When a solution of copper(II) sulfate is treated with a small volume of concentrated hydrochloric acid then the blue solution turns green as a new copper complex is formed in solution and water is released.
The reaction is readily reversible.
Describe a method to safely demonstrate that this reaction is reversible.
Answers
When copper[II] sulphate is dissolved in water, the solution is blue in colour as the cupric ion readily combines with water to form a complex ion called Hexaaqua copper ion or [Cu(H2O)6]+2 . When concentrated hydrochloric acid is added to a dilute copper sulphate, copper chloride and sulphuric acid are formed, the colour of copper chloride solution is actually also blue because of the presence of [Cu(H2O)6] +2 but the chloride ion being reactive, combines with Cu2+ ions to form another complex ion [Cu(Cl)4] -2 . This ion gives a green colour to the solution.
This reaction follows the Le Chatelier's principle and is entirely reversible. Once the [CuCl4] -2 ion is formed and the solution turns green, by adding more water to the solution and by diluting it, the concentration of Cl-1 decreases and the solution turns blue again as the Hexaaqua Cupric ion [Cu(H2O)6]+2 is reformed, if more concentrated hydrochloric acid is added, the solution turns green as the [CuCl4]-2 complex iron forms again, this shows that the reaction is entirely reversible and the concentration of Cl-1 ions or water determines the type of complex ion that is formed.