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£¨ÊµÑé°à×ö£©ÔÚ25 mL 0.1 mol/L NaOHÈÜÒºÖÐÖðµÎ¼ÓÈë0.2 mol/L CH3COOHÈÜÒº£¬ÇúÏßÈçÏÂͼËùʾ£¬ÓйØÁ£×ÓŨ¶È¹Øϵ±È½ÏÕýÈ·µÄ  £¨    £©

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  c(Na+) > c(CH3COO£­) > c(OH£­) > c(H+)

B£®ÔÚBµã£¬a > 12.5£¬ÇÒÓÐ

    c(Na+) = c(CH3COO£­) = c(OH£­) = c(H+)

C£®ÔÚCµã£ºc(CH3COO£­) = c(Na+) > c(H+) > c(OH£­)

D£®ÔÚDµã£ºc(CH3COO£­) + c(CH3COOH) = 2c(Na+)

 

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