Question

(a) In early studies it was observed that when the complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br}\) was placed in water, the electrical conductivity of a \(0.05 M\) solution changed from an initial value of \(191 \mathrm{ohm}^{-1}\) to a final value of \(374 \mathrm{ohm}^{-1}\) over a period of an hour or so. Suggest an explanation for the observed results. (See Exercise \(24.49\) for relevant comparison data.) (b) Write a balanced chemical equation to describe the reaction. (c) A 500-mL solution is made up by dissolving \(3.87 \mathrm{~g}\) of the complex. As soon as the solution is formed, and before any change in conductivity has occurred, a 25.00-mL portion of the solution is titrated with \(0.0100 \mathrm{M} \mathrm{AgNO}_{3}\) solution. What volume of \(\mathrm{AgNO}_{3}\) solution do you expect to be required to precipitate the free \(\mathrm{Br}^{-}(a q) ?\) (d) Based on the response you gave to part (b), what volume of \(\mathrm{AgNO}_{3}\) solution would be required to titrate a fresh \(25.00-\mathrm{mL}\) sample of \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br}\) after all conductivity changes have occurred?

Short answer

The increase in electrical conductivity is due to the dissociation of the complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br}\) into more ions in water. The balanced equation for the dissociation is \[\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br} \rightarrow [\mathrm{Co}(\mathrm{NH}_{3})_{4} \mathrm{Br}_{2}]^{+} + \mathrm{Br}^{-}\]. To precipitate the free \(\mathrm{Br}^{-}(a q)\) from the initial solution, 65.5 mL of 0.0100 M \(\mathrm{AgNO}_{3}\) is needed. After all conductivity changes have occurred, the required volume of \(\mathrm{AgNO}_{3}\) remains the same, which is 65.5 mL.

Step-by-step solution

(a) Explanation of the change in electrical conductivity

The observed increase in electrical conductivity suggests that the complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br}\) undergoes a reaction when placed in water, leading to the formation of more ions. These ions increase the charge carriers in the solution, thus increasing electrical conductivity.

(b) Balanced chemical equation for the reaction

The balanced equation for the dissociation of the complex into the cobalt coordination sphere and bromide ions is given below: \[\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br} \rightarrow [\mathrm{Co}(\mathrm{NH}_{3})_{4} \mathrm{Br}_{2}]^{+} + \mathrm{Br}^{-}\]

(c) Calculation of the volume of AgNO3 solution

To calculate the volume of AgNO3 solution required to precipitate the free \(\mathrm{Br}^{-}(a q)\), first determine the concentration of the \(\mathrm{Br}^{-}\) ion in the original solution: 1. Calculate the moles of complex in the 500-mL solution: Moles of complex = \(\frac{mass}{molar\ weight}\) = \(\frac{3.87\ \mathrm{g}}{{294.97\ \mathrm{g/mol}}}\) = 0.0131 mol 2. Calculate the concentration of \(\mathrm{Br}^{-}\) ions in the 0.05 M solution (500 mL): Concentration of \(\mathrm{Br}^{-}\) = \(\frac{Moles\ of\ complex}{Volume\ of\ solution}\) = \(\frac{0.0131\ \mathrm{mol}}{{0.5\ \mathrm{L}}}\) = 0.0262 M 3. Calculate the moles of \(\mathrm{Br}^{-}\) ions in 25.00 mL of 0.0262 M solution: Moles of \(\mathrm{Br}^{-}\) = 0.0262 M × 0.025 L = 0.000655 mol 4. Calculate the volume of 0.0100 M \(\mathrm{AgNO}_{3}\) solution needed to precipitate 0.000655 mol of \(\mathrm{Br}^{-}\) ions: Volume of \(\mathrm{AgNO}_{3}\) solution = \(\frac{moles\ of\ Br^{-}}{concentration\ of\ AgNO_{3}}\) = \(\frac{0.000655\ mol}{0.0100\ \mathrm{M}}\) = 0.0655 L = 65.5 mL So, 65.5 mL of 0.0100 M \(\mathrm{AgNO}_{3}\) solution is required to precipitate the free \(\mathrm{Br}^{-}(a q)\).

(d) Calculation of the volume of AgNO3 solution after all conductivity changes

Based on the balanced equation from part (b), we see that each mole of complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br}\) only releases 1 mole of \(\mathrm{Br}^{-}\). Consequently, after all conductivity changes have occurred, the dissociation is complete, and every mole of complex will contribute 1 mole of \(\mathrm{Br}^{-}\). Thus, the same moles of \(\mathrm{Br}^{-}\) are present in the solution, and the required volume of \(\mathrm{AgNO}_{3}\) would remain the same - 65.5 mL.

Key concepts

Coordination Chemistry

Coordination chemistry studies how metal ions form complexes by binding to other molecules or ions. These bindings occur at specific sites known as "coordination sites." In our exercise, we work with the complex \[ \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br} \] where cobalt (Co) is the central metal ion, and \[ \mathrm{NH}_{3} \] and \[ \mathrm{Br}^{-} \] act as ligands. Ligands are ions or molecules that can donate pairs of electrons to the metal to form what is called a coordination complex.

• The complex's geometry can influence its chemical properties.
• Measures such as conductivity can provide insights into how these compounds dissociate in water.

In the given problem, the complex's dissociation in water results in different ionic species, affecting the solution's conductivity over time. Understanding coordination chemistry helps explain why these transformations occur when a complex is put in a solution.

Chemical Equilibrium

Chemical equilibrium refers to a state where the reactants and products of a reaction are formed at the same rate. In the context of the given complex ion reaction, it signifies that the forward and reverse reactions reach a balance in an aqueous solution.
For the dissociation of our complex ion \[ \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right] \mathrm{Br} \] into its ionic components, the system aims to establish equilibrium between its dissociated ions and the associated complex form.

• At equilibrium, the rates of dissolution and recombination of the ions are equal.
• This state affects the concentration of ions, impacting the solution's electrical properties.

Initially, the solution's conductivity increases until equilibrium is reached. Understanding chemical equilibrium helps predict and explain the behavior of ionic concentrations over time.

Electrical Conductivity

Electrical conductivity measures a solution's ability to conduct electricity, closely linked to the movement of ions in the solution.
In our problem, the initial and final conductivity values indicate a reaction wherein the complex dissociates to produce more free ions. The process of dissociation changes the conductivity observed.

• Initially, the conductivity is lower due to fewer ions in solution.
• As the complex dissociates into individual constituents, like \[ \mathrm{Br}^{-} \] , conductivity increases as more charge carriers become present.

Observing changes in conductivity over time can thus provide evidence for the extent and speed of the complex's dissociation reaction.

Titration Calculations

Titration calculations involve determining the concentration of an analyte in a solution by reacting it with a titrant of known concentration.
In the exercise, we calculate how much \[ \mathrm{AgNO}_{3} \] solution is needed to precipitate bromide (\[ \mathrm{Br}^{-} \] ) ions. This helps measure how much of these ions are present in the solution before and after a reaction reaches equilibrium. Here’s how:

• Begin by figuring out the moles of bromide ions using the initial amounts given.
• Use stoichiometry to relate these moles to the volume of the given titrant.

For our example, the calculated change in bromide ion concentration before and after the conductivity change helps determine the titrant volume required for the equilibrium conditions. This technique is essential for analyzing reaction progress and final states in a quantitative way.