Question

table showing observation that occurred for each step. Give any indication about the amphoteric nature of the compounds formed in this section? why or why not? does ammonia or NaOH provide a more conclusive demonstration of amphoterism? explain

	Zn2+	Sn2+
Initial color	Clear	Slightly cloudy
Observation after adding 1M NH3	Slightly cloudy	Deep white cloudy
Observation after adding 6M NH3 and place it in the centrifuge	Clear	Reform cloudy white precipitate

	Zn2+	Sn2+
Initial color	Clear	Slightly cloudy
Observation after adding 1M NaOH	Slightly cloudy	Deep white cloudy
Observation after adding 6M NaOH and place it in the centrifuge	Clear	Clear

Answer 1

1. Zinc(II) ion reacts with 1 M aqueous ammonia to precipitate white gelatinous Zn(OH)2:

Zn2+(aq)+2NH3(aq)+2H2O(l)↽−−⇀Zn(OH)2(s)+2NH+4(aq)

2. The zinc(II) hydroxide precipitate dissolves in excess ammonia like 6 M as given here:

Zn(OH)2(s)+4NH3(aq)↽−−⇀[Zn(NH3)4]2+(aq)+2OH−(aq) hence solution become clear.

3. Sodium hydroxide also precipitates zinc(II) hydroxide:

Zn2+(aq)+2OH−(aq)↽−−⇀Zn(OH)2(s)

4. The zinc(II) hydroxide precipitate also dissolves in excess hydroxide:

Zn(OH)2(s)+2OH−(aq)↽−−⇀[Zn(OH)4]2−(aq)

similarly for Sn2+

1. Aqueous ammonia precipitates white Sn(OH)2 with tin(II)

[SnCl4]2−(aq)+2NH3(aq)+2H2O(l)↽−−⇀Sn(OH)2(s)+2NH+4(aq)+4Cl−(aq)

2. Sodium hydroxide also precipitates the hydroxides:

[SnCl4]2−(aq)+2OH−(aq)↽−−⇀Sn(OH)2(s)+4Cl−(aq)

These precipitates dissolve in excess hydroxide:

Sn(OH)2(s)+2OH−(aq)↽−−⇀[Sn(OH)4]2−(aq)