Question 3 of 7: Particles, Odour Control & Water Hardness

Question 3: Particles, Odour Control & Water Hardness (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.

3(i) — Colloidal particles: three properties driving technology selection. Take colloidal particles (roughly 0.001–1 µm):

1. Very small size / negligible settling velocity. Stokes-law settling scales with the square of diameter, so colloids will not settle in any practical detention time. This rules out plain sedimentation and forces destabilization then aggregation — i.e. coagulation to build settleable/filterable floc.
2. Surface charge (negative zeta potential). Colloids carry a net negative surface charge, so electrostatic repulsion keeps the suspension stable. Treatment must neutralize this charge — a coagulant (alum, ferric, or a cationic polymer) compresses the double layer and permits contact.
3. High specific surface area. The large area-to-volume ratio makes colloids reactive and gives them a high coagulant demand, but it also makes them amenable to adsorption/attachment onto floc and filter media.

Two suitable technologies: coagulation–flocculation followed by sedimentation/clarification, and granular-media (or membrane) filtration to capture the residual destabilized colloids.

3(ii) — Two methods to reduce H₂S / CH₄ / odorous sulfur gases. These gases are products of anaerobic, septic conditions, so control targets the conditions that create them:

1. Keep the system aerobic / prevent septicity. Maintain dissolved oxygen and reduce residence time in collection and primary units (aeration or pre-aeration, avoiding long stagnant force-mains and sludge storage). Sulfate-reducing and methanogenic bacteria are obligate anaerobes; maintaining oxidizing conditions suppresses sulfide and methane generation at the source.
2. Chemical dosing to suppress or bind sulfide. Adding an oxidant or metal salt — e.g. nitrate (provides an alternative electron acceptor and inhibits sulfate reduction), or iron salts that precipitate sulfide as FeS, or pH elevation — prevents the free H₂S from forming or escaping. (Downstream capture such as biofilters/scrubbers treats gas already formed, but the question asks to reduce production.)

3(iii) — Hardness as CaCO₃ and classification (7 marks)

Given. Lake-water divalent cations Ca²⁺ = 20 mg/L, Mg²⁺ = 30 mg/L, Cu²⁺ = 30 mg/L; atomic weights Ca = 40, Mg = 24 (Cu not supplied); CaCO₃ molar mass 100, equivalent weight 50.

Find. Total hardness expressed as mg/L CaCO₃, and the qualitative classification.

Approach. Convert each hardness-causing cation to its CaCO₃ equivalent through equivalent weights, $\;C_{\text{as CaCO}_3}=C_{\text{ion}}\times\dfrac{50}{EW_{\text{ion}}}\;$ with $EW=\text{AW}/2$, and sum. Hardness is conventionally the sum of Ca²⁺ and Mg²⁺; Cu²⁺ is treated separately.

1. Calcium contribution. $EW_{\text{Ca}}=40/2=20$, so $$H_{\text{Ca}} = 20\times\frac{50}{20} = 50\ \text{mg/L as CaCO}_3.$$
2. Magnesium contribution. $EW_{\text{Mg}}=24/2=12$, so $$H_{\text{Mg}} = 30\times\frac{50}{12} = 125\ \text{mg/L as CaCO}_3.$$ Magnesium out-contributes calcium here despite a similar mg/L, because its equivalent weight is smaller.
3. Total hardness and classification. $$\boxed{H = H_{\text{Ca}} + H_{\text{Mg}} = 175\ \text{mg/L as CaCO}_3\ \Rightarrow\ \textbf{hard}}$$ On the three-band scale (soft <75, moderately hard 75–150, hard >150 mg/L as CaCO₃) the water is hard.