Question

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

Short answer

(b) 0.01, 0.01

Step-by-step solution

Understanding the Mixture Composition

The mixture \( X \) consists of two coordination compounds: \([\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{SO}_4]\mathrm{Br}\) and \([\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{Br}]\mathrm{SO}_4\). Both compounds in \( X \) together have a total amount of \(0.02\) mol in \(2\) liters of solution.

Identifying Products Formed with AgNO₃

\([\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{SO}_4]\mathrm{Br}\) reacts with excess AgNO₃ to form AgBr as the precipitate, while \([\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{Br}]\mathrm{SO}_4\) does not precipitate Br⁻ ions when mixed with Ag⁺. Hence, only the Br⁻ ions from \([\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{SO}_4]\mathrm{Br}\) will react to form \( Y \).

Identifying Products Formed with BaCl₂

\([\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{Br}]\mathrm{SO}_4\) reacts with excess BaCl₂ to form BaSO₄ as the precipitate, while \([\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{SO}_4]\mathrm{Br}\) does not precipitate SO₄²⁻ ions. Hence, only the SO₄²⁻ ions from \([\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{Br}]\mathrm{SO}_4\) will react to form \( Z \).

Calculate Moles of Each Compound in Mixture X

Since the total moles of both compounds in the solution is \(0.02\) moles, and they are used equally in reactions with AgNO₃ and BaCl₂, it implies that both \([\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{SO}_4]\mathrm{Br}\) and \([\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{Br}]\mathrm{SO}_4\) are 0.01 mol each in 2 liters, therefore 0.005 mol each per liter.

Determining Moles of Y and Z in 1 Litre

For \(1\) liter of solution, \(0.005\) moles of \([\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{SO}_4]\mathrm{Br}\) reacts with AgNO₃ to form \( Y \), which corresponds to 0.005 mol of AgBr, and \(0.005\) moles of \([\mathrm{Co}(\mathrm{NH}_3)_5 \mathrm{Br}]\mathrm{SO}_4\) reacts with BaCl₂ to form \( Z \), corresponding to 0.005 mol of BaSO₄. Therefore, total moles of \( Y \) and \( Z \) are each 0.01 mol.

Key concepts

Coordination Chemistry

Coordination chemistry is the study of the complex compounds formed by the combination of metal ions with ligands. Ligands, such as ammonia (\(\text{NH}_3\) ), are molecules or ions that donate pairs of electrons to a metal ion, creating a coordination complex.
The coordination compounds in our exercise are two distinct complexes of cobalt: \([\text{Co}(\text{NH}_3)_5 \text{SO}_4]\text{Br}\) and \([\text{Co}(\text{NH}_3)_5 \text{Br}] \text{SO}_4\) .

Coordination compounds are characterized by their coordination number, which is the number of donor atoms bonded to the central metal ion.

• In our given compounds, cobalt is coordinated by five ammonia molecules, which act as neutral ligands.
• The sulfate and bromide ions are the counterions outside the coordination sphere in these compounds.

This structure gives rise to specific geometric arrangements, stability, and reactions that are unique to coordination chemistry.

Precipitation Reactions

Precipitation reactions are chemical reactions where two solutions are mixed, resulting in the formation of a solid, known as the precipitate. These reactions are essential for identifying and isolating specific ions in solution.
In the given exercise, the reaction of the coordination compounds with \( \text{AgNO}_3 \) and \( \text{BaCl}_2 \) demonstrates this concept.

• When \([\text{Co}(\text{NH}_3)_5 \text{SO}_4]\text{Br}\) is mixed with \( \text{AgNO}_3 \) , the bromide ion (\text{Br}^- ) reacts to form silver bromide (\text{AgBr}), which is a precipitate.
• In contrast, when \([\text{Co}(\text{NH}_3)_5 \text{Br}]\text{SO}_4\) reacts with \( \text{BaCl}_2 \) , the \(\text{SO}_4^{2-}\) ion forms barium sulfate(\text{BaSO}_4), another solid precipitate.

The distinct solubilities of \(\text{AgBr}\) and \(\text{BaSO}_4\) allow for the separation of the components in a mixture by selective precipitation.

Stoichiometry

Stoichiometry involves the calculation of the reactants and products in chemical reactions. It is an essential concept in chemistry that helps predict the amounts of substances consumed and produced in a reaction.

In this exercise, stoichiometry is used to determine the moles of products formed from the compounds in mixture \( X \).

• Each of the coordination compounds \([\text{Co}(\text{NH}_3)_5 \text{SO}_4]\text{Br}\) and \([\text{Co}(\text{NH}_3)_5 \text{Br}]\text{SO}_4\) contains 0.01 mol in a 2-liter solution.
• Therefore, in 1 liter of solution, both components react to give 0.005 mol of precipitate each.
• This knowledge allows us to calculate that 0.01 mol of each product \( Y \) and \( Z \) is formed.

Stoichiometry ensures that chemical equations are balanced and accurate, reflecting the conservation of mass and charge across reactions.