Problem 120 - Chemistry The Central Science

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: $$2 \mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$

Step-by-Step Solution

Answer

(a) The pH of pure water at 50°C is given by: \[ \mathrm{pH} = -\log\left(\sqrt{5.48 \times 10^{-14}}\right) \] (b) Based on the increase in $K_w$ with temperature, we can predict that ΔH for the autoionization reaction of water is positive (ΔH > 0), as the reaction is endothermic and becomes more favorable at higher temperatures.

Step 1: Find the concentration of H3O+ and OH- in pure water

In pure water, the ion-product constant Kw is given by: \[K_w = [\mathrm{H}_{3}\mathrm{O}^{+}][\mathrm{OH}^{-}]\] Since the concentration of H3O+ and OH- ions in pure water is equal, we can write: \[ K_w = [\mathrm{H}_{3}\mathrm{O}^{+}]^2 \] Now, given that at 50°C the ion-product constant is \(K_w = 5.48 \times 10^{-14}\), we can calculate the concentration of H3O+ ions: \[ [\mathrm{H}_{3}\mathrm{O}^{+}]^2 = 5.48 \times 10^{-14} \]

Step 2: Calculate the pH of the pure water

Now that we have the equation for the concentration of H3O+ ions at 50°C, we can calculate the pH of pure water. The pH is defined as the negative logarithm of the concentration of H3O+ ions: \[ \mathrm{pH} = -\log([\mathrm{H}_{3}\mathrm{O}^{+}]) \] First, solve for the concentration of H3O+ ions: \[ [\mathrm{H}_{3}\mathrm{O}^{+}] = \sqrt{5.48 \times 10^{-14}} \] Next, calculate the pH using the concentration of H3O+ ions: \[ \mathrm{pH} = -\log\left(\sqrt{5.48 \times 10^{-14}}\right) \] This will give you the pH of pure water at 50°C.

Step 3: Analyze the relationship between temperature and Kw

In this step, we will analyze the relationship between temperature and the ion-product constant, Kw. In general, the ion-product constant for water increases as temperature increases. This implies that the autoionization of water becomes more favorable at higher temperatures.

Step 4: Predict the sign of ΔH for the autoionization reaction of water

The increase in Kw as the temperature increases suggests that the autoionization reaction of water is endothermic since the reaction becomes more favorable at higher temperatures. Therefore, we can predict that the change in enthalpy (ΔH) for the autoionization reaction of water is positive, as the reaction requires heat to proceed: \(2 \mathrm{H}_{2} \mathrm{O}(l) \Longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)\) ΔH > 0 for this reaction.

Key Concepts

pH of Pure Water

Understanding the pH of pure water is essential as it's a fundamental concept in chemistry related to the acidity or basicity of a solution. The pH scale ranges from 0 to 14, with 7 being neutral, below 7 acidic, and above 7 basic. Pure water is considered neutral with a pH close to 7 at room temperature.

However, the pH of pure water changes with temperature. At 50°C, pure water has a pH lower than 7 due to increased ionization. To find this pH value, we take the square root of Kw to find the concentration of H3O+, then apply the pH equation:
\[ \text{pH} = -\log([\mathrm{H}_{3}\mathrm{O}^{+}]) \]
Calculating this for the given Kw at 50°C, \[ \text{pH} = -\log(\sqrt{5.48 \times 10^{-14}}) \]reveals the pH value of pure water at this elevated temperature.

Autoionization of Water

Autoionization or self-ionization of water is a process where water molecules react with each other to produce hydronium (H3O+) and hydroxide (OH−) ions. This is an equilibrium reaction described by:\[ 2 \mathrm{H}_{2} \mathrm{O}(l) \Longleftrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q) + \mathrm{OH}^{-}(a q) \]In pure water, under standard conditions, the concentrations of H3O+ and OH− ions are equal, leading to the ion-product constant, Kw. This constant changes with temperature, reflecting the inherent dynamic nature of this equilibrium process in aqueous solutions.

Change in Enthalpy (ΔH)

The change in enthalpy, ΔH, measures the amount of heat absorbed or released in a reaction. It's a crucial factor in assessing a reaction's spontaneity and determining whether it's exothermic (releases heat) or endothermic (absorbs heat).

For autoionization of water, we look at the sign of ΔH to understand the energetics of the process. An increase in Kw with temperature suggests that more energy is required to break the O-H bonds in water molecules, indicating that ΔH is positive. This positive sign aligns with the reaction being endothermic, where:\[ \Delta H > 0 \]In other words, the reaction absorbs heat from the surroundings to proceed.

Temperature Dependence of Kw

Temperature has a notable impact on the ion-product constant of water (Kw). When temperature increases, Kw also increases, signifying a greater propensity for water to autoionize. This dependence is due to the endothermic nature of the autoionization reaction where heat is consumed to generate more H3O+ and OH− ions.

The rise in Kw with temperature implies a shift in the equilibrium toward a higher concentration of ions, and consequently, an altered pH of the solution. Understanding this relationship helps explain the variations in pH seen in thermal environments and the behavior of reactions where water acts as a solvent.