pH=PKa2 +lg n(HPO42-)/n(H2PO4-)
8.0=7.2 +lg n(HPO42-)/n(H2PO4-)
n(HPO42-)/n(H2PO4-)=6.30------------(1)
若配制1L 0.1mol/L pH=8.0的磷酸缓冲液
n(HPO42-)+n(H2PO4-)=0.1mol----------------(2)
联立解得:n(NaH2PO4)= 0.014mol m(NaH2PO4)= 0.014mol * 120g/mol=1.68g
n(Na2HPO4)=0.086mol m(Na2HPO4)= 0.086mol * 142g/mol=12.21g
配制:分别称量,然后配成1L溶液