Question

How do you differentiate inorganic salts by precipitation with different anions?

Answer 1

Unfortunately, solubility is pretty random, and it's hard to give you a pattern on the periodic table for everything. You may have to memorize most, but there are some helpful exceptions!

Explanation:

In group 2A, Ca, Sr, and Ba (all right next to each other) cations coupled with `S^2-` and `OH^-` anions are still soluble, although most compounds with those anions are insoluble. Furthermore any group 1A element or ammonia coupled with the characteristic, common insoluble anions (`S^2-`, `CO_3^(2-)`, `PO_4^(3-)`, `OH^-`) are also soluble!

Honestly, rather than throw so much information that I memorized at you, you should search "Solubility Guidelines for Common Ionic Compounds in Water" and stare at and practice with that table for a few days to totally understand this. It's a lot!

Answer 2

`"How do you differentiate.......?"`

Explanation:

`"How else but by experiment.............?"`

For general rules of solubility we can advance the following generalities......mostly it is based on the solubilities where the COUNTERION, `"the gegenion"`, is an anion.........

All the salts of the alkali metals and ammonium are soluble.

All nitrates, and perchlorates are soluble.

All halides are soluble EXCEPT for # AgX, Hg_2X_2, PbX_2"#.

All sulfates are soluble EXCEPT for `PbSO_4, BaSO_4, HgSO_4`.

All carbonates and hydroxides are insoluble. All sulfides and oxides are insoluble; transition metal oxides, and main group metal oxides routinely tend to be as soluble as bricks.

The given rules follow a hierarchy. Alkali metal and ammonium salts tend to be soluble in all circumstances. The one exception to this rule is `K^(+)""^(-)BPh_4` and `NH_4^(+)""^(-)BPh_4`, both of which are as soluble as bricks. `Na^+""^(-)BPh_4`, the which has some aqueous solubility, is sold as `"kalignost"`, i.e. `"potassium recognizer"`.............on the basis of the following reaction.....

`Na^(+)BPh_4^(-)(aq)+KX(aq) rarr K^(+)BPh_4^(-)(s)darr + NaX(aq)`