Question

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+)

Answer 1

A

解析试题分析：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。
考点：离子浓度大小比较
点评：根据三大守恒定律和溶液的酸碱性判断。