Question

Glauber's salt, \(\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O},\) was used by \(\mathrm{J} . \mathrm{R}\). Glauber in the \(17^{\text {th }}\) century as a medicinal agent. At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=4.08 \times 10^{-25}\) for the loss of waters of hydration from Glauber's salt: $$ \mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+10 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) What is the vapor pressure of water at \(25^{\circ} \mathrm{C}\) in a closed container holding a sample of \(\mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}(s) ?\) (b) How do the following changes affect the ratio (higher, lower, same) of hydrated form to anhydrous form for the system above? (1) Add more \(\mathrm{Na}_{2} \mathrm{SO}_{4}(s)\) (2) Reduce the container volume. (3) Add more water vapor. (4) Add \(\mathrm{N}_{2}\) gas.

Short answer

(a) \( P_{H_2O} = 2.56 \times 10^{-3} \) atm (b1) Same (b2) Same (b3) Higher (b4) Same

Step-by-step solution

Identify the equilibrium expression

The given equation for the loss of waters of hydration is: \[ \mathrm{Na}_{2} \mathrm{SO}_{4} \cdot 10 \mathrm{H}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+10 \mathrm{H}_{2} \mathrm{O}(g) \] From this, the equilibrium constant expression can be written as: \[ K_{\mathrm{p}} = P_{\mathrm{H}_{2} \mathrm{O}}^{10} \] Here, \( K_{\mathrm{p}} \) is given as \( 4.08 \times 10^{-25} \).

Calculate the vapor pressure (a)

We need to find the vapor pressure of water, \( P_{\mathrm{H}_{2} \mathrm{O}} \). Using the given \( K_{\mathrm{p}} \): \[ 4.08 \times 10^{-25} = (P_{\mathrm{H}_{2} \mathrm{O}})^{10} \] To solve for \( P_{\mathrm{H}_{2} \mathrm{O}} \), take the 10th root of both sides: \[ P_{\mathrm{H}_{2} \mathrm{O}} = (4.08 \times 10^{-25})^{\frac{1}{10}} \approx 2.56 \times 10^{-3} \text{ atm} \]

Analyze effect of adding more \( \mathrm{Na}_{2} \mathrm{SO}_{4}(s) \) (b1)

Adding more \( \mathrm{Na}_{2} \mathrm{SO}_{4}(s) \) does not affect the equilibrium position since it is a solid and its concentration is constant. Therefore, the ratio of hydrated to anhydrous form remains the same.

Analyze effect of reducing the container volume (b2)

Reducing the container volume increases the pressure of gases. However, because it does not change the partial pressure of water vapor, the ratio of hydrated to anhydrous form remains the same.

Analyze effect of adding more water vapor (b3)

Adding more water vapor increases \( P_{\mathrm{H}_{2} \mathrm{O}} \), shifting the equilibrium to the left according to Le Chatelier's principle. Therefore, the ratio of hydrated form to anhydrous form increases.

Analyze effect of adding \( \mathrm{N}_{2} \) gas (b4)

Adding \( \mathrm{N}_{2} \) gas increases the total pressure but does not change the partial pressure of water vapor directly. Thus, the ratio of hydrated to anhydrous form remains the same.

Key concepts

Vapor pressure

Vapor pressure refers to the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a certain temperature. Imagine a closed container with water. Some water molecules will escape to the air (evaporation), creating a vapor above the liquid. When the rate of evaporation equals the rate of condensation, the system reaches equilibrium, and the pressure exerted by the vapor is called the vapor pressure.

In the case of Glauber's salt (\text{Na}_{2} \text{SO}_{4} \bullet 10 \text{H}_{2} \text{O}), the focus is on the water vapor pressure in equilibrium with the hydrated salt. The vapor pressure depends on the K_p value given for the reaction:\[ \text{Na}_{2} \text{SO}_{4} \bullet 10 \text{H}_{2} \text{O}(s) \rightleftharpoons \text{Na}_{2} \text{SO}_{4}(s) + 10 \text{H}_{2} \text{O}(g) \]From the equilibrium expression K_{\text{p}} = P_{\text{H}_{2} \text{O}}^{10} , we can calculate the water vapor pressure by taking the 10th root of K_{\text{p}} . This gives us the vapor pressure of water at the specified temperature.

Equilibrium constant

The equilibrium constant (\text{K}) is a value that expresses the ratio of concentrations of products to reactants at equilibrium. For gases, we use the partial pressures, represented as K_{\text{p}} . Each reaction has a unique K_{\text{p}} value at a given temperature.

Change in K_{\text{p}} signifies a shift in equilibrium. For Glauber's salt equilibrium:\[ \text{Na}_{2} \text{SO}_{4} \bullet 10 \text{H}_{2} \text{O}(s) \rightleftharpoons \text{Na}_{2} \text{SO}_{4}(s) + 10 \text{H}_{2} \text{O}(g) \]The K_{\text{p}} = 4.08 \times 10^{-25} is extremely small, indicating very little water vaporizes at equilibrium. This low value of K_{\text{p}} supports the predominance of the hydrated salt over free water vapor in the system.

Le Chatelier's principle

Le Chatelier's principle explains how a system at equilibrium responds to external changes. If a system under equilibrium is disturbed, the system will adjust to counteract the disturbance and re-establish equilibrium.

For our Glauber's salt system:\[ \text{Na}_{2} \text{SO}_{4} \bullet 10 \text{H}_{2} \text{O}(s) \rightleftharpoons \text{Na}_{2} \text{SO}_{4}(s) + 10 \text{H}_{2} \text{O}(g) \]1. **Adding more water vapor** shifts equilibrium to the left, increasing the hydrated form.
2. **Adding more Na_{2} \text{SO}_{4} (s)** or N_{2} (g) does not change the equilibrium ratio as they don't affect the vapor's partial pressure.
3. **Reducing container volume** increases overall pressure but doesn't change P_{\text{H}_{2} \text{O}} , leaving the ratio unchanged.

Hydration and dehydration reactions

Hydration refers to the process where water molecules are added to a substance, often forming a hydrate. Dehydration is the removal of water molecules.

In the context of Glauber's salt ( \text{Na}_{2} \text{SO}_{4} \bullet 10 \text{H}_{2} \text{O}) equilibrium, the reactions show a balance between the hydrated (water-added) and anhydrous (water-removed) forms:\[ \text{Na}_{2} \text{SO}_{4} \bullet 10 \text{H}_{2} \text{O}(s) \rightleftharpoons \text{Na}_{2} \text{SO}_{4}(s) + 10 \text{H}_{2} \text{O}(g) \]- **Hydration reactions** take on added water, driving the formation of hydrated salt.
- **Dehydration reactions** release water molecules, creating anhydrous form.

The direction of these reactions depends on external factors like temperature and pressure, aligning with equilibrium concepts and Le Chatelier's principle.