Question 1 of 7: Problem 1 — Material balance, reaction kinetics and microbiology

Question 1: Problem 1 — Material balance, reaction kinetics and microbiology (20 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.

Part (i) — Closed-system Henry's law partitioning (6 marks)

Given. A sealed 50 m³ tank holds equal volumes of water and air at 25 °C, into which a fixed quantity of glycol is charged; the glycol distributes itself between the two phases until Henry's law is satisfied.

Find. The equilibrium aqueous concentration $C$ (mol/L and mg/L) and the equilibrium partial pressure $p$ (atm) of the glycol in the head space.

Approach. Write a closed-system mole balance in which the single unknown is the partial pressure, expressing the moles held by each phase as that phase's capacity (mol/atm) multiplied by $p$, then solve one linear equation.

1. State the two equilibrium relations. Henry's law in the concentration-over-pressure form given by the question fixes the aqueous concentration, and the ideal gas law fixes the gas-phase concentration: $$C = K_H\,p \qquad\text{and}\qquad C_{\text{air}} = \frac{n_{\text{air}}}{V_a} = \frac{p}{RT}$$ Both phases are described by the same unknown $p$, which is what makes the problem a single equation rather than a system.
2. Impose conservation of mass on the sealed tank. Nothing enters or leaves after sealing, so the charge must be accounted for entirely by the two phases: $$n_T = n_{\text{aq}} + n_{\text{air}} = C\,V_w + \frac{p\,V_a}{RT} = p\left(K_H V_w + \frac{V_a}{RT}\right)$$ The bracket is the total capacity of the tank in moles per atmosphere.
3. Evaluate the two capacities separately. Working in litres, because $K_H$ is quoted per litre: $$K_H V_w = 156 \times 25\,000 = 3.900\times10^{6}\ \text{mol/atm}$$ $$\frac{V_a}{RT} = \frac{25\,000}{0.08206 \times 298.15} = \frac{25\,000}{24.466} = 1021.9\ \text{mol/atm}$$ The water can hold about 3800 times as much glycol per unit of driving pressure as the air can, so essentially the whole charge will dissolve. This ratio is worth forming first: it predicts the answer before the division is done.
4. Solve for the equilibrium partial pressure. $$p = \frac{n_T}{K_H V_w + V_a/RT} = \frac{500}{3.900\times10^{6} + 1021.9} = \boxed{1.28\times10^{-4}\ \text{atm}}$$ That is about 13 Pa, or roughly 0.13 ppm_v of the sealed atmosphere.
5. Back-substitute for the aqueous concentration. $$C = K_H\,p = 156 \times 1.2817\times10^{-4} = \boxed{2.00\times10^{-2}\ \text{mol/L}}$$ Converting with the molar mass of C₃H₈O₂, $M = 3(12.011)+8(1.008)+2(15.999) = 76.09$ g/mol: $$C = 0.02000 \times 76.09 \times 1000 = 1521\ \text{mg/L} \approx 1.52\ \text{g/L}$$
6. Close the mole balance as a check. The dissolved inventory is $0.02000 \times 25\,000 = 499.87$ mol and the head space holds $1.2817\times10^{-4} \times 1021.9 = 0.131$ mol. Their sum is 500.00 mol, recovering the charge exactly, and 99.97 % of the spill has partitioned to the water.

The engineering message is that a highly water-soluble spill of this class is almost invisible to head-space monitoring: an instrument sampling the tank atmosphere sees a fraction of a part per million while the liquor beneath it carries one and a half grams per litre. Emergency response to a glycol release must therefore sample the liquid, and the tank contents must be handled as a high-strength organic waste — 500 mol of propylene glycol exerts a theoretical oxygen demand of roughly 60 kg, which would overwhelm a small receiving works if it were simply drained.

Part (ii) — Total nitrogen as N (7 marks)

Given. A nitrogen speciation report expresses each species as the whole ion or molecule rather than as elemental nitrogen, so each result must be converted before the species can be summed.

Find. The total nitrogen concentration expressed as N, in mg/L.

Approach. Multiply each reported concentration by the mass fraction of nitrogen in the reported species, then add; organic-N is already on the N basis and passes through unchanged.

1. Form the molar masses from the given atomic weights. $$M_{\text{NH}_3} = 14 + 3(1) = 17 \qquad M_{\text{NO}_2^-} = 14 + 2(16) = 46 \qquad M_{\text{NO}_3^-} = 14 + 3(16) = 62\ \text{g/mol}$$ Each contains exactly one nitrogen atom, so the conversion factor to the N basis is simply $14/M$.
2. Convert the ammonia. The question reports the ammonia as NH₃, so the factor is 14/17 and not the 14/18 that would apply to a result reported as the ammonium ion: $$C_{\text{NH}_3\text{-N}} = 10 \times \frac{14}{17} = 8.24\ \text{mg/L as N}$$
3. Convert the oxidised species. $$C_{\text{NO}_2\text{-N}} = 3 \times \frac{14}{46} = 0.91\ \text{mg/L as N}$$ $$C_{\text{NO}_3\text{-N}} = 15 \times \frac{14}{62} = 3.39\ \text{mg/L as N}$$ Notice how heavily the oxygen atoms dilute the nitrate result: 15 mg/L of nitrate ion is worth only 3.4 mg/L of nitrogen.
4. Sum to total nitrogen. Organic-N is already reported as N, so it enters at its face value: $$\text{TN} = 8.235 + 0.913 + 3.387 + 5.000 = \boxed{17.5\ \text{mg/L as N}}$$
5. Report the analytical sub-total as well. Total Kjeldahl nitrogen is the reduced fraction that a Kjeldahl digestion actually measures — organic plus ammonia nitrogen: $$\text{TKN} = 8.235 + 5.000 = 13.2\ \text{mg/L as N}$$ and $\text{TN} = \text{TKN} + \text{NO}_2\text{-N} + \text{NO}_3\text{-N}$ recovers 17.5 mg/L, confirming the arithmetic by an independent route.

The speciation itself is diagnostic. Ammonia carries 47 % of the nitrogen and organic-N a further 29 %, while nitrite plus nitrate together account for less than a quarter. A ground water dominated by reduced nitrogen of this kind points to a recent, poorly nitrified source such as leaking sanitary sewer, septic-field breakthrough or manure storage, rather than to the aged agricultural nitrate plume that would show the opposite split. It also matters for compliance: the GCDWQ maximum acceptable concentration for nitrate is 45 mg/L as NO₃⁻ (10 mg/L as N), so at 15 mg/L as NO₃⁻ this water is well inside the nitrate limit but would still require treatment for ammonia before distribution, because ammonia consumes free chlorine and destroys the disinfectant residual.

Part (iii) — MPN and CFU enumeration of fecal indicator bacteria (7 marks)

The colony-forming unit (CFU) method is a direct-count, membrane-filtration or spread-plate technique. A measured volume of sample — typically 100 mL for a drinking-water or recreational-water compliance test — is drawn through a 0.45 µm membrane filter that retains the bacteria, and the filter is transferred to a selective differential agar such as m-Endo or m-FC. After incubation at the diagnostic temperature (35 °C for total coliforms, 44.5 °C for thermotolerant E. coli), each retained viable cell that is able to grow on that medium produces one visible colony, and the analyst counts the colonies directly, as in the plate shown with the question. The result is reported as CFU per 100 mL, obtained by dividing the colony count by the filtered volume; countable plates are conventionally restricted to roughly 20–80 colonies so that colonies neither merge nor become statistically sparse.

The most probable number (MPN) method is an indirect, statistical technique used where the sample is turbid, coloured or otherwise unfilterable, or where a presence–absence enzymatic substrate is preferred. The sample is divided among replicate tubes or into a sealed multi-well tray at several dilutions, each containing a growth medium whose response is binary: gas production in a lauryl-tryptose broth tube, or fluorescence of the MUG substrate in a Colilert tray under ultraviolet light. Each vessel is scored simply positive or negative after incubation. The pattern of positives across the dilution series is then converted, through a Poisson maximum-likelihood model tabulated in Standard Methods, into the bacterial density that was most probably present. The result is reported as MPN per 100 mL.

Two issues in interpreting reported laboratory data.

1. MPN and CFU are not interchangeable numbers, and both under-report the true bacterial population. MPN is a statistical estimate carrying wide and asymmetric 95 % confidence limits — for a standard five-tube series the upper limit can be three to four times the reported value — whereas CFU is an actual count with much narrower Poisson error. For the same water, MPN usually returns a somewhat higher figure than CFU because a clump of cells that yields a single colony still turns a whole tube positive. More fundamentally, both methods count only cells that are culturable on the chosen medium under the chosen conditions, so viable-but-non-culturable organisms and chlorine-injured cells are missed. A result must therefore always be quoted with its method, and a trend must not be assembled from a mixture of the two.
2. Sample handling, holding time and interferences govern whether the number means anything at all. Regulatory samples must be collected in sterile, sodium-thiosulphate-dechlorinated bottles, held at 4 °C, and analysed within the holding time specified in the provincial drinking-water regulation (six hours for regulatory samples under Ontario Regulation 170/03, with 24 h an absolute maximum) — bacteria die off or regrow outside that window. High background heterotrophic populations can overgrow a membrane filter and cause confluent growth, which is reported as “too numerous to count” or “overgrown” and is an invalid result, not a low one. High turbidity blinds the filter, which is precisely when MPN should be substituted. Finally, indicator organisms are a proxy: they evidence fecal contamination, and their absence does not by itself demonstrate the absence of chlorine-resistant protozoa such as Cryptosporidium or of viruses.