Question

\(\mathrm{PtCl}_{4} 6 \mathrm{H}_{2} \mathrm{O}\) can exist as a hydrated complex. \(1.0\) molal aqueous solution has depression in freezing point of \(3.72^{\circ} \mathrm{C}\). Assume \(100 \%\) ionization and \(K_{\ell}\) of water \(=1.86^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg}\). The complex is (a) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)\right] \mathrm{Cl}_{4}\) (b) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4} \mathrm{Cl}_{2}\right] \mathrm{Cl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (c) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3} \mathrm{Cl}_{3}\right] \mathrm{Cl} \cdot 3 \mathrm{H}_{2} \mathrm{O}\) (d) \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2} \mathrm{Cl}_{4}\right] \cdot 4 \mathrm{H}_{2} \mathrm{O}\)

Short answer

The complex is \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\mathrm{Cl}_{2}\right] \mathrm{Cl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (b).

Step-by-step solution

Determine the Molality of the Aqueous Solution

Given that the solution is 1.0 molal, it implies that there is 1.0 mole of the solute (hydrated complex) in 1 kilogram of water.

Determine the Van't Hoff Factor (i)

The depression in freezing point is given by the formula \( \Delta T_f = i \cdot K_f \cdot m \) where \( \Delta T_f \) is the depression in freezing point, \( i \) is the Van't Hoff factor, \( K_f \) is the cryoscopic constant, and \( m \) is the molality. We need to find \( i \) by rearranging the formula: \( i = \frac{\Delta T_f}{K_f \cdot m} \) Substituting the given values: \( i = \frac{3.72}{1.86 \cdot 1.0} = 2 \) Since we find \( i = 2 \) and we assume 100% ionization, it indicates that the complex dissociates into 2 ions in solution.

Determine the Correct Complex Formation

Given the Van't Hoff factor \( i = 2 \) and assuming full dissociation, we need to match this with the potential complexes. The correct complex must release 2 ions, one of which is the complex cation and the other is the counter ion. \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\mathrm{Cl}_{2}\right] \mathrm{Cl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) dissociates entirely into \(\left[\mathrm{Pt}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}\mathrm{Cl}_{2}\right]^2+ \) and 2 \(\mathrm{Cl}^-\), which aligns with the calculated Van't Hoff factor.

Key concepts

Van't Hoff factor

Understanding the Van't Hoff factor is crucial for solving any colligative properties problem. It is denoted by the symbol 'i' and represents the number of particles a compound produces when it dissolves in solution. In essence, it reflects the degree of ionization or dissociation of a solute. For instance, a non-electrolyte typically has a Van't Hoff factor of 1 because it doesn't dissociate into ions, while strong electrolytes can have an 'i' value greater than 1 depending on how many ions they break down into upon dissolution.

When a complex like \(\mathrm{PtCl}_{4} 6 \mathrm{H}_{2} \mathrm{O}\) dissociates, the 'i' value indicates not just the number of particles it breaks into, but also the extent to which it ionizes in the solution. This is essential to determine the compound's influence on the freezing point of the solvent, given that greater ionization means a larger effect on colligative properties.

Freezing point depression

Freezing point depression is one of the four colligative properties, which depend solely on the quantity and not the identity of solute particles. This particular property refers to the effect where the freezing point of a liquid (like water) is lowered by adding a solute, like a hydrated complex ion.

It is quantified using the equation \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) stands for the decrease in freezing point, 'i' is the Van't Hoff factor, \( K_f \) is the cryoscopic constant (a property of the solvent), and 'm' is the molality of the solution. Students often struggle with this concept because it requires recognizing how the interplay between the concentration of solute particles and their nature affects the overall solution properties. Hydrated complex ions add another layer of complexity, as they can release multiple particles into the solution, each contributing to the overall freezing point depression.

Molality

Molality is a measurement of a solution's concentration, specifically defining the amount of solute in moles per kilogram of solvent—this is different from molarity, which is moles per liter of solution. In the context of freezing point depression, molality is used because it doesn't change with temperature, unlike volume-based measurements.

When considering a specific example such as a \(1.0\) molal aqueous solution, it implies there's \(1.0\) mole of solute—here, the hydrated complex—per kilogram of water. Calculating molality can sometimes be tricky; however, it is a key component and must be accurately determined to correctly understand colligative effects like freezing point depression.

Hydrated complex ions

Hydrated complex ions are metal ions surrounded by water molecules, which can significantly influence the colligative properties of a solution. The number of water molecules associated with the metal ion can vary, leading to different degrees of hydration. These complexes are important in colligative properties problems because they dissociate in water, releasing ions that may further change the physical properties of the solution.

For example, when hydrating a metal ion like platinum, different complexes can form. In our exercise, the presence of \(6 \mathrm{H}_{2} \mathrm{O}\) molecules allows for various possible structures. It's only by using colligative property principles—such as the Van't Hoff factor—that we can deduce the correct form of the hydrated complex based on how it dissociates in the solution, which directly ties into its observed freezing point depression.