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

B£®ÔÚBµã£¬a>12.5£¬ÇÒÓÐC(Na+) = C(CH3COO£­) = C(OH£­) = C(H+)

C£®ÔÚCµã£¬C(CH3COO£­) > C(Na+) > C(OH£­) > C(H+)

D£®ÔÚDµã£¬C(CH3COO£­) + C(CH3COOH) = C(Na+)

 

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