[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·L1D．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) 题目和参考答案——青夏教育精英家教网—

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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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