Question 3 of 5: pH in Water Treatment; Carbonaceous BOD and Ultimate BOD

Question 3: pH in Water Treatment; Carbonaceous BOD and Ultimate BOD (25 marks: a 10, b 15)

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. pH: definition and significance (10 marks)

pH is the negative base-10 logarithm of the hydrogen-ion activity, $\text{pH}=-\log_{10}[H^+]$, a measure on a 0–14 scale of how acidic (pH < 7) or basic (pH > 7) water is, referenced to the ion product of water $K_w=[H^+][OH^-]=10^{-14}$ at 25 °C. For disinfection, pH controls the speciation of aqueous chlorine between hypochlorous acid ($HOCl$, $pK_a\approx7.5$) and hypochlorite ($OCl^-$): $HOCl$ is the far more effective disinfectant (roughly 80–100× more biocidal than $OCl^-$), so as pH rises above about 7.5 the equilibrium shifts toward the weaker $OCl^-$ form and a much larger free-chlorine residual is needed to achieve the same disinfection (CT) credit — utilities therefore disinfect at moderate pH (roughly 6.5–7.5) where practical. For coagulation–flocculation, pH governs the hydrolysis chemistry and surface charge of the metal coagulant: each coagulant (alum, ferric salts) has an optimum pH range (roughly 5.5–7 for alum) where it forms the insoluble, positively-charged hydroxide floc ($Al(OH)_3$, $Fe(OH)_3$) that neutralizes negatively-charged colloids and sweeps them out; outside that range the metal remains soluble (no floc forms) or re-dissolves as an anionic hydroxo-complex, so jar testing is always run across a pH range to locate the optimum before full-scale dosing.

b. Carbonaceous BOD (4-day) and ultimate BOD (15 marks)

Given. A raw sewage sample (3 mL) is diluted to 300 mL in a standard BOD bottle and incubated at 20 °C; the DO falls from 8.5 mg/L to 4.5 mg/L over 4 days, of which 5% of the observed depletion is attributed to the seed organisms already present in the sample (not the sewage itself).

Find. The carbonaceous BOD over the 4-day test ($\text{cBOD}_4$), the equivalent 5-day carbonaceous BOD ($\text{cBOD}_5$), and the ultimate carbonaceous BOD ($\text{BOD}_u$) of the undiluted sewage sample.

Approach. Scale the seed-corrected DO depletion by the dilution factor to get the 4-day BOD of the undiluted sample, then use the standard first-order BOD-exertion model to convert that single-duration reading to the 5-day and ultimate values.

1. Dilution fraction. $$P=\frac{V_\text{sample}}{V_\text{bottle}}=\frac{3}{300}=0.0100.$$
2. Seed-corrected 4-day depletion. The raw DO drop is $\Delta DO=8.5-4.5=4.0\ \text{mg/L}$; with 5% of that attributed to the seed, the sewage's own share is $$\Delta DO_\text{net}=4.0\times(1-0.05)=3.80\ \text{mg/L}.$$
3. 4-day cBOD of the undiluted sample. $$\text{cBOD}_4=\frac{\Delta DO_\text{net}}{P}=\frac{3.80}{0.0100}=\boxed{380\ \text{mg/L}}.$$
4. Ultimate BOD. The exam gives no deoxygenation rate constant, so a typical value for raw domestic sewage at 20 °C is assumed: $k_1=0.10\ \text{d}^{-1}$ (base-10; check assumption). The first-order model $\text{BOD}_t=\text{BOD}_u\left(1-10^{-k_1t}\right)$ at $t=4$ gives $$\text{BOD}_u=\frac{\text{cBOD}_4}{1-10^{-k_1(4)}}=\frac{380}{1-10^{-0.40}}=\frac{380}{0.6019}=\boxed{631.3\ \text{mg/L}}.$$
5. Equivalent 5-day cBOD. Evaluating the same first-order model at the standard $t=5\ \text{d}$, $$\text{cBOD}_5=\text{BOD}_u\left(1-10^{-k_1(5)}\right)=631.3\times\left(1-10^{-0.50}\right)=631.3\times0.6838=\boxed{431.7\ \text{mg/L}}.$$