PH=PKa2-lg[H2PO4-]/[HPO42-]lg[H2PO4-]/[HPO42-]=PKa2-PH=7.2-6.8=0.4[H2PO4-]/[HPO42-]=2.5/1即溶液中的磷酸二氢盐与磷酸一氢盐的物质的浓度比为2.5/1时即可得到PH=6.8的磷酸盐缓冲...