43.配制pH=5.05的缓冲溶液500ml,应取0.1 mol?L-1HAc及0.1 mol?L-1NaOH各多少ml?( pKa(HAc) = 4.75) (分值:

问题描述:

43.配制pH=5.05的缓冲溶液500ml,应取0.1 mol?L-1HAc及0.1 mol?L-1NaOH各多少ml?( pKa(HAc) = 4.75) (分值:

1.Ac- +H2O ----- HAc + OH- K12.HAc ----- H+ + Ac- KaKw = 1e-141+2式得K1*Ka=KwK1=Kw/Ka=[OH-]*[HAc]/[Ac-][OH-]=Kw*[Ac-]/(Ka*[HAc])POH = -lg([OH-])=14-5.05得[Ac-]/[HAc]=10^(5.05-4.75)=1.995=2根据电荷守恒...