问题
选择题
把0.02mol/LCH3COOH溶液和0.01mol/LNaOH溶液以等体积混和，则混合液中微粒浓度关系正确的是
A．c (CH3COO－)＞c (Na+)
B．c (CH3COOH)+c (CH3COO－)="0.02mol/L"
C．c (CH3COOH)＞c (CH3COO－)
D．c (OH－)＞c (H+)
答案
答案：A
题目分析：根据电荷守恒可知c (Na+)+ (H+)=" c" (CH3COO－)+ c (OH－)，溶液呈酸性，所以c (OH－)<c (H+)，则c (CH3COO－)＞c (Na+)。根据物料守恒c (CH3COOH)+c (CH3COO－)="2" c (Na+)=0.01mol/L，所以c (CH3COOH)<c (CH3COO－)。答案选A。
点评：根据三大守恒定律和溶液的酸碱性判断。