摘要:12.某溶液中仅含Na+.H+.OH-.CH3COO-四种离子.下列说法错误的是( ) A.溶液中四种粒子之间不可能满足:[Na+]>[OH-]>[CH3COO-]>[H+] B.若溶液中部分粒子间满足:[CH3COO-]=[Na+].则该溶液一定呈中性 C.若溶液中溶质仅为CH3COONa.则粒子间一定满足:[Na+]>[CH3COO-]>[OH-]>[H+] D.若溶液中的溶质为CH3COONa和CH3COOH.则溶液中粒子间一定满足:[CH3COO-]>[Na+]>[H+]>[OH-] [解析] 据电荷守恒.B正确,A中[OH-]>[H+]溶液显碱性.若溶质是CH3COONa.应符合C项.但若是NaOH和CH3COONa作溶质.就可以满足关系,D中思考的切入点.若在CH3COONa溶液中滴加少量CH3COOH时.溶液中[H+]也可能小于[OH-]. [答案] AD

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