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18-Env-A4 Water and Wastewater Engineering · Undated paper

Question 2 of 5: Alkalinity Titration; Ozonation vs. Chlorination

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

Notes on this paper

National Exams — May 2019 — 18-Env-A4 / Water and Wastewater Engineering. 3 hours duration; closed book with one double-sided aid sheet; approved Casio/Sharp calculator permitted. Question 1 is compulsory; the paper instructs candidates to attempt any three of the remaining four (100 marks total); all five are solved below for completeness.

Reference texts. Metcalf & Eddy, Wastewater Engineering: Treatment and Resource Recovery (5th ed.) — BOD kinetics, activated-sludge/clarifier design, anaerobic digestion; Davis & Cornwell, Introduction to Environmental Engineering (6th ed.) — alkalinity chemistry, disinfection, coagulation/flocculation; MWH’s Water Treatment: Principles and Design (3rd ed.) — ozonation, turbidity; Standard Methods for the Examination of Water and Wastewater — alkalinity titration (2320B).

Question 2: Alkalinity Titration; Ozonation vs. Chlorination (25 marks: a 15, b 10)

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. Alkalinity titration and speciation (15 marks)

Given. A 20 mL water sample is titrated with 0.02N $H_2SO_4$: 3 mL to reach the phenolphthalein endpoint (pH 8.3), and a further cumulative total of 5 mL to reach the Bromocresol Green endpoint (pH 4.5) — i.e. the standard double-titration where the phenolphthalein reading $P$ and the total reading $T$ are both measured from the start of the titration.

QuantityValue
Sample volume $V_s$20 mL
Titrant normality $N$0.02 eq/L
Volume to pH 8.3 (phenolphthalein), $P$3 mL
Volume to pH 4.5 (Bromocresol Green), $T$5 mL

Find. The alkalinity indicated by each endpoint (P-alkalinity, T-alkalinity), and the additional alkalinity type(s) recoverable from the P/T relationship, all as mg/L CaCO₃.

Approach. Convert each titrant volume to mg/L as CaCO₃ with the standard normality formula, then apply the Standard-Methods P/T decision table to split total alkalinity into its hydroxide, carbonate and bicarbonate components.

  1. Phenolphthalein alkalinity. The pH 8.3 endpoint marks the point where all carbonate ($CO_3^{2-}$) has converted to bicarbonate and any hydroxide has been fully neutralized — this is phenolphthalein (P) alkalinity. $$P_{alk}=\frac{P\times N\times 50{,}000}{V_s}=\frac{3\times0.02\times50{,}000}{20}=\boxed{150\ \text{mg/L as }CaCO_3}.$$
  2. Total alkalinity. The pH 4.5 endpoint marks the point where essentially all bicarbonate has converted to carbonic acid/$CO_2$ — this is total (T) alkalinity. $$T_{alk}=\frac{T\times N\times 50{,}000}{V_s}=\frac{5\times0.02\times50{,}000}{20}=\boxed{250\ \text{mg/L as }CaCO_3}.$$
  3. Which species table row applies. Compare $P$ to $T/2=2.5$ mL: since $P=3\ \text{mL} > T/2$, the sample falls in the "hydroxide + carbonate present, no bicarbonate" regime of the Standard Methods P/T table.
  4. Other alkalinity values calculable — hydroxide and carbonate alkalinity. $$OH_{alk}=2P_{alk}-T_{alk}=2(150)-250=\boxed{50\ \text{mg/L as }CaCO_3},$$ $$CO_{3,alk}=2\left(T_{alk}-P_{alk}\right)=2(250-150)=\boxed{200\ \text{mg/L as }CaCO_3}.$$ Bicarbonate alkalinity is $T_{alk}-OH_{alk}-CO_{3,alk}=250-50-200=0$, consistent with the P>T/2 regime.
Check: the P>T/2 result (both hydroxide and carbonate present, zero bicarbonate) is unusual for a natural water/typical wastewater — it is the signature of a strongly basic sample (e.g. downstream of lime dosing or a caustic industrial discharge). The arithmetic and the Standard-Methods classification are both confirmed by the check $OH_{alk}+CO_{3,alk}=50+200=250\ \text{mg/L}=T_{alk}$.
QuantityResult
Phenolphthalein alkalinity, $P_{alk}$150 mg/L as CaCO₃
Total alkalinity, $T_{alk}$250 mg/L as CaCO₃
Hydroxide alkalinity, $OH_{alk}$50 mg/L as CaCO₃
Carbonate alkalinity, $CO_{3,alk}$200 mg/L as CaCO₃
Bicarbonate alkalinity, $HCO_{3,alk}$0 mg/L as CaCO₃

b. Ozonation vs. chlorination for disinfection (10 marks)

Ozone ($O_3$) disinfects by direct oxidative attack on microbial cell walls, enzymes and nucleic acids, exploiting its very high oxidation potential (2.07 V, second only to fluorine among practical oxidants); because it is unstable, ozone must be generated on site by passing dry oxygen or air through a high-voltage electrical discharge (corona discharge) and is then bubbled into a contact chamber through fine diffusers, where it decomposes within minutes back to oxygen, so it cannot be stored or piped like chlorine and leaves no persistent residual once it has reacted.

Advantages of ozonation over chlorination: (1) ozone is a far stronger and faster-acting oxidant/disinfectant, achieving effective inactivation of chlorine-resistant protozoan cysts such as Giardia and, especially, Cryptosporidium (which is highly resistant to chlorine at practical CT values) with much lower CT requirements; (2) ozone does not form chlorinated disinfection by-products (trihalomethanes, haloacetic acids) when it reacts with natural organic matter, avoiding that regulated by-product family entirely (though it can form its own by-product, bromate, in bromide-rich source waters).

Disadvantages of ozonation over chlorination: (1) ozone leaves no persistent residual in the distribution system, so a secondary disinfectant (typically chlorine or chloramine) is still required downstream to protect against regrowth and recontamination between the plant and the tap — ozone alone cannot fulfil this role; (2) ozonation is substantially more capital- and energy-intensive, requiring on-site generation equipment (air-prep/oxygen supply, high-voltage generators), dedicated contact chambers, and an off-gas destruction system, and it demands more operator training/monitoring than the comparatively simple chlorine feed systems, making it a materially more expensive technology to install and operate.