pH相同；[H+][HPO42-]/[H2PO4-]=Ka2反应后有：[HPO42-]=[H2PO4-]C