2 H2O(l) ⇄ H3O+(aq) + OH-(aq) The autoionization of water is represented by the equation above. Valu — Equilibrium Chemistry Question

STEPS:

1. Analyze the relationship between $p\text{K}_w$ and the equilibrium constant ($\text{K}_w$): The water autoionization constant is related to $p\text{K}_w$ by the logarithmic definition $p\text{K}_w = -\log \text{K}_w$, which can be rewritten as $\text{K}_w = 10^{-p\text{K}_w}$. Because of the negative sign in the definition, a smaller $p\text{K}_w$ value corresponds to a larger equilibrium constant, $\text{K}_w$.
2. Observe the trend in the provided data table: As the temperature increases from $0^\circ\text{C}$ to $40^\circ\text{C}$, the value of $p\text{K}_w$ steadily decreases from $14.9$ to $13.5$. This indicates that as temperature increases, $\text{K}_w$ increases (from $10^{-14.9} \approx 1.26 \times 10^{-15}$ at $0^\circ\text{C}$ to $10^{-13.5} \approx 3.16 \times 10^{-14}$ at $40^\circ\text{C}$).
3. Determine the enthalpy of the reaction: Since the equilibrium constant $\text{K}_w$ increases as the temperature increases, the forward reaction (autoionization/dissociation of water) is favored at higher temperatures. According to Le Chatelier's Principle, an increase in temperature shifts a reaction in the endothermic direction to absorb the added thermal energy. Therefore, the dissociation of water is an endothermic process ($\Delta H > 0$), which disproves Option A.
4. Relate $\text{K}_w$ to the ion concentrations in pure water: Pure water must be neutral, which means that the concentration of hydronium ions ($\text{H}_3\text{O}^+$) is stoichiometrically equal to the concentration of hydroxide ions ($\text{OH}^-$):
$$[\text{H}_3\text{O}^+] = [\text{OH}^-] \quad$$
Substituting this neutrality condition into the equilibrium expression for $\text{K}_w$ gives:
$$\text{K}_w = [\text{H}_3\text{O}^+][\text{OH}^-] = [\text{H}_3\text{O}^+]^2 \quad$$
Taking the square root of both sides, we find the concentration of hydronium ions in pure water:
$$[\text{H}_3\text{O}^+] = \sqrt{\text{K}_w} = \left(10^{-p\text{K}_w}\right)^{1/2} = 10^{-p\text{K}_w / 2} \quad$$
5. Formulate the pH of pure water as a function of $p\text{K}_w$: Taking the negative logarithm of the hydronium ion concentration gives the pH:
$$\text{pH} = -\log [\text{H}_3\text{O}^+] = -\log\left(10^{-p\text{K}_w / 2}\right) = \frac{p\text{K}_w}{2} \quad$$
6. Calculate the pH of pure water at different temperatures to identify the correct trend:
* At $0^\circ\text{C}$: $\text{pH} = 14.9 / 2 = \mathbf{7.45}$
* At $20^\circ\text{C}$: $\text{pH} = 14.2 / 2 = \mathbf{7.10}$
* At $40^\circ\text{C}$: $\text{pH} = 13.5 / 2 = \mathbf{6.75}$
Thus, as the temperature increases, the pH of pure water decreases, which confirms Option D is the correct statement.

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WHY_OTHERS_WRONG:

• A is incorrect: The dissociation of water is an endothermic process, not exothermic. Because $p\text{K}_w$ decreases as temperature increases, the equilibrium constant $\text{K}_w$ increases, meaning that higher temperatures favor the forward, ion-producing reaction.
• B is incorrect: The pH of pure water is only 7.00 at $25^\circ\text{C}$, where $\text{K}_w = 1.0 \times 10^{-14}$ and $p\text{K}_w = 14.0$. At any other temperature, the pH of pure water deviates from 7.00 (such as $7.45$ at $0^\circ\text{C}$ or $6.75$ at $40^\circ\text{C}$) due to the temperature dependence of $\text{K}_w$. However, the water remains neutral at all temperatures because $[\text{H}_3\text{O}^+]$ remains stoichiometrically equal to $[\text{OH}^-]$.
• C is incorrect: As temperature increases, the concentration of $\text{H}_3\text{O}^+$ increases, which mathematically causes the pH to decrease, not increase.