Solid Copper (II) Nitrate Synthesis Lab Report
http://www.chem-toddler.com/chemical-equilibrium/chromatedichromate.html http://ibchem.com/IB/ibnotes/full/aab_htm/18.3.htm In the experiment part (i) (a), the solid copper(II) nitrate is used to dissolve in a test tube filled with half distilled water. The solid copper(II) nitrate ionize in water to form copper(II) ions and nitrate ion. Hence, a blue solution is formed due to the blue copper(II) ion present in the solution. When excess solid is added, the solution will become saturated and do not allow any solid to dissolve, so excess solid will remain in the solution. This is shows that the equilibrium between solid and aqueous copper(II) nitrate is achieved. When the test tube is being placed in the water bath of 60 , the level of
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solid copper(II) nitrate tends to decrease to ionize in the solution but the blue intensity of the solution remains the same. The dissolution of copper(II) nitrate is an endothermic reaction. So, the equilibrium constant of the reaction increases as the temperature of the solution increases which allow more products are formed. The equilibrium position is shift to the left in the equilibrium equation below:
Cu(NO3)2 (aq) Cu2+ (aq) + NO3- (aq)
Eventually, more copper(II) ions and nitrate ion are formed at the high temperature due to the equilibrium effect. Oppositely, the level of solid copper(II) nitrate is increases at the 0 . When the test tube is placed in an ice bath, the low temperature causes the equilibrium constant of the solution decreases and hence the formation of solid copper(II) nitrate is increases. The equilibrium position shifts to the left at the condition of 0 . Next, the cobalt(II) chloride is added with hydrochloric acid in this experiment in order to investigate its equilibrium at different temperature.
The cobalt (II) chloride is added with hydrochloric acid, which a pink solution is formed. When the test tube is placed in a water bath with 60 , the solution turns from light pink to dark pink. This is because the cobalt(II) chloride dissolve in solution is an endothermic reaction which it tend s to form blue cobalt (II) chloride in the solution. The dark pink solution is formed instead of light pink because the dark pink colour of solution is the mixture of pink and blue. The equilibrium position shift to right in the equilibrium equation below:
[Co[H2O]6]2+ (aq) + 4Cl-(aq) [CoCl4]2- (aq) + 6H2O (l)
Since the equilibrium equation above is an endothermic reaction, the position of equilibrium tends to shift to right at high temperature, 60 in this experiment. The increase in equilibrium constant leads to more products is formed. So, the colour changes to dark pink due to the amount of cobalt(II) chloride formed is increases. On the other hand, the equilibrium position shifts to the left at the temperature of 0 . When the equilibrium shifts to left, the concentration of hydrated cobalt ion form and hence the solution forms at a light pink
colour.
In the experiment part (ii) (a), the potassium dichromate solution is orange colour while the potassium chromate solution is appears in yellow colour. When there is addition of two drops of 0.2M hydrochloric acid,HCl, the colour of potassium chromate solution turns from yellow to orange due to the equilibrium effect. The equilibrium equation between chromate ion and dichromate ion is
2CrO42- (aq) + 2H+ (aq) Cr2O72- (aq) + H2O (l)
When HCl is being added, the acid will ionize to become H+ ion and Cl- ion. The chromate ion, CrO42- ion accept the excess H+and shift to the right to become dichromate ion, Cr2O72- which is orange colour ion. But, when two drops of sodium hydroxide ion,
NaOH is added, the OH- ion is ionize and tends to combine with H+to become water and hence the concentration of H+ ion in the solution decreases. The dichromate ion, Cr2O72-will release H+ ion to form chromate ion, CrO42- in the solution, thus the colour change to yellow colour. When second time of HCl is added, the chromate ion, CrO42-will accept the H+ion to dichromate ion, Cr2O72- form if the H+ ion concentration is excess.
In the part (ii) (b) experiment, the thymolphthalein is undergoes reduction which its blue colour solution is decolorized. When the blue thymolphthalein is exposed to the air, the carbon dioxide in the air will react with the thymolphthalein since it is a basic solution. Initially, dilute sodium hydroxide solution is added into the thymolphthalein when preparing the thymolphthalein indicator in order to maintain it pH in basic condition. If hydroxide ion, OH- is excess in the solution, the blue colour can be perceived. However, when thymolphthalein is exposed to the air, the sodium hydroxide will tends to react with carbon dioxide to form sodium carbonate and water as shown in the chemical equation below:
At the same time, the carbon dioxide will react with water in the solution to form carbonic acid, H2CO3. Due to the partial ionization of carbonic acid, the hydronium ion, H+ is produced which caused the condition become more and more acidic (pH value dropping).
The ionized H+ ion will react with the thymolphthalein which driving the equilibrium shift to the right to form colourless Hln in the solution.
Hln (aq) H+ (aq) + ln- (aq)
The equilibrium between ammonia molecule and ammonium ion is studied in this experiment. When phenolphthalein is added into the ammonia solution, the colour present in the solution is light purple colour. This is because the ammonia solution is a weak acid with pH value of 11.6. The phenolphthalein turns from colourless to light purple when it is added into the solution with pH 11.6. When solid ammonium chloride is added, the ammonium chloride dissolves quickly in the solution to form ammonium ions and chloride ions. The ammonium chloride dissociates completely in the solution. The formation of ammonium ions in the solution combine with hydroxide ions and hence shifts the equilibrium position to left to form more ammonia and water as shown below:
NH3 (aq) + H2O (l) NH4+ (aq) OH- (aq)
At the same time, the pH value of the solution decreases due to the concentration of the hydroxide ions is reduced by the reaction. As a result, the phenolthalein in the solution is decolourized due to the drops in pH of the solution.
solid copper(II) nitrate tends to decrease to ionize in the solution but the blue intensity of the solution remains the same. The dissolution of copper(II) nitrate is an endothermic reaction. So, the equilibrium constant of the reaction increases as the temperature of the solution increases which allow more products are formed. The equilibrium position is shift to the left in the equilibrium equation below:
Cu(NO3)2 (aq) Cu2+ (aq) + NO3- (aq)
Eventually, more copper(II) ions and nitrate ion are formed at the high temperature due to the equilibrium effect. Oppositely, the level of solid copper(II) nitrate is increases at the 0 . When the test tube is placed in an ice bath, the low temperature causes the equilibrium constant of the solution decreases and hence the formation of solid copper(II) nitrate is increases. The equilibrium position shifts to the left at the condition of 0 . Next, the cobalt(II) chloride is added with hydrochloric acid in this experiment in order to investigate its equilibrium at different temperature.
The cobalt (II) chloride is added with hydrochloric acid, which a pink solution is formed. When the test tube is placed in a water bath with 60 , the solution turns from light pink to dark pink. This is because the cobalt(II) chloride dissolve in solution is an endothermic reaction which it tend s to form blue cobalt (II) chloride in the solution. The dark pink solution is formed instead of light pink because the dark pink colour of solution is the mixture of pink and blue. The equilibrium position shift to right in the equilibrium equation below:
[Co[H2O]6]2+ (aq) + 4Cl-(aq) [CoCl4]2- (aq) + 6H2O (l)
Since the equilibrium equation above is an endothermic reaction, the position of equilibrium tends to shift to right at high temperature, 60 in this experiment. The increase in equilibrium constant leads to more products is formed. So, the colour changes to dark pink due to the amount of cobalt(II) chloride formed is increases. On the other hand, the equilibrium position shifts to the left at the temperature of 0 . When the equilibrium shifts to left, the concentration of hydrated cobalt ion form and hence the solution forms at a light pink
colour.
In the experiment part (ii) (a), the potassium dichromate solution is orange colour while the potassium chromate solution is appears in yellow colour. When there is addition of two drops of 0.2M hydrochloric acid,HCl, the colour of potassium chromate solution turns from yellow to orange due to the equilibrium effect. The equilibrium equation between chromate ion and dichromate ion is
2CrO42- (aq) + 2H+ (aq) Cr2O72- (aq) + H2O (l)
When HCl is being added, the acid will ionize to become H+ ion and Cl- ion. The chromate ion, CrO42- ion accept the excess H+and shift to the right to become dichromate ion, Cr2O72- which is orange colour ion. But, when two drops of sodium hydroxide ion,
NaOH is added, the OH- ion is ionize and tends to combine with H+to become water and hence the concentration of H+ ion in the solution decreases. The dichromate ion, Cr2O72-will release H+ ion to form chromate ion, CrO42- in the solution, thus the colour change to yellow colour. When second time of HCl is added, the chromate ion, CrO42-will accept the H+ion to dichromate ion, Cr2O72- form if the H+ ion concentration is excess.
In the part (ii) (b) experiment, the thymolphthalein is undergoes reduction which its blue colour solution is decolorized. When the blue thymolphthalein is exposed to the air, the carbon dioxide in the air will react with the thymolphthalein since it is a basic solution. Initially, dilute sodium hydroxide solution is added into the thymolphthalein when preparing the thymolphthalein indicator in order to maintain it pH in basic condition. If hydroxide ion, OH- is excess in the solution, the blue colour can be perceived. However, when thymolphthalein is exposed to the air, the sodium hydroxide will tends to react with carbon dioxide to form sodium carbonate and water as shown in the chemical equation below:
At the same time, the carbon dioxide will react with water in the solution to form carbonic acid, H2CO3. Due to the partial ionization of carbonic acid, the hydronium ion, H+ is produced which caused the condition become more and more acidic (pH value dropping).
The ionized H+ ion will react with the thymolphthalein which driving the equilibrium shift to the right to form colourless Hln in the solution.
Hln (aq) H+ (aq) + ln- (aq)
The equilibrium between ammonia molecule and ammonium ion is studied in this experiment. When phenolphthalein is added into the ammonia solution, the colour present in the solution is light purple colour. This is because the ammonia solution is a weak acid with pH value of 11.6. The phenolphthalein turns from colourless to light purple when it is added into the solution with pH 11.6. When solid ammonium chloride is added, the ammonium chloride dissolves quickly in the solution to form ammonium ions and chloride ions. The ammonium chloride dissociates completely in the solution. The formation of ammonium ions in the solution combine with hydroxide ions and hence shifts the equilibrium position to left to form more ammonia and water as shown below:
NH3 (aq) + H2O (l) NH4+ (aq) OH- (aq)
At the same time, the pH value of the solution decreases due to the concentration of the hydroxide ions is reduced by the reaction. As a result, the phenolthalein in the solution is decolourized due to the drops in pH of the solution.