A.任何溶液都遵循电中性原则,则有n(Na + )+n(H + )=n(CH 3 COO - )+n(OH - ),故A正确;
B.n(NaOH)=0.01L×0.1moL/L=0.001mol,n(CH 3 COOH)=0.01L×0.2moL/L=0.002mol,混合后醋酸过量,溶液呈酸性,则有n(CH 3 COO - )>n(Na + )>n(H + )>n(OH - ),故B正确;
C.反应后生成n(CH 3 COO - )=0.001mol,剩余n(CH 3 COOH)=0.001mol,醋酸过量,溶液呈酸性,则n(CH 3 COO - )>n(Na + ),故C错误;
D.根据物料守恒,n(NaOH)=0.01L×0.1moL/L=0.001mol,n(CH 3 COOH)=0.01L×0.2moL/L=0.002mol,可知混合后,
2n(Na + )=n(CH 3 COO - )+n(CH 3 COOH),故D正确.
故选C.
本文如未解决您的问题请添加抖音号:51dongshi(抖音搜索懂视),直接咨询即可。