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

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 $O {H}^{-}$ 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 (${S}^{2} -$, $C {O}_{3}^{2 -}$, $P {O}_{4}^{3 -}$, $O {H}^{-}$) 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

$\text{How do you differentiate.......?}$

Explanation:

$\text{How else but by experiment.............?}$

For general rules of solubility we can advance the following generalities......mostly it is based on the solubilities where the COUNTERION, $\text{the gegenion}$, 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 $P b S {O}_{4} , B a S {O}_{4} , H g S {O}_{4}$.

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 \text{kalignost}, i.e. $\text{potassium recognizer}$.............on the basis of the following reaction.....

$N {a}^{+} B P {h}_{4}^{-} \left(a q\right) + K X \left(a q\right) \rightarrow {K}^{+} B P {h}_{4}^{-} \left(s\right) \downarrow + N a X \left(a q\right)$