问题 选择题

下列表达不正确的是(  )

A.NaHS在水中的电离方程式为:NaHS═Na++HS-和HS-═H++S2-

B.同物质的量浓度的氨水和盐酸反应至中性时所用体积:V(NH3•H2O)>V(HCl)

C.Na2SO3溶液中:c(H+)+c(HSO3-)+2c(H2SO3)=c(OH-

D.同浓度的下列溶液中,c(CH3COO-)的大小:CH3COONa>CH3COONH4>CH3COOH

答案

A.NaHS为强电解质,溶液中电离出钠离子和硫氢根离子,硫氢化钠在水中的电离方程式为:NaHS═Na++HS-,故A错误;

B.同物质的量浓度的氨水和盐酸反应,如果等体积混合,反应生成氯化铵,溶液为酸性,所以加入的氨水应该过量,溶液才能为中性,即:V(NH3•H2O)>V(HCl),故B正确;

C.Na2SO3溶液中,根据质子守恒可得:c(H+)+c(HSO3-)+2c(H2SO3)=c(OH-),故C正确;

D.相同浓度的三种溶液中,醋酸溶液中只能部分电离出醋酸根离子,其醋酸根离子浓度最小;醋酸铵溶液中,铵根离子水解促进了醋酸根离子的水解,溶液中醋酸根离子浓度小于醋酸钠,所以溶液中c(CH3COO-)的大小为:CH3COONa>CH3COONH4>CH3COOH,故D正确;

故选A.

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阅读理解。

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