Experiment 4 Rdr - Chemical Equilibrium

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Iron-Silver Equilibrium

In the first phase of the experiment, iron sulfate was mixed with silver nitrate, and the reaction produced solid silver and iron nitrate, which is formally written as,

Fe2+ (aq) + Ag+ (aq) ⇄ Ag (s) + Fe3+ (aq)

The mixture was then placed in a centrifuge in order for the solid silver to settle and separate from the supernate. The supernate was then tested for the presence of Fe2+, Fe3+ and Ag+ by placing K3Fe(CN)6, KSCN and HCl to 3 separate samples of the supernate. The result of the tests was as follows:

1.Addition of K3Fe(CN)6

After K3Fe(CN)6 was added to the supernate, a Prussian blue precipitate was formed, more formally written as,

Fe2+ (aq) + Fe(CN)63- (aq) + K+ (aq) → K∙Fe2(CN)6 (s)

This reaction thus proves that the supernate contains Fe2+.

2.Addition of KSCN

After KSCN was added to the supernate, a blood red complex was formed, more formally written as,

Fe3+ (aq) + SCN- (aq) → FeSCN2+ (aq)

This reaction thus proves that the supernate contains Fe3+.

3.Addition of HCl

The last test was the addition of HCl to the supernate. This produced a white precipitate, which we can formally write as,

Ag+ (aq) + Cl- (aq) → AgCl (s)

This reaction thus proves that the supernate conatins Ag+.

The 3 tests showed that all 3 ions were present in the supernate. This goes to show that the reaction between iron sulfate and silver nitrate was in a state of equilibrium since both the ions in the product and reactant side were present, meaning to say that the forward and reverse reactions were proceeding at the same rate.

Furthermore, the range of the equilibrium constant (Keq) for the reaction of iron sulfate and silver nitrate is from 10-10 to 1010 [1].

Copper-Ammonia Equilibrium

The second phase of the experiment dealt with the reaction between copper sulfate and ammonia, more formally written as,

CuSO4 (aq) + 2 NH4OH (aq) → Cu(OH)2 (aq) + (NH4)2SO4 (aq)

The pale blue precipitate formed at the beginning of the reaction of copper sulfate and ammonia was the Cu(OH)2.

From the pale blue color, it turned into a deep cerulean blue when 11 drops of ammonia was added. The solution then went back to the pale blue color after 4 drops of hydrochloric acid was added. The addition of hydrochloric acid added more H+ ions to the solution; therefore, drawing the equilibrium back to the reactant side. This equilibrium reaction is formally stated as,

[Cu(H2O)6]2+ (aq) + 4 NH3 (aq) ⇄ [Cu(NH3)6]2+ (aq) + H2O (l)

As observed, it took almost 3 times the number of drops of ammonia to change the pale blue color to a deep cerulean blue as compared to the number of drops of hydrochloric acid that changed the deep cerulean blue back to pale blue. This means that the reverse reaction was more spontaneous than the forward reaction.

Chromate-Dichromate Equilibrium

In the third phase of the experiment, the chromate and dichromate solutions were observed. Chromate had a yellow color, while dichromate had an orange color. When sulfuric acid was added to a sample of chromate and dichromate solutions, the yellow chromate solution turned orange, while the dichromate solution remained orange. The equation for the chromate’s change in color is as follows,

2 CrO42- (aq) + 2 H+ (aq) → H2O (l) + Cr2O7- (aq)

When sodium hydroxide was added to a sample of chromate and dichromate, the chromate solution remained yellow, while the orange dichromate solution turned yellow, formally written as,

2 OH- (aq) + Cr2O7- (aq) → 2 CrO42- (aq) + H2O (l)

The change in color of chromate as hydrochloric acid was added and the change in color of dichromate as sodium hydroxide was added was due to the instability of the reactions, causing a shift in the equilibrium.

The acid H2SO4 (sulfuric acid) was used in the reaction since it’s a strong acid, and strong acids dissociate more. The added H+ ions increase the concentration, therefore, shifting the...
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