酸、碱、盐在水溶液中的反应,以离子反应为特征.(1)醋酸钠水解的离子方程式为______;升高温度可以___

2024-11-08 09:48:11
推荐回答(1个)
回答(1):

(1)醋酸钠为强碱弱酸盐,醋酸根离子水解导致其溶液呈碱性,水解方程式为CH3COO-+H3O+?CH3COOH+H2O,盐类水解是吸热反应,升高温度促进盐类水解,故答案为:CH3COO-+H3O+?CH3COOH+H2O;促进;
(2)硝酸是强酸,pH=1的硝酸溶液中氢离子浓度为0.1mol/L,n(HNO3)=0.1mol/L×0.048L=0.0048mol,n(KOH)=0.4mol/L×0.012=0.0048mol,二者恰好反应生成强酸强碱盐,没有水解的离子,所以其溶液呈中性;
酸式滴定管含有活塞、碱式滴定管下端有橡胶管,所以乙是碱式滴定管,
故答案为:中性;乙;
(3)一定温度和压强下,足量的锌与酸反应,生成氢气体积与酸的物质的量成正比,等物质的量浓度等体积的醋酸和盐酸,二者的物质的量相等,且都是一元酸,所以生成氢气的体积在同温同压下相等;反应速率与氢离子浓度成正比,氯化氢是强电解质,醋酸是弱电解质,氯化氢完全电离、醋酸部分电离,所以相同浓度的醋酸和盐酸中,醋酸中氢离子浓度小于盐酸,则反应过程中盐酸与锌反应产生气泡的速率较快,
故答案为:相等;盐酸中氢离子浓度大于醋酸,导致盐酸反应速率快.

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