The value of Kw at 40°C is 3.0 × 10^−14. What is the pH of pure water at 40°C? — Acids and Bases Chemistry Question

STEPS:

1. Understand the autoionization of water and $K_w$:
Pure water undergoes autoionization according to the equilibrium equation:
$$\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)$$
The equilibrium constant expression for this process, known as the ion-product constant of water ($K_w$), is:
$$K_w = [\text{H}^+][\text{OH}^-]$$

2. Establish ion equality in pure water:
In pure water, every single water molecule that dissociates produces exactly one hydronium/hydrogen ion ($\text{H}^+$) and one hydroxide ion ($\text{OH}^-$). Because there are no external acid or base solutes added, the concentration of hydrogen ions must be exactly equal to the concentration of hydroxide ions:
$$[\text{H}^+] = [\text{OH}^-]$$

3. Substitute into the $K_w$ equation to solve for $[\text{H}^+]$:
Substitute the equality $[\text{H}^+] = [\text{OH}^-]$ into the $K_w$ expression to obtain:
$$K_w = [\text{H}^+]^2$$
At $40^\circ\text{C}$, the problem states that $K_w = 3.0 \times 10^{-14}$. Setting these equal allows us to solve for $[\text{H}^+]$:
$$[\text{H}^+]^2 = 3.0 \times 10^{-14}$$
$$[\text{H}^+] = \sqrt{3.0 \times 10^{-14}} = \sqrt{3.0} \times \sqrt{10^{-14}} = \sqrt{3.0} \times 10^{-7}\text{ M}$$
Without a calculator, estimate the value of $\sqrt{3.0}$. Because $1^2 = 1$ and $2^2 = 4$, the square root of $3.0$ must lie between $1$ and $2$ (specifically, it is $\approx 1.73$). Thus:
$$[\text{H}^+] \approx 1.7 \times 10^{-7}\text{ M}$$

4. Calculate or estimate the pH:
The formula for pH is:
$$\text{pH} = -\log[\text{H}^+]$$
Substitute the estimated hydrogen ion concentration:
$$\text{pH} = -\log(1.7 \times 10^{-7}) = 7 - \log(1.7)$$
Because $\log(1) = 0$ and $\log(10) = 1$, the value of $\log(1.7)$ is a very small positive decimal (specifically, $\approx 0.23$). Subtracting this small decimal from $7$ yields:
$$\text{pH} \approx 6.8$$
This identifies Option B as the correct choice.

5. Double-check using qualitative thermodynamic reasoning:
The autoionization of water is an endothermic process. As the temperature is raised from $25^\circ\text{C}$ to $40^\circ\text{C}$, the equilibrium shifts to the right to absorb the added heat (in accordance with Le Châtelier's principle), which increases the value of $K_w$ from $1.0 \times 10^{-14}$ to $3.0 \times 10^{-14}$. Because water ionizes to a greater extent at this higher temperature, the concentration of $\text{H}^+$ in pure water increases, which means the pH must be lower than 7.0. The only option slightly below $7.0$ is 6.8 (Option B).

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WHY_OTHERS_WRONG:

• Option A is incorrect (3.0): This is a distractor for students who confuse the exponent math or mistakenly assume that the coefficient "3.0" in $3.0 \times 10^{-14}$ directly dictates the pH value. A pH of $3.0$ corresponds to an extremely acidic solution ($[\text{H}^+] = 1.0 \times 10^{-3}\text{ M}$), which is chemically impossible for pure neutral water.
• Option C is incorrect (7.0): This is the pH of pure water at $25^\circ\text{C}$, where $K_w = 1.0 \times 10^{-14}$. Because $K_w$ is temperature-dependent, the autoionization of water shifts with temperature, meaning pure water does not maintain a pH of $7.0$ at temperatures other than $25^\circ\text{C}$.
• Option D is incorrect (7.2): This value is greater than $7.0$. Because water autoionizes more at elevated temperatures, the concentration of $\text{H}^+$ ions is higher than at $25^\circ\text{C}$, which requires the pH of the pure neutral water to drop below $7.0$, not rise above it.