Reference

Topics: Ion, Sodium hydroxide, Ammonia Pages: 3 (1070 words) Published: June 18, 2013
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 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...
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