16-Civ-B5 Water Supply and Wastewater Treatment · May 2016

Question 2 of 5: Chlorine Demand, the Breakpoint Curve and the pH Dependence of Chlorination

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 2: Chlorine Demand, the Breakpoint Curve and the pH Dependence of Chlorination (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.

Given. The mechanism is explained against one representative water so that the shape of the curve can be attached to real doses.

Find. (a) Why the residual–dose relationship is not a straight line, what each of the four regions of the breakpoint curve represents chemically, and what dose this water actually needs; (b) how the distribution of free chlorine between HOCl and OCl⁻ changes with pH and what that does to the contact time required.

Approach. Follow the applied chlorine through the reactions it undergoes in order of decreasing thermodynamic favourability — reduced inorganics first, then ammonia to chloramines, then chloramines to nitrogen gas, and only then free chlorine — and read each stage off the corresponding zone of the residual-versus-dose curve; then treat the free residual as a two-species equilibrium and apply the acid–base distribution law.

(a) Chlorine demand and the chlorination curve (15 marks)

1. Define demand as the difference between what is applied and what survives. Chlorine gas or hypochlorite hydrolyses on contact with water to hypochlorous acid, $\mathrm{Cl_2 + H_2O \rightarrow HOCl + H^{+} + Cl^{-}}$. The chlorine demand of a water at a stated contact time is $$\text{demand} \;=\; \text{applied dose} \;-\; \text{measured residual}$$ If the water contained nothing reactive, the residual would rise one-for-one with the dose along the broken 1:1 line of Figure 2.1. Every departure of the real curve below that line is a reaction that consumed chlorine.
2. Zone 1 — the immediate demand of reduced inorganic species and reactive organics. Hypochlorous acid is a strong oxidant and attacks the most reduced species first, essentially instantaneously and irreversibly: $$\mathrm{Fe^{2+}},\ \mathrm{Mn^{2+}},\ \mathrm{H_2S},\ \mathrm{NO_2^{-}} \;\xrightarrow{\ \mathrm{HOCl}\ }\; \mathrm{Fe^{3+}},\ \mathrm{MnO_2},\ \mathrm{S^{0}/SO_4^{2-}},\ \mathrm{NO_3^{-}}$$ Reactive natural organic matter is attacked in parallel, by substitution as well as oxidation, and it is this pathway that generates trihalomethanes and haloacetic acids. Throughout this zone the measured residual stays at zero: the chlorine is being reduced to chloride as fast as it is added. In the representative water this accounts for the first 2.0 mg/L of dose.
3. Zone 2 — the ammonia reacts and combined residual appears. Once the fast reductants are gone, hypochlorous acid substitutes onto ammonia to make chloramines: $$\mathrm{NH_3 + HOCl \rightarrow NH_2Cl + H_2O} \qquad \mathrm{NH_2Cl + HOCl \rightarrow NHCl_2 + H_2O}$$ Chloramines are oxidants and are detected by the standard DPD test, so the measured residual now rises — but as combined chlorine, roughly two orders of magnitude weaker as a biocide than free chlorine. Monochloramine predominates while the applied chlorine-to-ammonia-nitrogen mass ratio remains below about 5:1, which is the 1:1 molar ratio. The residual climbs almost linearly to a local maximum, the peak visible in Figure 2.1.
4. Zone 3 — the chloramines are destroyed and the curve falls to the breakpoint. Adding chlorine past the peak no longer forms chloramines, it oxidises them. The classical pathway proceeds through dichloramine and an unstable intermediate to nitrogen gas, $$\mathrm{2\,NH_2Cl + HOCl \rightarrow N_2\uparrow + 3\,HCl + H_2O}$$ with the overall stoichiometry $$\mathrm{2\,NH_3 + 3\,Cl_2 \rightarrow N_2\uparrow + 6\,HCl}$$ Because the products are gaseous nitrogen and chloride, the measured residual falls even though chlorine is still being added — the counter-intuitive feature of the curve. The minimum of the curve is the breakpoint.
5. Quantify the breakpoint from the overall stoichiometry. Three moles of Cl₂ are needed for every two moles of nitrogen, so the theoretical requirement is $$\frac{m_{\mathrm{Cl_2}}}{m_{\mathrm{N}}} \;=\; \frac{3\times 70.906}{2\times 14.007} \;=\; \boxed{7.59\ \text{g Cl}_2\ \text{per g NH}_3\text{-N}}$$ In practice, side reactions to nitrate and to organic chloramines push the requirement to roughly 8–10:1, which is why plants dose above the theoretical value and confirm the breakpoint by test.
6. Apply the stoichiometry to the representative water. The ammonia alone consumes $$7.59\times 3.0 = 22.8\ \text{mg/L as Cl}_2$$ and adding the 2.0 mg/L consumed in Zone 1 gives a total chlorine demand at the breakpoint of 24.8 mg/L. To leave the required 0.8 mg/L of free residual the plant must apply $$D \;=\; 22.8 + 2.0 + 0.8 \;=\; \boxed{25.6\ \text{mg/L as Cl}_2}$$ An operator who reasoned only from the target residual and dosed 3 mg/L would sit near the top of Zone 2, believe from the DPD reading that the water was chlorinated, and in fact be applying a weak combined residual with essentially no virucidal capacity.
7. Zone 4 — free residual at last. Beyond the breakpoint the ammonia is gone, and further chlorine survives as free available chlorine, so the residual again rises one-for-one with the dose, parallel to the ideal line but offset from it by the total demand. This zone is where free-chlorine disinfection actually operates. Overshooting it carries its own penalties: dichloramine and nitrogen trichloride formed in the trough are responsible for the characteristic “swimming-pool” taste and odour, and disinfection by-product formation continues to grow with dose and contact time.

(b) The pH dependence of chlorination efficiency (10 marks)

8. Write the equilibrium that free chlorine obeys. Free available chlorine is not one substance but two in equilibrium: $$\mathrm{HOCl} \rightleftharpoons \mathrm{H^{+}} + \mathrm{OCl^{-}}, \qquad K_a = 2.9\times 10^{-8}\ \text{at}\ 20\,{}^{\circ}\text{C}\ \ (\mathrm{p}K_a = 7.54)$$ The fraction present as the acid follows directly: $$f_{\mathrm{HOCl}} = \frac{1}{1+10^{\,(\mathrm{pH}-\mathrm{p}K_a)}}$$
9. Evaluate the fraction across the operating range. Substituting pH values into the distribution law gives 97.2 per cent HOCl at pH 6.0, 77.6 per cent at pH 7.0, 52.3 per cent at pH 7.5, 25.7 per cent at pH 8.0 and only 3.4 per cent at pH 9.0. Figure 2.2 plots the pair of curves; they cross at the $\mathrm{p}K_a$.
10. Connect speciation to germicidal efficiency. Hypochlorous acid is a small, uncharged molecule that passes through the microbial cell wall and attacks enzymes and nucleic acids inside; hypochlorite carries a negative charge, is repelled by the similarly charged cell surface, and is between 80 and 100 times less effective for the same measured concentration. Disinfection follows the Chick–Watson relation $$\ln\frac{N}{N_0} = -\,\Lambda\,C^{\,n} t$$ in which the effective $C$ is the hypochlorous acid concentration, not the total free residual reported by the DPD test.
11. Quantify the penalty of drifting one pH unit. Holding the total free residual constant and moving from pH 7.0 to pH 8.0, $$\frac{f_{\mathrm{HOCl}}(7.0)}{f_{\mathrm{HOCl}}(8.0)} = \frac{0.776}{0.257} = \boxed{3.0}$$ so the contact time — or equivalently the CT value, or the residual carried — must be about three times larger at pH 8 to achieve the same log inactivation. This is exactly the pattern of the CT tables in the Canadian and USEPA disinfection guidance, in which the CT required for a given Giardia credit rises steadily with pH.
12. Draw the operating conclusions. Three follow immediately. Chlorine should be applied where the pH is low, which in a lime-softening plant means before recarbonation rather than after, since at pH 10.5 barely 1 per cent of the free residual is HOCl and the disinfection is nearly worthless. Where corrosion control forces the finished water to pH 8 or above, the contact tank must be sized for that condition, not for the laboratory jar test. And the choice is a genuine trade-off, because lowering the pH raises the haloacetic-acid yield even as it lowers the trihalomethane yield, so the pH that optimises disinfection is not automatically the pH that minimises by-products. Chloramination sidesteps the pH sensitivity, monochloramine being the stable species from pH 7 to 9, but at the cost of a far weaker and slower disinfectant suited only to maintaining a distribution residual.