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【题目】[双选]对于常温下pH为2的盐酸,叙述正确的是

A.c(H+)=c(Cl)+c(OH)

B.与等体积pH=12的氨水混合后所得溶液显酸性

C.由H2O电离出的c(H+)=1.0×1012 mol·L1

D.与等体积0.01 mol·L1乙酸钠溶液混合后所得溶液中:c(Cl)=c(CH3COO)

【答案】AC

【解析】盐酸中离子只有H+、Cl、OH,根据电荷守恒,A正确。pH=12的氨水的物质的量浓度远大于pH=2的盐酸,两者等体积混合后氨水过量,溶液呈碱性,B项错误。由于盐酸电离产生的c(H+)=1.0×102 mol·L1,则c(OH)=1.0×1012 mol·L1,即水电离产生的氢离子的浓度为1.0×1012 mol·L1,C正确。CH3COONa与盐酸浓度相等,等体积混合后H+与CH3COO结合生成CH3COOH,CH3COOH发生微弱电离,故c(Cl)>c(CH3COO),D错误。

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