Problem 120 - Chemistry The Central Science In Si Units

Question

At \(50^{\circ} \mathrm{C}\), the ion-product constant for \(\mathrm{H}_{2} \mathrm{O}\) has the value \(K_{w}=5.48 \times 10^{-14}\). (a) What is the pH of pure water at \(50^{\circ} \mathrm{C} ?\) (b) Based on the change in \(K_{w}\) with temperature, predict whether \(\Delta H\) is positive, negative, or zero for the autoionization reaction of water: $$ 2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) $$

Step-by-Step Solution

Answer

(a) The pH of pure water at \(50^{\circ} \text{C}\) is approximately 6.63. (b) \( \Delta H \) is positive because \( K_w \) increases with temperature.

Step 1: Understand the ion-product constant

The ion-product constant of water, denoted as \( K_w \), is the equilibrium constant for the autoionization reaction of water: \[ 2 ext{H}_2 ext{O} (l) \rightleftharpoons ext{H}_3 ext{O}^+ (aq) + ext{OH}^- (aq) \]At any specific temperature, \( K_w = [ ext{H}_3 ext{O}^+][ ext{OH}^-] \). Given \( K_w = 5.48 \times 10^{-14} \) at \( 50^{\circ} \text{C} \), this is the product of the concentrations of \( ext{H}_3 ext{O}^+ \) and \( ext{OH}^- \) ions in pure water.

Step 2: Calculate [H3O+] and [OH−]

In pure water, the concentration of hydronium ions \([\text{H}_3\text{O}^+]\) is equal to the concentration of hydroxide ions \([\text{OH}^-]\). Thus, we can set \( [\text{H}_3\text{O}^+] = [\text{OH}^-] = x \). From the equation \( K_w = x^2 \), we solve for \( x \):\[ x^2 = 5.48 \times 10^{-14} \] \[ x = \sqrt{5.48 \times 10^{-14}} \] \[ x \approx 2.34 \times 10^{-7} \text{ M} \] Thus, \([\text{H}_3\text{O}^+]\) is approximately \(2.34 \times 10^{-7} \text{ M}\).

Step 3: Calculate pH of pure water

The pH is calculated using the formula:\[ \text{pH} = -\log [\text{H}_3\text{O}^+] \] Substituting for \([\text{H}_3\text{O}^+]\):\[ \text{pH} = -\log(2.34 \times 10^{-7}) \]\[ \text{pH} \approx 6.63 \] Therefore, the pH of pure water at \(50^{\circ} \text{C}\) is approximately 6.63.

Step 4: Discuss the effect of temperature on ΔH

Since \( K_w \) increases with an increase in temperature (from \( 25^{\circ} \text{C} \) \( K_w \approx 1.0 \times 10^{-14} \) to \( K_w = 5.48 \times 10^{-14} \) at \( 50^{\circ} \text{C} \)), the autoionization reaction of water is endothermic. According to Le Chatelier's principle, an increase in temperature favors the endothermic reaction, causing \( \Delta H \) to be positive. Thus, \( \Delta H \) for the reaction is positive.

Key Concepts

pH calculation

To find the pH of pure water at a given temperature, we rely on the concept of the ion-product constant for water, known as \( K_w \). The equation \( K_w = [\text{H}_3\text{O}^+][\text{OH}^-] \) describes this balance, where \([\text{H}_3\text{O}^+]\) is the concentration of hydronium ions and \([\text{OH}^-]\) is the concentration of hydroxide ions. For pure water in equilibrium, the concentrations of both ions are equal. This means we can solve for their concentrations using \( K_w \) which is provided as \( 5.48 \times 10^{-14} \) at \( 50^{\circ} \text{C}\). The solution simplifies to: \( x^2 = 5.48 \times 10^{-14} \), where \( x \) is both \([\text{H}_3\text{O}^+]\) and \([\text{OH}^-]\).

Using the square root, \( x \approx 2.34 \times 10^{-7} \), we find the concentration of hydronium ions. pH is then determined by the formula \( \text{pH} = -\log [\text{H}_3\text{O}^+] \), leading to \( \text{pH} \approx 6.63 \). At this temperature, water is slightly more acidic than at room temperature, where pH would be nearly neutral or 7.

• Remember, pH indicates the acidity or basicity of a solution.
• Even a small change in \( K_w \) due to temperature significantly affects the pH.

temperature effect on equilibrium

The ion-product constant of water, \( K_w \), changes with temperature. This variance is due to the nature of chemical equilibrium, particularly for reactions like the autoionization of water. At \( 25^{\circ} \text{C} \), \( K_w \) is about \( 1.0 \times 10^{-14} \), but it increases to \( 5.48 \times 10^{-14} \) as temperature reaches \( 50^{\circ} \text{C} \). Such an increase suggests that more products (\([\text{H}_3\text{O}^+]\) and \([\text{OH}^-]\)) are formed at higher temperatures.

When equilibrium is shifted by temperature changes, it naturally points towards a reaction being either endothermic or exothermic. For water's autoionization reaction, the increased \( K_w \) with temperature suggests the reaction is endothermic. Le Chatelier's Principle supports this, indicating that an increase in temperature favors the forward reaction, leading to more dissociation of water molecules into ions.

• An increase in temperature leads to a greater \( K_w \) value.
• The shift towards endothermic reactions with higher temperatures reflects the bonds in water absorbing heat.

endothermic reaction

Understanding the nature of the autoionization of water as endothermic clarifies why such reactions favor product formation at higher temperatures. In an endothermic reaction, heat is absorbed as a necessary condition to drive the reactants toward the products. Thus, increasing temperature adds energy to the system, making it easier for the water molecules to dissociate into \(\text{H}_3\text{O}^+\) and \(\text{OH}^-\). This is characterized by a positive enthalpy change, \( \Delta H > 0 \).

The equation in question is: \[ 2 \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq) \] In this, the absorption of heat permits an increased level of ionization of water molecules. This absorption translates to increased \( K_w \) with rising temperatures, telling us that with each degree of temperature increase, more energy is embedded in water molecules to enhance dissociation.

• Endothermic reactions are characterized by heat absorption throughout the bond-breaking process.
• A positive change in enthalpy is indicative of these reactions as energy input is crucial.