怎么鉴别氯离子,碳酸根离子,氨根离子,硫酸根离子

写出化学方程式,所用药品,反应类型
2025-04-05 15:14:33
推荐回答(2个)
回答1:

氯离子加硝酸银生成氯化银不溶于酸沉淀,复分解。碳酸跟加盐酸有二氧化碳气体(气体通澄清石灰水鉴别)。铵根加氢氧化钠加热有碱性气体(气体用湿润石蕊试纸鉴别)。硫酸根加氯化钡溶液有硫酸钡不溶于酸沉淀。都是复分解,方程式写不下了…自己推下就出来了…

回答2:

碳酸根:加入过量稀盐酸,将生成的气体(无气体则表明没有碳酸根)通入澄清石灰水,石灰水变浑表明有碳酸根(反之亦然)。
硫酸根:先加入过量稀盐酸(防止银离子的干扰,有沉淀需加至无沉淀继续生成后过滤),再加入氯化钡溶液(不可用硝酸钡,因为硝酸根在酸性环境下会把亚硫酸根等低价硫的含氧酸根氧化为硫酸根,造成误检),如有白色沉淀生成则表明有硫酸根(已先用盐酸酸化,此处不必再酸化)。
氯离子:加硝酸钡沉淀硫酸根等杂质离子。加硝酸酸化的硝酸银溶液(可排除氨的干扰),产生白色沉淀(无沉淀表明无氯离子)。取白色沉淀加稀氨水(防止其它难溶银盐的干扰),沉淀溶解,表明原溶液含氯离子。
氢离子、氢氧根离子在水中必定存在,用ph试纸可以粗测它们的量。

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