A.任何溶液都遵循电中性原则,则有n(Na+)+n(H+)=n(CH3COO-)+n(OH-),故A正确;
B.n(NaOH)=0.01L×0.1moL/L=0.001mol,n(CH3COOH)=0.01L×0.2moL/L=0.002mol,混合后醋酸过量,溶液呈酸性,则有n(CH3COO-)>n(Na+)>n(H+)>n(OH-),故B正确;
C.反应后生成n(CH3COO-)=0.001mol,剩余n(CH3COOH)=0.001mol,醋酸过量,溶液呈酸性,则n(CH3COO-)>n(Na+),故C错误;
D.根据物料守恒,n(NaOH)=0.01L×0.1moL/L=0.001mol,n(CH3COOH)=0.01L×0.2moL/L=0.002mol,可知混合后,
2n(Na+)=n(CH3COO-)+n(CH3COOH),故D正确.
故选C.