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

2 Answers
Jun 24, 2017

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!

Jun 24, 2017

#"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)#