Question 2 of 5: Disinfection Equilibrium, BOD, and Atmospheric Lapse Rate

Problem 2: Disinfection Equilibrium, BOD, and Atmospheric Lapse Rate (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Undissociated Fraction of Hypochlorous Acid

Given. HOCl ⇌ H⁺ + OCl⁻, pK_a = 7.54; dose = 15 mg/L HOCl; solution pH = 7.0 at 25°C.

Find. The percent of the added HOCl that remains undissociated at pH 7.0.

Approach. Combine the acid-dissociation equilibrium with the Henderson–Hasselbalch relation to get the undissociated fraction directly as a function of pH − pK_a, then apply it to the 15 mg/L dose.

1. Write the equilibrium and solve for the species ratio. $$K_a = \frac{[\text{H}^+][\text{OCl}^-]}{[\text{HOCl}]} \ \Rightarrow\ \frac{[\text{OCl}^-]}{[\text{HOCl}]} = \frac{K_a}{[\text{H}^+]} = 10^{\,\text{pH}-\text{p}K_a}$$ Substituting pH = 7.0 and pK_a = 7.54: $[\text{OCl}^-]/[\text{HOCl}] = 10^{-0.54} = 0.2884$.
2. Convert to the undissociated fraction. With total hypochlorite $C_T = [\text{HOCl}]+[\text{OCl}^-]$, $$f_{\text{HOCl}} = \frac{[\text{HOCl}]}{C_T} = \frac{1}{1 + 10^{\,\text{pH}-\text{p}K_a}} = \frac{1}{1+0.2884} = 0.7762$$ $$\%\ \text{undissociated} = \boxed{77.6\%}$$
3. Apply to the 15 mg/L dose. The undissociated (HOCl) concentration remaining at pH 7.0 is $$C_{\text{HOCl}} = f_{\text{HOCl}}\times 15\ \text{mg/L} = 0.7762\times 15 = \boxed{11.6\ \text{mg/L}}$$

(b) Meaning and Measurement of BOD

Biochemical Oxygen Demand (BOD) is the mass of dissolved oxygen consumed by micro-organisms as they aerobically stabilize (oxidize) the biodegradable organic matter in a water or wastewater sample, expressed in mg O₂/L. It is the standard surrogate measure of a discharge's oxygen-demanding strength — the same property responsible for the river oxygen-sag behaviour discussed in Problem 1(a) — and, unlike a direct organic-carbon assay, it measures demand in the same currency (dissolved oxygen) that the receiving water actually loses. It is measured by the standard 5-day, 20°C dilution test (BOD₅): a sample (diluted with aerated, nutrient- and seed-organism-amended dilution water so that oxygen is not exhausted before day 5) is incubated in a sealed BOD bottle in the dark for exactly 5 days at 20°C, and the dissolved oxygen is measured at the start (DO₀) and again after 5 days (DO₅). The BOD₅ of the diluted sample is DO₀ − DO₅, corrected for any oxygen consumed by the seed alone (a seed blank) and back-calculated through the dilution factor to the strength of the original sample. Because 5 days captures only a fraction of the ultimate (long-term) demand, BOD₅ is understood as a standardized, reproducible index rather than a complete accounting of oxidizable matter — the underlying first-order model, $\text{BOD}_t = L_0(1-e^{-k_1 t})$, is used to extrapolate BOD₅ to the ultimate BOD, L₀, when the full oxygen demand is needed for stream oxygen-sag modelling. Continuous-recording respirometers, which track dissolved oxygen uptake directly and avoid the dilution step, are an increasingly common alternative that give the full time-course of demand rather than a single 5-day endpoint.

(c)(i)–(ii) Atmospheric Lapse Rate and Stability

Given. T₁ = 20.2°C at z₁ = 4 m; T₂ = 19.1°C at z₂ = 224 m.

Find. (i) The existing (environmental) lapse rate; (ii) whether the atmosphere is stable, neutral, or unstable.

Approach. Compute the environmental lapse rate from the two measured points and compare it against the dry adiabatic lapse rate (DALR ≈ 0.98°C/100 m, the rate at which a rising, unsaturated parcel cools on its own).

1. Environmental lapse rate. $$\Gamma_{env} = \frac{T_1-T_2}{z_2-z_1} = \frac{20.2-19.1}{224-4}\ {}^{\circ}\text{C/m} = \frac{1.1}{220} = 0.0050\ {}^{\circ}\text{C/m}$$ $$\Gamma_{env} = \boxed{0.50\ {}^{\circ}\text{C}/100\ \text{m}\ (5.0\ {}^{\circ}\text{C/km})}$$
2. Compare against the dry adiabatic lapse rate. The DALR is $\Gamma_d = g/c_p \approx 0.98\ {}^{\circ}\text{C}/100\ \text{m}$. Since $\Gamma_{env}$ (0.50°C/100 m) < $\Gamma_d$ (0.98°C/100 m), a parcel displaced upward cools adiabatically faster than the surrounding air is already cooling, so it becomes cooler (denser) than its surroundings and sinks back toward its starting level.
3. Conclusion. The atmosphere is stable (sub-adiabatic environmental lapse rate) — vertical motion is suppressed and pollutants released near the surface disperse slowly.