Question

向0.1mol/L草酸溶液中滴加NaOH溶液至pH=6.00，求溶液中H2C2O2,HC2O4-,和C2O4^2-的浓度?

已知H2C2O4的K1=5.4x10^(-2), K2=5.4x10^(-5)
需要解答过程

Answer 1

用分布系数来解决
C(H2C2O4)=0.1mol/L* [H+]^2 /{ [H+]^2 + [H+]*K1+ K1*K2}
=0.1mol/L*{ (10^-6)^2 /[ (10^-6)^2 + (10^-6)*5.4x10^(-2)+5.4x10^(-2)*5.4x10^(-5)]}
=0.1mol/L*3.36*10^-7
=3.36*10^-8mol/L

C(HC2O4-)=0.1mol/L* {[H+]*K1} /{ [H+]^2 + [H+]*K1+ K1*K2}
=0.1mol/L*{ (10^-6)*5.4x10^(-2) /[ (10^-6)^2 + (10^-6)*5.4x10^(-2)+5.4x10^(-2)*5.4x10^(-5)]}
=0.1mol/L*1.82^-2
=1.82*10^-3 mol/L

C(C2O42-)=0.1mol/L* (K1*K2) /{ [H+]^2 + [H+]*K1+ K1*K2}
=0.1mol/L*{ 5.4x10^(-2)*5.4x10^(-5) /[ (10^-6)^2 + (10^-6)*5.4x10^(-2)+5.4x10^(-2)*5.4x10^(-5)]}
=0.1mol/L*9.81*10^-1
=9.81*10^-2 mol/L

Answer 2

用分布分数乘以总浓度C(C=0.1mol/L)即可
[H2C2O4]=([H+]^2/([H+]^2+[H+]*Ka1+Ka1*Ka2))*C
[HC2O4^-]=([H+]*Ka1/([H+]^2+[H+]*Ka1+Ka1*Ka2))*C
[C2O4^-2]=(Ka1*Ka2/([H+]^2+[H+]*Ka1+Ka1*Ka2))*C