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?

题目

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)

答案

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