Chemical Equilibrium Experimental Results

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Chemical Equilibrium



A. Iron-Silver Equilibrium

In studying equilibrium between iron and silver, 0.10 M FeSO4 and 0.10 M AgNO3 were used. The balanced equation for the reaction is:

FeSO4 (aq) + 2 AgNO3 (aq) ↔ Fe(NO3)2 (aq) + Ag2SO4 (s)

It has a net equation of:

Fe2+(aq) + Ag+(aq) ↔ Fe3+(aq) + Ag(s)
This part of the experiment required centrifugation of the solution. This was done to completely separate the precipitate from the supernate for easier decantation process.

In determining the presence of Fe2+ in the supernate, 0.10 M of K3Fe(CN)6 was used . The resulting product had a Prussian blue precipitate. The reaction has a balanced equation of:

3 Fe(NO3)2 (aq) + K3Fe(CN)6 (aq) ↔ 3 Fe(CN)2 + K3Fe(NO3)6

It has a net ionic equation of:

Fe2+(aq) + K3Fe(CN)6(aq) ↔ KFe(III)Fe(II)(CN)6(s) + 2K+(aq)
To determine the presence of Fe3+, 0.10 M of KSCN was used. The product was a blood red solution. The reaction’s balanced equation is:

Fe(NO3)2 (aq) + 2 KSCN (aq) Fe(SCN)2 (aq) + 2 KNO3 (aq)

It has a net ionic equation of:

Fe3+(aq) + SCN-(aq) → FeSCN2+(aq)
1.00 M HCl, meanwhile, was used to determine the presence of Ag+ in the resulting supernate. This resulted to formation of white precipitate. The reaction has a balanced equation of:

Ag+(aq) + Cl-(aq) → AgCl(s)

The table below shows the summary of the observed confirmatory test results:

Table 1. Data for Iron(II)-Silver Ions System

Test Reagent| Visible Result|
K3Fe(CN)6| Prussian blue|
KSCN| Blood red|
HCl| White ppt|

The results observed on the three separate confirmatory tests done proved the existence of the system reaching equilibrium and that the reaction between iron and silver did not reach absolute completion. The results of the first and second confirmatory tests show that after the (reaction of Fe2+ + Ag+) reaction not all of the reactants were consumed while the last test shows that products were formed. This is an indication that after some of the products were formed, some of these reformed the reactants until reaching a point of equilibrium.

B. Copper-Ammonia Equilibrium

To study the copper-ammonia equilibrium, 0.1 M CuSO4 and 1M NH3 were used. The balanced equation for the reaction is:

CuSO4 (aq) + 4 NH3 (aq) [Cu(NH3)4]SO4 (aq)

The reaction has a net ionic equation of:

Cu2+(aq) + 4NH3(aq) → Cu(NH3)42+(aq)
At the beginning of the reaction between Cu2+ and NH3, a pale blue precipitate formed. Ammonia is a weak base and forms a few ammonium and hydroxide ions in solution NH3(g) + H2O(l) ⇌ NH4+(aq) + OH-(aq)

The hexaaquacopper(II) ions react with hydroxide ions to form a precipitate. This involves deprotonation of two of the water ligand molecules. The reaction resulted to a pale blue precipitate due to the formation of Cu(OH)2 (aq) which is represented by the equation:

[Cu(H2O)6]2+(aq)(pale blue) + 2OH-(aq) [Cu(H2O)4(OH)2](s)(pale blue precipitate) + 2H2O(l)

or simply,

Cu2+(aq) + 2OH-(aq) Cu(OH)2 (s)

When the precipitate dissolved through the addition of excess ammonia, the resulting solution was dark blue. The copper (II) hydroxide precipitate reacts with the ammonia (NH3) molecules which was responsible for the formation of tetraamminediaquacopper (II) ions. This involves ligand exchange: [Cu(H2O)4(OH)2](s) + 4NH3(aq) ⇌ [Cu(NH3)4(H2O)2)]2+(aq) + 2OH-(aq) + 2H2O(l)

After the addition of HCl to the dark blue solution, the solution decolorized back to pale blue. This is because the acid introduces\d H+ ions, which reacted with NH3 molecules to form NH4+ ions, and this draws the equilibrium back to the left-hand side,...
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