摘要：把0．02mol/LCH3COOH和0．01mol/LNaOH以等体积混和后溶液呈酸性.则混合液中微粒浓度关系正确的为: A．c(CH3COO-)＜c(Na+) B．c(CH3COOH)+c(CH3COO-)=0．01mol/L C．c(CH3COOH)＞c(CH3COO-) D．无法判断
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把0.02mol/LCH3COOH和0.01mol/LNaOH以等体积混和，则混合液中微粒浓度关系正确的为（ ）
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| A．（CH3COO-）＜C（Na+） |
| B．C（CH3COOH）+C（CH3COO-）=0.01mol/L |
| C．C（CH3COOH）＞C（CH3COO-） |
| D．无法判断 |
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 |
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+) |