Preparation of Hexaamminecolbalt (III) Chloride, Chemistry tutorial


Cobalt consists of the outer electronic configuration 3d74s2 and its highest significant oxidation state is +4. This reflects the trend towards decreased stability of the incredibly high oxidation state on moving across the transition metal series. Cobalt (I) forms several complexes, most by p-bonding ligands. The chemistry of cobalt (I) is better characterized than any other unipositive oxidation state of the first transition series apart from copper. The two most significant oxidation states are cobalt (II) [Ar]3d7 and cobalt(III) [Ar]3d6, and in an aqueous solution having no complexing agent cobalt(III) is simply reduced to cobalt(II). Though, cobalt (III) is more stable in the presence of a complexing agent like ammonia as represented by the electrode potentials. 

[Co(H2O)6]3+ + e- = [Co(H2O)6]2+ Eo = 1.85 V. [Co(NH3)6]3+ + e- = [Co(NH3)6]2+ Eo = 0.1 V

Cobalt(II) forms both octahedral and tetrahedral complexes however they are labile and they encompass a strong tendency to be oxidized via molecular oxygen. The complexes are generally prepared in an inert atmosphere. In aqueous solution, the cobalt (II) ion is pale pink as the absorption is weak and takes place in the blue area of the visible. Tetrahedral cobalt (II) complexes are often highly colored owing to their lower order of symmetry relative to the octahedral complexes. The spectrum of [CoCl4]2- exhibits a big absorption in the visible part of the spectrum, that accounts for its deep-blue color.  

Cobalt (III) salts are hard to prepare as the ion is a strong oxidizing agent and the chemistry of this oxidation state is largely that of coordination compounds. Cobalt (III) generally forms octahedral complexes and it consists of a strong affinity for nitrogen donors like ammonia, amines (example: ethylenediamine), nitro groups and nitrogen bonded-SCN groups and also water molecules and halide ions.

Preparation of [Co(NH3)6]Cl3

Hexaamminecobalt (III) salts might be made up through any of three methods that is based on oxidation of cobalt(II) ion in the ammoniacal solution:

1) Air oxidation, by the formation of the pentamine ion that is transformed to the hexamine via heating by aqueous ammonia under pressure.

2) Oxidation by an agent like hydrogen peroxide, iodine, potassium permanganate, lead oxide or hypochlorite solution;

3) Oxidation in the presence of a catalyst which lets equilibrium between the pentamine and hexamine ions to be set up at room temperature and atmospheric pressure. Such compounds might as well be prepared indirectly from other hexaminecobalt(III) salts.

In the best of catalytic processes diamminesilver ion or decolorizing the charcoal is employed as a catalyst. The process devised by J.Bjerrum in which decolorizing charcoal is catalyst is simple, provides high yields of pure product, and is not time-consuming. The high concentration of ammonium salt is enough to stabilize the hexaminecobalt(III) ion, and the carbon serves just to establish the equilibrium. Air is employed as oxidant apart from when the cobalt (II) compound is slightly soluble in the ammoniacal solution, as in the preparation of [Co(NH3)6]Br3, for which the hydrogen peroxide is preferable.

4CoCl2 + 4NH4Cl + 20NH3 + O2 → 4[Co(NH3)6]Cl3 + 2H2O

Materials needed: CoCl2.6H2O, NH4OH, NH4Cl, HCl, charcoal 


240g (1mol) of cobalt(II) chloride 6-hydrate and 160g (3mol) of ammonium chloride are added to the 200ml of water. The mixture is shaken till almost of the salts are dissolved. Then 4g of activated decolorizing charcoal and 500ml of concentrated ammonia are added. Air is bubbled vigorously via the mixture till the red solution becomes yellowish brown (generally about 4 hours). The air inlet tube is of fairly large bore (10mm) to prevent the clogging by the precipitated hexamminecobalt(III) salt.

The crystals and carbon are filtered on a Buchner funnel and then added to the solution of 15 to 30ml. of concentrated hydrochloric acid in 1500ml of water; adequate acid  reaction. The mixture is heated on the hot plate to affect the complete solution and is filtered hot. The hexamminecobalt(III) chloride is precipitated by adding 400ml of concentrated hydrochloric acid and slowly cooling to 0oC. The precipitate is filtered, washed first by 60% and then by 95% alcohol, and dried at 80 to 100oC.

Preparation of [Co(en)3]Cl3:

Ethylenediamine coordinates by metallic ions via both nitrogen atoms. The five-membered chelate rings which are therefore formed are very stable. Most of the cobalt (III) ammines are transformed by the aqueous ethylenediamine to ethylenediamine cobalt (III) chloride. Therefore, Jorgensen prepared the salt by heating [Co(NH3)5Cl]Cl2 by aqueous ethylenediamine. Grossman and Schuck obtained the salt through oxidizing a mixture of cobalt(II) chloride, ethylenediamine, and water. The procedure illustrated below has been developed from the latter suggestion. 

Material required: en, HCl, CoCl2.6H2


61g of 30% ethylenediamine is partially neutralized by 17 ml of 6N hydrochloric acid and the resultant mixture poured to a solution of 24 g of CoCl2.6H2O in 75 ml. of water. The cobalt is oxidized through bubbling a vigorous stream of air via the solution for three hours. The solution allowed evaporating on a steam bath till a crust starts to form over the surface (that is, the volume will be around 15 to 20 ml.); then 15 ml of concentrated hydrochloric acid and 30 ml of ethyl alcohol are added. After cooling, the crystals of [Co(en)3]Cl3 are filtered and washed by alcohol till the washings are colorless. They are then washed by ether or dried in an oven. 


[Co(en)3]Cl3 crystallizes in orange-yellow needles that are readily soluble in water however insoluble in the usual organic solvents. Its solubility in 6 N hydrochloric acid is around 3 %. It is stable at temperatures as high as 200o and is decomposed only slowly through hydrogen sulphide and sodium hydroxide.

Concept of Limiting Reactant:

Most of the chemical reactions need two or more reactants. Generally, one of the reactants is used up before the other, at which time the reaction stops. The chemical which is used up is termed as the limiting reactant whereas the other reactant is present in excess. If both the reactants are present in precisely the right amount to react completely, without either in surplus, the amounts of reactants are stated to be in the stoichiometric ratio to one other. The stoichiometric ratio is the mole ratio of the reactants, or reactants to products, as determined through the coefficients in the balanced chemical equation. As the limiting reactant will find out the amount of product that can be generated throughout a reaction, it is significant to be capable to compute which reactant is the limiting reactant. There are some ways to do this, however each begins with a balanced chemical equation in such a way that the Stoichiometry of the reaction is known.


Most of the times an analysis will be based on the fact which you encompass a limiting reactant comprised in the chemical reaction. The limiting reactant determines the percent yield of product. As chemicals react stoichiometrically, just a limited amount of product forms from given amounts of reactants. For illustration, the analysis of a sample having soluble sulphate salt like sodium sulphate can be  performed by dissolving the sample in water and adding  a  solution  having barium chloride to form an insoluble barium sulphate salt and sodium chloride that is soluble.

Na2SO4 (aq) + BaCl2 (aq) → BaSO4 (s) + 2NaCl (aq)

As all of the salts, apart from BaSO4, are soluble, the total ionic equation for the chemical reaction is Ba+2 (aq) + SO4-2 (aq) → BaSO4 (s)

One mole of barium ion, from BaCl2.2H2O (244.2g) in solution, reacts by 1 mole of sulphate ion, from 1 mole of Na2SO4 (142.1g) in solution, to generate 1 mole of barium sulphate precipitate (233.4g) if the reaction occurs fully. The reaction is termed to as being quantitative as it will let the determination of the amount of sodium sulphate in the sample. Sodium sulphate is the limiting reactant and to make sure that a total, complete reaction takes place a surplus of barium chloride solution is added. 

A 1.000 g sample of unknown salt mixture having sodium sulphate is weighed to the nearby 1mg. The number of moles of sodium sulphate is computed as if this were a pure sample of the sodium sulphate,

1.000g Na2SO4 x (1 mole Na2SO4/142.1g Na2SO4) = 7.030 x 10-3 moles Na2SO4

Therefore representing the maximum number of moles of sodium sulphate which might be present in the sample. By employing the balanced equation:

1.00 mole SO4-2 generates 1.00 mole of BaSO4,


= 7.030 x 10-3 moles SO4-2 generates 7.030 x 10-3 moles BaSO4


7.030 x 10-3 moles BaSO4 x (233.4g BaSO4/1 mole BaSO4) = 1.641 g BaSO4

Therefore, 1.641g BaSO4 symbolizes the maximum amount of precipitate which can be made up if the sample is 100% pure. This is termed to as the theoretical yield of the product. As the sample is not pure sodium sulphate, you will get less than the theoretical amount that we will call the experimental value. The percentage of sodium sulphate in the unknown cans be computed by using the percent yield formula that is stated as:

Percent yield = (experimental yield/theoretical yield) x 100

Most of the industrial products like food ingredients, deodorants, mineral waters and cosmetics would use this method for analysis of sulphate ion.

In this experiment, an unknown salt mixture having Na2SO4 is examined for the percentage of Na2SO4 by adding surplus BaCl2 solution to make sure completeness of reaction. The other ingredient in the mixture is a non-reactive, soluble substance that will not interfere with the quantitative reaction. The precipitate, BaSO4, will be vacuum filtered, dried and weighed to get experimental yield of BaSO4. The theoretical yield will be based on 100% activity of the salt mixture sample weighed. 


Weigh a clean, empty 400 ml beaker to the nearby 0.001 g. Add around 1.0 g of the unknown salt mixture to the beaker and re-weigh the beaker to the nearby 0.001 g.

Compute the number of moles of sodium sulphate which theoretically are contained in this sample weight. As well compute the theoretical yield of BaSO4 as illustrated in the theory part of the experiment.

Add around 200 ml of distilled water to the beaker and stir by a stirring rod. In the hood, measure 1 ml of concentrated HCl in a small graduated cylinder and by the help of a stirring rod, add to the solution in the beaker.

CAUTION: If spilled, concentrated HCl (hydrochloric acid) is a severe skin irritant-flush affected area having large amounts of water. 

In graduated cylinder, measure 50 ml of BaCl2 0.5 M and add to the solution in the beaker and stir for some minutes by a stirring rod. Let the precipitate to settle. Confirm that the amount of barium chloride added to the solution does, however, represent a surplus of the reactant.

By the stirring rod in the beaker, cover the beaker by a watch glass and place the whole apparatus on a hot plate. Heat at low setting to maintain the solution temperature between 80 to 90oC for 1 hour. Check the temperature of the solution by your thermometer occasionally. 

Ignore boiling the solution. Heat around 100ml of distilled water in a beaker to be employed as the washing solvent.

As the solution is heating, assemble the vacuum filtration apparatus. Two kinds of filters will be employed: a Buchner funnel by a paper filter and a fritted glass filtering crucible. 

The benefit of vacuum filtration is the speed of isolation for a solid product from solution. One trial will be completed in recovering the precipitate on the paper and a second trial will recover the solid in the filtering crucible. It will need the recorded weight of both, paper and crucible. The order in which such two isolation processes are done is not an important factor.

Before the filtration method is started, tip the beaker and allow the precipitate to settle beneath the lip of the beaker. As the solution is still hot, by employing beaker tongs, decant the supernatant liquid trying to reduce solid transfer to the filter. Wash the precipitate several times by warm distilled water, by employing a rubber policeman to get rid of any solid adhering to the sides of the beaker. By using the wash bottle, transfer the precipitate to the filtering medium and rinse sometimes with the warm distilled water. Allow air to be drawn via the filtering medium for some minutes after the last rinse. Break the vacuum seal at the filtering flask and take out the filtering medium to a clean, dry, preweighed, marked watch glass. Put the watch glass and filtering medium in drying oven for some hours or overnight. Whenever the precipitate is dry, let it to cool and weigh watch glass and filtering medium.

Compute the weight of BaSO4 isolated. Compute the percent yield of the reaction. Report the average percentage yield on two trials.

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