Salt analysis (cation) [closed]








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I'm reading about salt analysis and I've got some questions:


[![enter image description here](https://i.stack.imgur.com/4rIyT.jpg)](https://i.stack.imgur.com/4rIyT.jpg)
[![enter image description here](https://i.stack.imgur.com/HqgAl.jpg)](https://i.stack.imgur.com/HqgAl.jpg)
Image source: [pdf](https://ncert.nic.in/pdf/publication/sciencelaboratorymanuals/classXII/chemistry/lelm107.pdf)


Why are some sulphides (group II) precipitated in presence of HCl while others (group IV) in presence of ammonia solution? Can NaOH be used in place of ammonia solution? What's the role of NH4Cl in testing for group III? Will Na2CO3 work for group V?


(The groups here are not related to groups of periodic table.)




Let me answer your three questions.


1. Some sulphides are soluble in acidic solutions (group IV). Some are insoluble in the same sort of acidic solution (group II). It is a question of solubility product. It can be taken as an experimental result, a fact.
2. If no $\ce{NH4Cl}$ were present in the solution tested for Group III, the precipitates $\ce{Al(OH)3}$ and $\ce{Fe(OH)3}$ may remain in a colloidal state, and will cross the filter paper. This is a serious drawback. In the presence of enough ions, like $\ce{NH4^+}$ and $\ce{Cl-}$, the colloid is transformed into a precipitate.
3. $\ce{NaOH}$ can replace ammonia solution, but it will prevent the precipitation of $\ce{Al(OH)3}$ in group III and $\ce{Zn(OH)2}$ in group IV. These precipitates will redissolve in $\ce{NaOH}$ to form a solution of aluminate $\ce{Al(OH)4^-}$ and zincate $\ce{[Zn(OH)4]^{2-}}$. So the method would not detect $\ce{Al}$ or $\ce{Zn}$.


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