答案:
分析:A.草酸是二元中强酸,V(NaOH)=0时,NaHC2O4不能完全电离出H+;
B.V(NaOH)<10mL时,溶液中溶质为NaHC2O4、Na2C2O4;
C.V(NaOH)=10 mL时,溶液中溶质为Na2C2O4;
D.V(NaOH)>10 mL时,溶液中溶质为NaOH、Na2C2O4.
解答:解:A.草酸是二元中强酸,V(NaOH)=0时,NaHC2O4不能完全电离出H+,则c(H+)<1×10-2mol•L-1,故A错误;
B.V(NaOH)<10mL时,溶液中溶质为NaHC2O4、Na2C2O4,当电离等于水解,即c(H+)=c(OH-),所以存在c(Na+)=2c(C2O42-)+c(HC2O4- ),故B错误;
C.V(NaOH)=10 mL时,溶液中溶质为Na2C2O4,溶液显碱性,则c(H+)<1×10-7mol•L-1,故C错误;
D.V(NaOH)>10 mL时,溶液中溶质为NaOH、Na2C2O4,C2O42-水解生成HC2O4-,则离子浓度为c(Na+)>c(C2O42-)>c(HC2O4- ),故D正确;
故选D.