Aqueous equilibria — At room temperature, the ionic product of water is [H+][OH−] = 10^-14 (mol/L)^2. If a solution has [OH−] = 10^-6 mol/L, what will be the pH of the solution?

Introduction / Context:
pH and pOH are cornerstone ideas in aqueous chemistry and environmental engineering. At a given temperature (commonly 25°C, i.e., room temperature), the self-ionization of water imposes a fixed product [H+][OH−] known as the ionic product (Kw). This question checks your fluency in moving between hydroxide concentration, hydrogen ion concentration, and pH using Kw and logarithms, which is essential for process calculations in water treatment, biochemical reactions, and acid–base titrations.

Given Data / Assumptions:

• At room temperature: Kw = [H+][OH−] = 1.010^-14 (mol/L)^2.
• The solution has [OH−] = 1.010^-6 mol/L.
• Activities are approximated by concentrations (dilute solution assumption).

Concept / Approach:

The relationships are: pH = −log10[H+], pOH = −log10[OH−], and pH + pOH = 14 (at 25°C). Given [OH−], we can compute either [H+] via Kw or directly compute pOH and then pH. Both routes should agree. Ensure unit consistency (mol/L) and base-10 logarithms.

Step-by-Step Solution:

Verification / Alternative check:

Both methods yield the same pH = 8.00, confirming internal consistency. This also matches the expectation that [OH−] above 10^-7 mol/L leads to a basic solution (pH > 7) at 25°C.

Why Other Options Are Wrong:

6: Would imply [H+] = 10^-6 mol/L, which contradicts the given [OH−]. 10 or 12: These correspond to much larger [OH−] (10^-4 or 10^-2 mol/L). The “Not defined” choice is incorrect since Kw at room temperature is well-established.

Common Pitfalls:

Forgetting that pH + pOH = 14 only strictly holds at 25°C; using natural logs instead of base-10; confusing units; or assuming pH 7 is always neutral regardless of temperature (Kw varies with temperature).

Final Answer:

8