Question

Mixture of \((\mathrm{X})=0.02\) moles of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}\) and \(0.02\) mole of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4}\) was prepared in 2 litre of solution 1 litre of mixture \((\mathrm{X})+\) excess \(\mathrm{AgNO}_{3} \longrightarrow(\mathrm{Y})\) 1 litre of mixture \((\mathrm{X})+\) excess \(\mathrm{BaCl}_{2} \longrightarrow(\mathrm{Z})\) (a) \(0.01,0.01\) (b) \(0.02,0.01\) (c) \(0.01,0.02\) (d) \(0.02,0.02\)

Short answer

(a) 0.01, 0.01

Step-by-step solution

Identify Components of the Mixture

We are given two components: 0.02 moles of \( \left[ \mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br} \) and 0.02 moles of \( \left[ \mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4} \) in 2 litres of solution. This means each component's concentration in the 2-litre mixture is \(\frac{0.02}{2} = 0.01\) M.

Reaction of Mixture X with AgNO3

When the 1-litre solution mixture is reacted with \( \mathrm{AgNO}_{3} \), the complex \( \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}\right] \mathrm{SO}_{4} \) will release Br^- ions to form \( \mathrm{AgBr} \). The other complex does not release Br^- ions. Thus, 0.01 moles of \( \mathrm{AgBr} \) precipitate is formed.

Reaction of Mixture X with BaCl2

The reaction with \( \mathrm{BaCl}_{2} \) will precipitate \( \mathrm{BaSO}_{4} \) from \( \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Br}\). Again, 0.01 moles of \( \mathrm{BaSO}_{4} \) will be formed as only the sulfate complex releases \( \mathrm{SO}_{4}^{2-} \) ions.

Key concepts

Complex Ion Reactions

Complex ion reactions are fascinating because they involve compounds known as coordination complexes where a metal atom is bonded to surrounding molecules or ions, called ligands. In the exercise above, we observe cobalt complexes with slightly different arrangements of ligands: one has sulfate, while the other has bromide, within its coordination sphere. These differences in ligands determine the type of ion the complex will release into a solution and, consequently, what kind of reaction it will participate in.

Coordination compounds play key roles in reactions because:

• They can form precipitates with specific reagents.
• The arrangement of ligands influences the reactivity of the complex.
• They can participate in selective ion release based on the ligand's type.

By understanding how different ligands behave, one can predict and control the outcome of reactions involving complex ions. This knowledge is crucial in various applications, from industrial processes to biochemical systems.

Stoichiometry in Reactions

Stoichiometry is the branch of chemistry that deals with quantitatively measuring this precise relationship between reactants and products. It's like the recipe of the chemical world, ensuring that everything is accounted for down to the last molecule.

In our context, consider the mixture of cobalt complexes. Each solution component's concentration is determined using stoichiometric calculations. Initially, both complexes were present at 0.02 moles in a 2-litre solution, giving each a concentration of 0.01 M. This precise mixture helps predict how much product (like precipitates) will form when the mixture reacts with another reagent:

• For each mole of \([\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}] \mathrm{SO}_{4}\), 0.01 moles of \( \mathrm{AgBr} \) will form.
• For each mole of \([\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}] \mathrm{Br}\), 0.01 moles of \( \mathrm{BaSO}_{4} \) will form.

Understanding stoichiometry is essential for effectively predicting and understanding the outcomes of chemical reactions, especially those involving complex ions.

Precipitation Reactions

Precipitation reactions occur when two soluble substances in a solution react to form an insoluble product known as a precipitate. These reactions are a visual marvel in chemistry as they change the solution's appearance as solid particles suddenly appear.

In the provided exercise, we encounter two separate precipitation reactions:

• The cobalt complex with bromide forms \( \mathrm{AgBr} \) upon encountering \( \mathrm{AgNO}_3 \). \( \mathrm{AgBr} \) is an insoluble salt, appearing as a precipitate in the solution.
• Similarly, the cobalt complex with sulfate forms \( \mathrm{BaSO}_4 \) when \( \mathrm{BaCl}_2 \) is introduced. This leads to the visible formation of a solid that settles out of the solution.

Understanding precipitation reactions helps chemists easily separate and identify compounds based on their solubility properties. This knowledge is widely used in fields like environmental chemistry, water treatment, and even mineral extraction. Knowing how, when, and why these reactions occur can enhance both theoretical knowledge and practical laboratory skills.