Question 2 of 5: Chlorination Chemistry; Taste, Odour and Hardness

Question 2: Chlorination Chemistry; Taste, Odour and Hardness (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.

Part (a) — The chlorination curve and its four concepts (15 marks). When chlorine gas or hypochlorite is added to water it hydrolyses immediately to hypochlorous acid, which dissociates according to pH:

$$\mathrm{Cl_2} + \mathrm{H_2O} \rightleftharpoons \mathrm{HOCl} + \mathrm{H}^+ + \mathrm{Cl}^-, \qquad \mathrm{HOCl} \rightleftharpoons \mathrm{H}^+ + \mathrm{OCl}^-,\ \ \mathrm{p}K_a = 7.54\ \text{at }20\ {}^{\circ}\text{C}$$

What an operator then measures is not the dose but the residual, and the relationship between the two is the chlorination curve below. Plotting measured residual against applied dose for a water containing ammonia produces four distinct regions, and the four concepts the question names are simply four features of that one curve.

Chlorine demand is the difference between the chlorine applied and the chlorine remaining as a measurable residual after a stated contact time:

$$D = C_{\text{applied}} - C_{\text{residual}}$$

It is not a property of the chlorine but of the water, and it is meaningless unless the contact time and temperature are stated with it, because the reactions that consume chlorine are kinetic. In zone 1 of the curve the dose is destroyed outright with no residual at all: ferrous iron, manganese, sulphide, nitrite and readily oxidisable organic matter reduce the chlorine as fast as it is added. Only when this immediate demand is satisfied does any residual appear.

Combined residual chlorine is the chlorine bound as chloramines. Beyond zone 1 the chlorine meets ammonia and reacts stepwise, the first step dominating at ordinary pH and at chlorine-to-nitrogen ratios below the breakpoint:

$$\mathrm{NH_3} + \mathrm{HOCl} \rightarrow \mathrm{NH_2Cl} + \mathrm{H_2O} \quad (\text{monochloramine})$$

$$\mathrm{NH_2Cl} + \mathrm{HOCl} \rightarrow \mathrm{NHCl_2} + \mathrm{H_2O} \quad (\text{dichloramine})$$

Because one mole of chlorine binds one mole of nitrogen in the first step, $70.90/14.007 = 5.06$ mg of Cl₂ are consumed per mg of NH₃-N, and the residual measured over zone 2 rises almost in step with the dose. Combined residual is a genuine disinfectant but a weak and slow one — roughly two orders of magnitude less potent than free chlorine against bacteria and far weaker still against viruses — which is exactly why it is valued as a secondary disinfectant: chloramine is stable, persists to the ends of a distribution system, produces far less trihalomethane and haloacetic acid than free chlorine, and does not generate the chlorophenolic tastes that free chlorine creates with trace phenols.

The residual reaches a local maximum — the hump — and then falls through zone 3 even though the dose keeps rising. This is the destruction of the chloramines already formed, by chlorine now present in excess of the 1:1 ratio, ultimately to nitrogen gas:

$$2\,\mathrm{NH_3} + 3\,\mathrm{Cl_2} \rightarrow \mathrm{N_2} \uparrow + 6\,\mathrm{HCl}$$

The minimum of that descent is the breakpoint: the dose at which the ammonia has been oxidised away and the chloramine residual has collapsed to near zero. Its stoichiometry follows from the equation above, three moles of chlorine per two moles of nitrogen:

$$\frac{3 \times 70.90}{2 \times 14.007} = 7.59 \approx 7.6\ \text{mg Cl}_2 \text{ per mg NH}_3\text{-N}$$

which is the origin of the familiar 7.6:1 rule of thumb; in practice side reactions with organic nitrogen push the observed requirement to 8 or 10:1.

Given. A source water at NH₃-N 1.8 mg/L with an additional immediate demand of 1.2 mg/L from reduced inorganic species, requiring a free residual of 0.5 mg/L leaving the contact tank. Find. The design dose and the chlorine demand at the breakpoint. The stoichiometric ammonia requirement is $1.8 \times 7.59 = 13.7$ mg/L, so

$$\boxed{C_{\text{applied}} = 13.7 + 1.2 + 0.5 = 15.4\ \text{mg/L as Cl}_2, \qquad D = 15.4 - 0.5 = 14.9\ \text{mg/L}}$$

Free residual chlorine is what appears in zone 4: chlorine present as HOCl and OCl⁻, unbound to nitrogen. Past the breakpoint every additional milligram of dose yields very nearly a milligram of free residual, so the curve rises at unit slope parallel to the no-demand line, displaced by the demand. Free chlorine is the strong, fast disinfectant, and its potency depends sharply on speciation because HOCl is some 80 times more germicidal than the hypochlorite ion. At pH 7.5,

$$f_{\mathrm{HOCl}} = \frac{1}{1 + 10^{\,\mathrm{pH} - \mathrm{p}K_a}} = \frac{1}{1 + 10^{\,7.5 - 7.54}} = 0.52$$

whereas at pH 8.5 the same calculation gives 0.099 — one pH unit costs more than a five-fold loss of the active species, which is why disinfection is done before, not after, lime softening. Compliance is assessed on the CT product; the 0.5 mg/L residual held for a 30-minute baffled contact time gives $CT = 15$ mg·min/L, comfortably above the requirement for 4-log virus inactivation at 10 °C and pH 7.

The practical lesson of the curve is that a partial dose is worse than none: dosing to the hump gives a large but feeble combined residual and maximum taste-and-odour complaints, whereas the same water dosed past the breakpoint gives a smaller number on the residual meter that is a far stronger disinfectant. An operator who responds to a taste complaint by reducing the dose can walk the plant backwards into zone 3 and make the problem worse.

Part (b) — Taste and odour, and hardness (10 marks). Taste and odour in a source water arise from four families of cause. The dominant one in Canadian surface supplies is biological: cyanobacterial and actinomycete blooms release geosmin and 2-methylisoborneol, whose earthy-musty odours are detectable at 5 to 10 ng/L, four orders of magnitude below any health-based concern. Second are industrial and agricultural organics — phenols, chlorophenols, solvents, petroleum products — that are odorous themselves and, worse, form intensely medicinal compounds on chlorination. Third are dissolved inorganics: hydrogen sulphide from anoxic groundwater at its rotten-egg threshold near 0.05 mg/L, iron and manganese giving metallic and bitter tastes above roughly 0.3 and 0.05 mg/L, and chloride and sulphate giving salty and bitter tastes at a few hundred milligrams per litre. Fourth is treatment itself: free chlorine, chloramine and their by-products. The significance is that taste and odour are almost always aesthetic rather than toxicological objectives — and are treated with complete seriousness precisely for that reason, because they are the only water-quality attributes a consumer can perceive. Loss of confidence drives people to untreated and possibly unsafe alternatives such as roadside springs and private wells, and it makes every other message from the utility harder to deliver. Treatment is therefore aimed at removal, not masking: powdered activated carbon during bloom season, ozone or advanced oxidation for geosmin and MIB, aeration and oxidation followed by filtration for sulphide, iron and manganese, and reservoir management to prevent the blooms in the first place.

Given. A groundwater at Ca²⁺ 78 mg/L and Mg²⁺ 19 mg/L, with the 118.9 mg/L alkalinity as CaCO₃ computed in Question 1(iii). Find. Total hardness and its carbonate and non-carbonate split. Hardness is the sum of the multivalent cations, in practice calcium and magnesium, expressed as CaCO₃:

$$\text{TH} = 78 \times \frac{50.04}{20.04} + 19 \times \frac{50.04}{12.15} = 194.8 + 78.2 = 273\ \text{mg/L as CaCO}_3$$

Since alkalinity (118.9) is less than total hardness, the carbonate hardness equals the alkalinity and

$$\boxed{\text{CH} = 118.9,\qquad \text{NCH} = 273 - 118.9 = 154\ \text{mg/L as CaCO}_3}$$

which places the water in the "very hard" class (above 180 mg/L). The causes of hardness are geological: rainwater made acidic by dissolved carbon dioxide percolates through limestone, dolomite and gypsum, dissolving them as bicarbonates and sulphates. Groundwaters in carbonate terrain — much of southern Ontario, the Prairies and the interior valleys of British Columbia — are consequently hard, while surface waters draining the granitic Shield and the Coast Mountains are soft. The significance is economic and operational rather than health-related: hard water precipitates carbonate scale in water heaters, boilers and hot-water mains, wasting fuel and eventually blocking pipe; it consumes soap in forming insoluble curd; and it provokes complaints about spotted glassware and stiff laundry. The carbonate/non-carbonate split matters because it determines the treatment chemistry — carbonate hardness is removed by lime alone, non-carbonate hardness needs lime plus soda ash, or ion exchange, or nanofiltration. Softening also has a cost of its own: it raises pH, sodium and total dissolved solids, and over-softening produces a corrosive water, so recarbonation and a target of roughly 80 to 120 mg/L as CaCO₃ — not zero — is the usual objective. Soft waters, correspondingly, are the aggressive ones, and the ones that most need corrosion control.