Question

配制pH=5.05的缓冲溶液500ml,应取0.1mol.L-1HAc及0.1mol.L-1NaOH各多少ml?(pKa

Answer 1

pH=pKa+lg[c(Ac-)/c(HAc)]

5.05=4.75+lg[c(Ac-)/c(HAc)]

lg[c(Ac-)/c(HAc)]=0.3

c(Ac-)/c(HAc)=2

设 应取0.1 mol•L-1HAc及0.1 mol•L-1NaOH的体积分别是 XmL ,Y mL

X +Y=500

0.1Y/(0.1X-0.1Y)=2

X=300 Y=200