Chapter 23: Problem 97

(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 23.69 for relevant comparison data.) (b) Write a balanced chemical equation to describe the reaction. (c) \(A 500\)-mL solution is made up by dissolving 3.87g 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 AgNO \(_{3}\) solution do you expect to be required to precipitate the free \(\operatorname{Br}^{-}(a q) ?(\mathbf{d})\) 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 -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

Expert verified

The complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right]\mathrm{Br}\) undergoes a reaction in water that releases \(\mathrm{Br}^-\) ions, increasing the conductivity over time: \[\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right]\mathrm{Br} (s) \rightarrow \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{H}_{2}\mathrm{O}\mathrm{Br}\right]^{+} (aq) + \mathrm{Br}^{-} (aq)\] Before any conductivity change, 0.125 L of 0.0100 M \(\mathrm{AgNO}_{3}\) solution is needed to titrate the free \(\mathrm{Br}^-\) ions in a 25.00 mL sample of the solution. After all conductivity changes have occurred, the same volume of \(\mathrm{AgNO}_{3}\) solution (0.125 L) is needed to titrate a fresh 25.00 mL sample of the complex.

Step by step solution

When the complex \(\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right]\mathrm{Br}\) is placed in water, initially, the complex dissolves and the conductivity increases due to the presence of \(\mathrm{Br}^-\) ions. Over time, the coordination sphere of cobalt(III) changes due to the replacement of one or more \(\mathrm{Br}^{-}\) ions by water molecules. This leads to an increase in conductivity as more \(\mathrm{Br}^{-}\) ions are released into the solution. The complex would undergo the reaction: \[\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right]\mathrm{Br} \rightarrow \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{H}_{2}\mathrm{O}\mathrm{Br}\right]^{+} + \mathrm{Br}^{-}\] #b) Write a balanced chemical equation to describe the reaction#

Based on the explanation for the change in conductivity, the balanced chemical equation for the reaction is: \[\left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}\right]\mathrm{Br} (s) \rightarrow \left[\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{H}_{2}\mathrm{O}\mathrm{Br}\right]^{+} (aq) + \mathrm{Br}^{-} (aq)\] #c) Calculate the volume of AgNO3 solution needed to precipitate the free Br-(aq)#

To calculate the volume of \(\mathrm{AgNO}_{3}\) solution needed, first determine the moles of free \(\mathrm{Br}^-\) ions in the 25.00 mL aliquot: 1. Calculate the moles of the complex in 500 mL of the 0.05 M solution: \[\text{moles of complex} = \text{M} \times \text{volume} = 0.05 \ M \times 0.5\ \text{L} = 0.025\ \text{moles}\] 2. Calculate the moles of the complex in the 25.00 mL aliquot: \[\text{moles of complex} = \frac{0.025\ \text{moles}}{500\ \text{mL}} \times 25\ \text{mL} = 0.00125\ \text{moles}\] Since there is one mole of free \(\text{Br}^-\) per mole of complex, there are 0.00125 moles of free \(\text{Br}^-\) ions in the 25.00 mL aliquot. 3. Finally, calculate the volume of 0.0100 M \(\mathrm{AgNO}_{3}\) solution needed to precipitate the free \(\mathrm{Br}^-\) ions: \[\text{volume} = \frac{\text{moles of } \text{Br}^-}{\text{M}} = \frac{0.00125\ \text{moles}}{0.0100\ \text{M}} = 0.125\ \text{L}\] #d) Calculate the volume of AgNO3 solution needed after all conductivity changes have occurred#

After all conductivity changes have occurred, one mole of \(\text{Br}^-\) will be released per mole of the complex: \[\text{moles of total } \text{Br}^- = 0.00125\ \text{moles}\] Calculate the volume of 0.0100 M \(\mathrm{AgNO}_{3}\) solution needed to precipitate both the free and released \(\mathrm{Br}^-\) ions: \[\text{volume} = \frac{\text{moles of total } \text{Br}^-}{\text{M}} = \frac{0.00125\ \text{moles}}{0.0100\ \text{M}} = 0.125\ \text{L}\] After all conductivity changes have occurred, the same volume of \(\mathrm{AgNO}_{3}\) solution (0.125 L) is needed to titrate a fresh 25.00 mL sample of the complex.