题目

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为( )A.c(CH3COO－)<c (Na+)B.c(OH－)>c (H+)C.c(CH3COOH)>c (CH3COO－)D.c(CH3COOH)+c (CH3COO－)=0.01mol/L 答案：【答案】D【解析】0.02mol/LCH3COOH溶液和0.01mol/LNaOH溶液以等体积混和后，溶质为等物质的量的CH3COONa和CH3COOH，溶液中存在CH3COOH的电离和CH3COO-的水解，据此分析解答。A．CH3COOH的电离大于CH3COO-的水解，所以c(CH3COO－)>c(Na+)，A错误；B．CH3COOH的电离大于CH3COO-的水解，所以溶液显酸性，c (OH－)<c(H+)，B错误—How about fried chicken? —Oh, it ___ lots of fatA.remainsB.containsC.holdsD.includes