Which of the following accounts for the observation that the pH of pure water at 37°C is 6.8 ? — Acids and Bases Chemistry Question

STEPS:

1. Understand the definition of pH and neutrality: The pH of a solution is defined by the mathematical relationship $\text{pH} = -\log[\text{H}^+]$. A solution is chemically *neutral* if and only if the concentration of hydronium ions equals the concentration of hydroxide ions ($[\text{H}^+] = [\text{OH}^-]$). At the standard reference temperature of $25^\circ\text{C}$, pure water has $[\text{H}^+] = 1.0 \times 10^{-7}\text{ M}$, which yields a neutral $\text{pH} = 7.0$.
2. Analyze the chemical equation for water autoionization: Water molecules naturally undergo self-ionization to a very small extent:
$$\text{H}_2\text{O}(l) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{OH}^-(aq) \quad \Delta H > 0$$
This chemical process is endothermic ($\Delta H > 0$) because it requires a substantial input of energy to break strong covalent $\text{O}-\text{H}$ bonds in order to separate the neutral water molecules into charged ions.
3. Apply Le Chatelier's Principle to temperature changes: According to Le Chatelier's Principle, if a system at equilibrium is subjected to an increase in temperature, the equilibrium will shift to counteract that disturbance by favoring the endothermic (forward) direction to absorb the excess thermal energy.
4. Determine the effect on the equilibrium constant ($K_w$): Because the autoionization of water is endothermic, raising the temperature from $25^\circ\text{C}$ to $37^\circ\text{C}$ drives the equilibrium forward, causing water to dissociate to a greater extent. This mathematically increases the value of the autoionization constant, $K_w = [\text{H}^+][\text{OH}^-]$ (from $1.0 \times 10^{-14}$ at $25^\circ\text{C}$ to approximately $2.4 \times 10^{-14}$ at $37^\circ\text{C}$).
5. Relate the larger $K_w$ to the lower pH of 6.8:
* Since water always autoionizes in a $1:1$ stoichiometric ratio, the concentrations of hydronium and hydroxide remain equal:
$$[\text{H}^+] = [\text{OH}^-] = \sqrt{K_w}$$
* At $37^\circ\text{C}$, because $K_w$ is larger, the concentration of $[\text{H}^+]$ must also be larger:
$$[\text{H}^+] = \sqrt{2.4 \times 10^{-14}} \approx 1.55 \times 10^{-7}\text{ M} \quad (\text{which is } > 1.0 \times 10^{-7}\text{ M})$$
* Because $[\text{H}^+]$ has increased, the negative log calculation results in a lower pH:
$$\text{pH} = -\log(1.55 \times 10^{-7}) \approx \mathbf{6.8}$$
Therefore, Option B is the correct explanation.

*

WHY_OTHERS_WRONG:

• Option A is incorrect: Although the pH of pure water at $37^\circ\text{C}$ is $6.8$, the water is not acidic. Acidity requires $[\text{H}^+] > [\text{OH}^-]$. Because pure water always dissociates into equal concentrations of $[\text{H}^+]$ and $[\text{OH}^-]$, the sample remains chemically neutral despite having a lower pH value.
• Option C is incorrect: While the density of liquid water does decrease slightly when heated from $25^\circ\text{C}$ to $37^\circ\text{C}$, this minor physical expansion of volume is far too small to account for the increase in hydronium concentration. The shift in $[\text{H}^+]$ is governed entirely by the chemical autoionization equilibrium shift, not density.
• Option D is incorrect: This option claims that water ionizes to a *lesser* extent at $37^\circ\text{C}$. Because autoionization is an endothermic process, increasing the temperature drives the reaction forward, meaning water ionizes to a greater extent at higher temperatures.