Water chemistry – computing pH from hydroxide ion concentration at 25 °C A wastewater sample contains [OH⁻] = 10^(-5.6) mmol/L at 25 °C. Calculate the pH of the sample. (Assume pure-water ionic product Kw = 1.0 × 10^(-14) at 25 °C and carefully convert units from mmol/L to mol/L.)

Introduction:Understanding how to move between hydroxide concentration, pOH, and pH is fundamental in water and wastewater chemistry. This item tests careful unit handling (mmol/L versus mol/L) and the relationship pH + pOH = 14 at 25 °C. Many quick errors arise from forgetting to convert milli- to base units before taking logarithms.

Given Data / Assumptions:

• Temperature: 25 °C (so Kw ≈ 1.0 × 10^(-14)).
• Hydroxide concentration reported as 10^(-5.6) mmol/L.
• Definition: pOH = −log10[OH⁻] when [OH⁻] is in mol/L; pH + pOH = 14.

Concept / Approach:

The key is unit conversion. The symbol “mmol/L” means millimoles per litre. Since 1 mmol = 10^(-3) mol, a concentration reported in mmol/L must be multiplied by 10^(-3) to convert to mol/L before applying logarithms. From there, compute pOH and finally pH using the 25 °C identity.

Step-by-Step Solution:

Verification / Alternative check:

As a sense check, [OH⁻] = 10^(−8.6) mol/L is far below neutral [OH⁻] of 10^(−7) mol/L; therefore pOH > 7 and pH < 7. The obtained pH = 5.4 is consistent with this qualitative expectation for an acidic solution.

Why Other Options Are Wrong:

8.6 and 8.4 treat 10^(−5.6) as mol/L, not mmol/L, yielding an alkaline pH. 5.6 confuses the exponent with pH. 7.0 assumes neutrality and ignores the given hydroxide level.

Common Pitfalls:

Forgetting to convert from mmol/L to mol/L; applying pH = −log[H⁺] directly to [OH⁻] without using Kw; using pH + pOH ≠ 14 at non-standard temperatures (here, 25 °C is given).

Final Answer:

5.4