Question

A sample containing 0.0500 mole of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is dissolved in enough water to make 1.00 L of solution. This solution contains hydrated \(\mathrm{SO}_{4}^{2-}\) and \(\mathrm{Fe}^{3+}\) ions. The latter behaves as an acid: $$\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(a q) \rightleftharpoons \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{OH}^{2+}(a q)+\mathrm{H}^{+}(a q)$$ a. Calculate the expected osmotic pressure of this solution at \(25^{\circ} \mathrm{C}\) if the above dissociation is negligible. b. The actual osmotic pressure of the solution is 6.73 atm at \(25^{\circ} \mathrm{C} .\) Calculate \(K_{\mathrm{a}}\) for the dissociation reaction of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+} .\) (To do this calculation, you must assume that none of the ions go through the semipermeable membrane. Actually, this is not a great assumption for the tiny \(\mathrm{H}^{+}\) ion.)

Short answer

The expected osmotic pressure of the solution without dissociation is 6.13 atm. However, the actual osmotic pressure is 6.73 atm, indicating that some dissociation occurs. By taking the change in osmotic pressure into account, we can calculate the dissociation constant \(K_\mathrm{a}\) for the reaction of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) which is found to be \(1.20 \times 10^{-4}\).

Step-by-step solution

Calculate the osmotic pressure without dissociation

Without dissociation, the \(1.0 \ \text{L}\) solution contains \(0.0500\) mole of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). Each formula unit of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) can dissociate in water to form \(2\ \mathrm{Fe}^{3+}\) ions and \(3\ \mathrm{SO}_{4}^{2-}\) ions. Therefore, \(0.0500\) mole of \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is equivalent to a total of \(2(0.0500) + 3(0.0500) = 0.250\) mole of ions. The osmotic pressure \(\Pi\) can be calculated using the van't Hoff equation: \[\Pi = n_\text{ions} \frac{RT}{V}\] Where \(n_\text{ions}\) is the moles of ions in the solution, \(R = 0.0821 \frac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}}\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(V\) is the volume of the solution in liters. At \(25^{\circ} \mathrm{C}\), the temperature is \(T = 25 + 273.15 = 298.15 \ \text{K}\). The osmotic pressure can be calculated as follows: \[\Pi = \frac{(0.250\ \text{mol})(0.0821 \frac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}})(298.15 \ \text{K})}{1.0\ \text{L}}\]

Calculate the expected osmotic pressure

Solve the equation from step 1 to find the osmotic pressure without dissociation: \[\Pi = 6.13\ \text{atm}\]

Calculate change in osmotic pressure due to dissociation

The actual osmotic pressure of the solution is given as \(6.73\ \text{atm}\). This is higher than the expected osmotic pressure calculated in step 2, indicating that some of the \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) ions have dissociated, releasing \(\mathrm{H}^{+}\) ions and increasing the total number of ions in the solution. The difference in osmotic pressures is: \[\Delta \Pi = 6.73\ \text{atm} - 6.13\ \text{atm} = 0.60\ \text{atm}\]

Calculate the concentration of dissociated ions to find \(K_\mathrm{a}\)

Since the increase in osmotic pressure is due to the dissociation of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}(aq)\) into \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\mathrm{OH}^{2+}(aq)\) and \(\mathrm{H}^{+}(aq)\), we can calculate the concentration of these ions using the van't Hoff equation. For the dissociated ions, we get: \[\Delta \Pi = \frac{n_{\mathrm{H}^+}RT}{V}\] Solving for the moles of dissociated \(\mathrm{H}^{+}\) ions: \[n_{\mathrm{H}^{+}} = \frac{\Delta \Pi \cdot V}{RT} = \frac{(0.60\ \text{atm})(1.0\ \text{L})}{(0.0821 \ \frac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}})(298.15\ \text{K})} = 0.00245\ \text{mol}\] Now we know that the total moles of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(0.0500\ \text{mol}\) and the moles of dissociated ions are \(0.00245\ \text{mol}\). The concentration \([\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}] = (0.0500 - 0.00245) \ \text{M}\), \([\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\mathrm{OH}^{2+}] = [\mathrm{H}^{+}] = 0.00245 \ \text{M}\). The dissociation constant \(K_\mathrm{a}\) can be calculated as: \[K_\mathrm{a} = \frac{[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}\mathrm{OH}^{2+}][\mathrm{H}^{+}]}{[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}]}\]

Calculate the dissociation constant, \(K_\mathrm{a}\)

Substitute the concentrations into the equation from step 4 to find \(K_\mathrm{a}\): \[K_\mathrm{a} = \frac{(0.00245 \ \text{M})(0.00245 \ \text{M})}{(0.0500 - 0.00245) \ \text{M}} = 1.20 \times 10^{-4}\] The dissociation constant, \(K_\mathrm{a}\), for the dissociation reaction of \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is \(1.20 \times 10^{-4}\).

Key concepts

Dissociation Constant

When a compound like \( \mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6^{3+} \) dissolves in water, it can dissociate into ions. The extent of this dissociation is characterized by the dissociation constant, \( K_a \). This constant indicates how much of the original compound breaks apart in a solution.
The dissociation constant is crucial in understanding the balance between the intact molecules and their dissociated ions:

• **Higher \( K_a \) values:** More dissociation, meaning stronger acid.
• **Lower \( K_a \) values:** Less dissociation, suggesting a weaker acid.

In our context, \( K_a \) for the dissociation of \( \mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6^{3+} \) is calculated using concentrations of the ions in the equation: \[K_a = \frac{[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_5\mathrm{OH}^{2+}][\mathrm{H}^+]}{[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6^{3+}]}\] This provides valuable insight into the acidity and reactivity of the solution.

van't Hoff Equation

The van't Hoff equation is essential in calculating osmotic pressure, an important property in understanding how solutions behave. Osmotic pressure, denoted by \( \Pi \), is the pressure required to stop the flow of solvent into a solution through a semipermeable membrane. It's calculated using the formula: \[\Pi = n_{\text{ions}} \frac{RT}{V}\]

• **\( n_{\text{ions}} \):** Number of moles of ions in the solution.
• **\( R \):** Ideal gas constant \((0.0821 \frac{\text{atm}\cdot\text{L}}{\text{mol}\cdot\text{K}})\).
• **\( T \):** Temperature in Kelvin.
• **\( V \):** Volume of the solution in liters.

For the problem, we calculated the osmotic pressure assuming no dissociation to find a baseline of \(6.13 \text{ atm} \). The actual pressure deviates due to partial dissociation, revealing more ions present in the solution.

Ion Concentration

Ion concentration is a vital aspect of solution chemistry, influencing properties like osmotic pressure and pH. When substances like \( \mathrm{Fe}_2(\mathrm{SO}_4)_3 \) dissolve in water, they can produce multiple ions. For example, a single formula unit can create two \( \mathrm{Fe}^{3+} \) ions and three \( \mathrm{SO}_4^{2-} \) ions, impacting the solution's characteristics.
To calculate ion concentrations:

• Determine the number of moles of compound dissolved.
• Consider the number of ions produced per mole.

In the exercise, the dissociation of \( \mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6^{3+} \) led to additional \( \mathrm{H}^+ \) ions, changing the overall ion concentration and resulting in a higher osmotic pressure than initially calculated. Understanding ion concentration helps in predicting and controlling the behavior of solutions in various contexts, from industrial applications to biological systems.