Analytical Chemistry - Identification of Cations and Amphoteric Metals

Vocabulary flashcards testing cation identification reactions with sodium hydroxide and ammonium hydroxide, and the chemical equations for amphoteric metals, oxides, and hydroxides.

FeCl2 Formation Method

Can only be formed by precipitation reaction.

FeCl3 Formation Method

Can only be formed by combination reaction.

ZAL

Refers to the amphoteric metals Zinc ($Zn$), Aluminium ($Al$), and Lead ($Pb$), which react with both acids and bases.

Ca2+ Reaction with NaOH

Forms a white precipitate according to $Ca(NO_3)_2 + 2NaOH \rightarrow Ca(OH)_2 + 2NaNO_3$, which is insoluble in excess $NaOH$.

Ca2+ Reaction with NH4OH

Shows no reaction.

Fe2+ Reaction with NaOH

Forms a dirty green precipitate according to FeCl_2 + 2NaOH \rightarrow Fe(OH)_2\blank + 2NaCl, which is insoluble in excess $NaOH$.

Fe2+ Reaction with NH4OH

Forms a dirty green precipitate according to $FeSO_4 + 2NH_4OH \rightarrow Fe(OH)_2 + (NH_4)_2SO_4$, which is insoluble in excess $NH_4OH$.

Fe3+ Reaction with NaOH

Forms a reddish brown precipitate according to $FeCl_3 + 3NaOH \rightarrow Fe(OH)_3 + 3NaCl$, which is insoluble in excess $NaOH$.

Fe3+ Reaction with NH4OH

Forms a reddish brown precipitate according to $FeCl_3 + 3NH_4OH \rightarrow Fe(OH)_3 + 3NH_4Cl$, which is insoluble in excess $NH_4OH$.

Cu2+ Reaction with NaOH

Forms a pale blue precipitate according to $CuSO_4 + 2NaOH \rightarrow Cu(OH)_2 + Na_2SO_4$, which is insoluble in excess $NaOH$.

Cu2+ Reaction with NH4OH

Forms a pale blue precipitate in drops, which turns into an inky blue solution in excess $NH_4OH$.

Zn2+ Reaction with NaOH

Forms a gelatinous white precipitate ($Zn(OH)_2$) that is soluble in excess $NaOH$, yielding a clear solution of sodium zincate ($Na_2ZnO_2$) via $Zn(OH)_2 + 2NaOH \rightarrow Na_2ZnO_2 + 2H_2O$.

Zn2+ Reaction with NH4OH

Forms a white precipitate according to $Zn(NO_3)_2 + 2NH_4OH \rightarrow Zn(OH)_2 + 2NH_4NO_3$, which is soluble in excess $NH_4OH$.

Pb2+ Reaction with NaOH

Forms a chalky white precipitate according to $Pb(NO_3)_2 + 2NaOH \rightarrow Pb(OH)_2 + 2NaNO_3$, which is soluble in excess $NaOH$.

Pb2+ Reaction with NH4OH

Forms a chalky white precipitate according to $Pb(NO_3)_2 + 2NH_4OH \rightarrow Pb(OH)_2 + 2NH_4NO_3$, which is insoluble in excess $NH_4OH$.

Reaction of Zn with NaOH

$$Zn + 2NaOH \rightarrow Na_2ZnO_2 + H_2\rightarrow$$

Reaction of Zn with KOH

$$Zn + 2KOH \rightarrow K_2ZnO_2 + H_2\rightarrow$$

Reaction of Pb with NaOH

$$Pb + 2NaOH \rightarrow Na_2PbO_2 + H_2\rightarrow$$

Reaction of Pb with KOH

$$Pb + 2KOH \rightarrow K_2PbO_2 + H_2\rightarrow$$

Reaction of Al with NaOH

$$2Al + 2NaOH \rightarrow 2NaAlO_2 + 3H_2\rightarrow$$

Reaction of Al with KOH

$$2Al + 2KOH \rightarrow 2KAlO_2 + 3H_2\rightarrow$$

Reaction of ZnO with NaOH

$$ZnO + 2NaOH \rightarrow Na_2ZnO_2 + H_2O$$

Reaction of ZnO with KOH

$$ZnO + 2KOH \rightarrow K_2ZnO_2 + H_2O$$

Reaction of PbO with NaOH

$$PbO + 2NaOH \rightarrow Na_2PbO_2 + H_2O$$

Reaction of PbO with KOH

$$PbO + 2KOH \rightarrow K_2PbO_2 + H_2O$$

Reaction of Al2O3 with NaOH

$$Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2 + H_2O$$

Reaction of Al2O3 with KOH

$$Al_2O_3 + 2KOH \rightarrow 2KAlO_2 + H_2O$$

Reaction of Zn(OH)2 with NaOH

$$Zn(OH)_2 + 2NaOH \rightarrow Na_2ZnO_2 + 2H_2O$$

Reaction of Zn(OH)2 with KOH

$$Zn(OH)_2 + 2KOH \rightarrow K_2ZnO_2 + 2H_2O$$

Reaction of Pb(OH)2 with NaOH

$$Pb(OH)_2 + 2NaOH \rightarrow Na_2PbO_2 + 2H_2O$$

Reaction of Pb(OH)2 with KOH

$$Pb(OH)_2 + 2KOH \rightarrow K_2PbO_2 + 2H_2O$$

Reaction of Al(OH)3 with NaOH

$$Al(OH)_3 + NaOH \rightarrow NaAlO_2 + 2H_2O$$

Reaction of Al(OH)3 with KOH

$$Al(OH)_3 + KOH \rightarrow KAlO_2 + 2H_2O$$