Acid–base calculations – relation between pH and pOH For an aqueous solution at 25°C with pH = 8, determine the corresponding pOH value using the water ion-product relationship.

Introduction / Context:
pH and pOH are complementary measures of acidity and basicity in water. At 25°C, their sum is fixed by the ionic product of water. This quick relation is used constantly in environmental engineering, bioprocessing, and analytical chemistry.

Given Data / Assumptions:

• Temperature 25°C (298 K).
• pH = 8 given.
• Pure water reference where Kw = 1.0 * 10^-14 at 25°C.

Concept / Approach:
By definition, pH = -log10[H+] and pOH = -log10[OH-]. The ion product is Kw = [H+]*[OH-] = 1.0 * 10^-14 at 25°C, which leads to pH + pOH = 14. Hence knowing one immediately gives the other by subtraction.

Step-by-Step Solution:

Verification / Alternative check:
[H+] = 10^-8 M → [OH-] = Kw / [H+] = 10^-14 / 10^-8 = 10^-6 M → pOH = 6. Consistent.

Why Other Options Are Wrong:

• 1, 7, 10: Do not satisfy pH + pOH = 14 for pH = 8.

Common Pitfalls:
Using pH + pOH = 14 at temperatures far from 25°C without adjusting Kw; being careless with base-10 logarithms.

Final Answer:
6