Anhydrous copper sulphate, CuSO4, is white. When it is dissolved in water, the resulting solution is — Physical Chemistry — Kinetics Chemistry Question
Preparation of the Complex Salt Cu(NH3)4SO4 . H2O
Anhydrous copper sulphate, CuSO4, is white. When it is dissolved in water, the resulting solution is sky blue because of the formation of the complex ion [Cu(H2O)6]2+, or [Cu(H2O)4(H2O)2]2+ or Cu2+(aq). The six water molecules are not equivalent due to the Jahn–Teller effect.
The hydrated solid salt of copper sulfate, CuSO4 . 5 H2O which may be written as [Cu(H2O)4]SO4·H2O is also blue.
If a solution of NH3 is added to a solution of Cu2+(aq), the colour becomes intensely blue because of the formation of a new complex:
Cu2+(aq) + 4 NH3 <-> [Cu(NH3)4]2+ + water
In NH3 solutions with a concentration of 0.01 to 5 mol dm–3 the complex [Cu(NH3)4]2+ is mainly formed. In lower concentrations of NH3 formation of complexes containing fewer NH3 molecules is favored, that is [Cu(NH3)3(H2O)]2+, [Cu(NH3)2(H2O)2]2+ and [Cu(NH3)(H2O)3]2+. In concentrations of NH3 higher than 5 mol dm–3, [Cu(NH3)5(H2O)]2+ is also formed. Under these conditions the predominant complex is [Cu(NH3)4]2+.
Kform = [[Cu(NH3)4]2+] / ([Cu2+(aq)][NH3]4)
Kform has a large value, that is the equilibrium is shifted to the right, while Kinst, which is defined as 1/Kform, is small, hence the complex [Cu(NH3)4]2+ is stable.
The equilibrium is established quickly, that is, the complex [Cu(NH3)4]2+ is labile.
Complexes in which the corresponding equilibrium is established slowly are called inert.
Due to the lability of the complex [Cu(NH3)4]2+ the NH3 molecules that are bound to the central ion Cu2+ are exchanged quickly and continuously with non-complexed NH3 molecules, which are present in the solution as well as with molecules of the solvent (water).
Experiment
1. 6.25 g of hydrated copper sulfate CuSO4 . 5 H2O are dissolved in a mixture of 10 cm3 of concentrated NH3 solution and 6 cm3 of distilled water. The intensely blue solutions of the complex [Cu(NH3)4]2+ will be formed according to the previous equilibrium.
- The precipitated salt is filtered under vacuum and washed sequentially by (a) a mixture of equal volumes of ethanol and concentrated solution of NH3, (b) pure ethanol and (c) finally ether.
- The so obtained crystals are placed in a desiccator. If a drying compound is used that can react with NH3, e.g. CaCl2, gas phase NH3 will be bound and the complex will decompose in order to maintain the solid – gas equilibrium. A compound not reacting with NH3 must be used, like CaO.
The complex salt [Cu(NH3)4]SO4 . H2O is less soluble in a mixture of ethanol-water than in water. (Explain why). By adding 10 cm3 of ethanol to the aqueous solution and cooling, a precipitate is formed. Is the dissolution in a mixture of ethanol-water endothermic or exothermic?
The binding of Cu(II) with NH3 can be shown qualitatively as follows:
0.3 g of the starting CuSO4· 5 H2O are dissolved in 10 cm3 water, a few drops of Na2CO3 solution (2 mol dm–3) are added. Blue precipitate of CuCO3 is formed. A similar solution of [Cu(NH3)4]SO4· H2O does not give the previous reaction since Cu(II) is in the form of [Cu(NH3)4]2+.
Under which conditions formation of CuCO3 would be possible from the solution of the complex salt [Cu(NH3)4]SO4· H2O?
[Cu(NH3)4]2+ <-> Cu2+(aq) + 4 NH3
Removal of NH3 would shift the equilibrium to the right:
(a) by heating
(b) by addition of CaCl2
(c) by addition of HCl.
Why is the complex salt more soluble in water than in ether?