6Â¥: Originally posted by
robertno at 2014-11-09 10:24:57
ÒÔÅäÖÃ1L 0.1mol/L pH7.8µÄÁ×ËáÑλº³åÈÜҺΪÀý:
»º³åÈÜÒº¼ÆË㹫ʽ£º
pH=pKa2-lgc(HPO4^2-) /c(H2PO4^-)
°ÑpH=7.8ºÍpKa2=7.2´úÈ룬4c(H2PO4^-)/c(HPO4^2-)=4/1
Á×ËáÑÎŨ¶ÈΪ0.1mol/L£¬¾ÍÊÇc(H2PO4^-)+c(HPO4 ...