Question 2 of 7: Particle Characteristics, Solution Chemistry and Gases

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) — Integrating sedimentation and filtration; turbidity as an indicator (9 marks)

Particles in wastewater span some six orders of magnitude in size, and no single process removes them all economically. The rational design strategy is therefore a treatment train in which each unit is matched to the particle class it removes most cheaply, and each unit protects the one that follows it. Sedimentation and filtration are the two workhorses of that train, and they are complementary rather than alternative.

Sedimentation exploits the density difference between particle and water and is governed by Stokes' law, in which the terminal settling velocity varies with the square of particle diameter. That quadratic dependence is the key to the whole arrangement: settling is extremely efficient for large, dense particles and becomes uneconomic very quickly as particles get smaller. Grit chambers remove sand and other discrete particles above roughly 200 µm in a minute or two of detention. Primary clarifiers, with an overflow rate of the order of 30 to 50 m³/m²·d and two hours of detention, remove the settleable organic solids, typically 50 to 70 % of the influent suspended solids and 25 to 40 % of the BOD. Secondary clarifiers perform a different duty again, separating the biological floc grown in the aeration basin, and here the particles are low-density flocculent aggregates that settle as a hindered blanket rather than as discrete grains. Colloids, in the range from about 1 nm to 1 µm, will not settle at all in any practical detention time, because they carry a negative surface charge that keeps them mutually repelled and because Brownian motion overwhelms gravity at that scale. They must first be destabilised by coagulation with alum or ferric salts and aggregated by gentle flocculation into settleable floc — the step that converts an unsettleable particle into a settleable one.

Filtration takes over precisely where settling ceases to be economic. Granular media filters — typically dual-media anthracite over sand, run at 5 to 15 m/h — remove the residual floc, the small carry-over solids from the secondary clarifier and the particles that never formed settleable aggregates, bringing effluent suspended solids from perhaps 20 to 30 mg/L down to below 5 to 10 mg/L. Removal is not simple straining: transport mechanisms of interception, sedimentation onto the grain surfaces and diffusion bring particles to the media, and attachment then holds them there, which is why filter performance depends on upstream coagulant conditioning. Where a still finer barrier is required, membrane filtration extends the range — microfiltration to about 0.1 µm and ultrafiltration to about 0.01 µm, the latter providing an absolute barrier to protozoan cysts and most bacteria.

The integration is what makes the train work. Sedimentation is placed first because it is inexpensive per unit mass removed and because it strips out the bulk solids load that would otherwise blind a filter within minutes; a filter asked to remove 200 mg/L of suspended solids would spend most of its time backwashing. Filtration is placed last because it is the polishing step, and because it is the only unit capable of producing the low, stable particle count that downstream disinfection requires. That last point is the crucial engineering linkage: particles shield micro-organisms from chlorine and from ultraviolet light, so the disinfection process cannot achieve its design log-removal unless the filter has first delivered a low-turbidity water. Where the effluent is destined for reuse, the same train is extended with coagulation ahead of the filter and membranes or reverse osmosis after it.

Turbidity is a good indicator of how well this train is performing for three related reasons. First, it is a direct optical measure of the very thing the particle-removal processes are meant to control — it quantifies light scattered by suspended and colloidal matter, and it responds to the fine particles that dominate the risk, since scattering is most sensitive to particles in the 0.1 to 10 µm range that carry pathogens and resist settling. Second, it is a strong surrogate for several quantities that are slow or expensive to measure: it correlates well with suspended solids, and hence with a substantial part of the effluent BOD, and it correlates with the shielding of micro-organisms that undermines disinfection. Third, and decisively for operations, it can be measured continuously, on-line, in real time by a nephelometer costing a small fraction of a laboratory programme, so a filter breakthrough or a clarifier upset is detected within minutes instead of the day or more that a suspended-solids or coliform result requires. A rising turbidity trend is thus an early warning that gives the operator time to act before a permit limit is exceeded. The caveats are that turbidity is a surrogate and not the quantity of interest itself — the relationship to suspended solids in mg/L is site-specific and must be calibrated locally, and dissolved colour or a change in particle size distribution can move the reading without any change in mass concentration.

Part (ii) — Hardness as CaCO₃ (6 marks)

Given. Average divalent-cation concentrations for the lake water, together with the atomic weights the paper directs the candidate to use.

Find. The total hardness expressed in mg/L as CaCO₃, and the resulting classification of the water as soft or hard.

Approach. Hardness is the sum of the multivalent metallic cations expressed on a common equivalent basis, so convert each ion to its CaCO₃ equivalent through equivalent weights — not through molar mass — and sum.

1. Establish the CaCO₃ reference. Using the atomic weights the question supplies, the molar mass of calcium carbonate is $$M_{\mathrm{CaCO_3}} = 40 + 12 + 3(16) = 100\ \text{g/mol}$$ Carbonate carries two charges, so the equivalent weight is $$EW_{\mathrm{CaCO_3}} = \frac{100}{2} = 50\ \text{g/eq}$$
2. Compute the equivalent weight of each hardness-causing ion. Each of the three cations is divalent, so its equivalent weight is half its atomic weight: $$EW_{\mathrm{Ca}} = \frac{40}{2} = 20, \qquad EW_{\mathrm{Mg}} = \frac{24}{2} = 12, \qquad EW_{\mathrm{Fe}} = \frac{56}{2} = 28\ \text{g/eq}$$
3. Convert each ion to the CaCO₃ basis. The general conversion is $$C_{\text{as CaCO}_3} = C_{\text{ion}} \times \frac{EW_{\mathrm{CaCO_3}}}{EW_{\text{ion}}}$$ Applying it to each ion in turn: $$\mathrm{Ca}: \ 50 \times \frac{50}{20} = 125.0 \ \text{mg/L as CaCO}_3$$ $$\mathrm{Mg}: \ 20 \times \frac{50}{12} = 83.3 \ \text{mg/L as CaCO}_3$$ $$\mathrm{Fe}: \ 30 \times \frac{50}{28} = 53.6 \ \text{mg/L as CaCO}_3$$
4. Sum to obtain the total hardness. The conventional total hardness counts calcium and magnesium: $$TH_{\mathrm{Ca+Mg}} = 125.0 + 83.3 = 208.3$$ $$\boxed{TH = 208\ \text{mg/L as CaCO}_3}$$ Ferrous iron is a divalent metallic cation and strictly contributes to hardness as well, so the inclusive total is $$TH_{\mathrm{Ca+Mg+Fe}} = 208.3 + 53.6 = \boxed{262\ \text{mg/L as CaCO}_3}$$
5. Classify the water. On the three-band scale implied by the question's wording (soft below 75, moderately hard 75 to 150, hard above 150 mg/L as CaCO₃) the water is hard. On the four-band scale used with the Guidelines for Canadian Drinking Water Quality (soft below 60, moderately hard 60 to 120, hard 120 to 180, very hard above 180) it is very hard on either total. Both conventions place it well into the range where softening would be considered.

A point worth drawing out is that magnesium punches above its weight. It is reported at 20 mg/L against calcium's 50, yet it contributes 83 mg/L as CaCO₃ against calcium's 125 — that is, each milligram of magnesium carries 1.67 times the hardness of a milligram of calcium, because its equivalent weight is 12 against calcium's 20. Converting through molar mass instead of equivalent weight is the single most common error on this question and it understates magnesium badly. The distinction has an operational consequence too: in lime softening, calcium is precipitated as CaCO₃ at about pH 10.3, whereas magnesium requires excess lime to reach pH 11 or above before Mg(OH)₂ will precipitate, so a magnesium-rich water costs disproportionately more in lime and produces more sludge than its total hardness alone would suggest.

Part (iii) — Hydronium and hydroxide concentrations in acidic leachate (5 marks)

Given. A landfill leachate at pH 3.0, at ambient temperature where the ion product of water is $K_w = 1.0 \times 10^{-14}$.

Find. The hydronium ion concentration $[\mathrm{H_3O^+}]$ and the hydroxide ion concentration $[\mathrm{OH^-}]$, both in moles per litre.

Approach. Invert the definition of pH to obtain the hydronium concentration, then use the ion product of water to obtain the hydroxide concentration.

1. Invert the definition of pH. By definition $\mathrm{pH} = -\log_{10}[\mathrm{H_3O^+}]$, so $$[\mathrm{H_3O^+}] = 10^{-\mathrm{pH}} = 10^{-3.0}$$ $$\boxed{[\mathrm{H_3O^+}] = 1.0 \times 10^{-3}\ \text{mol/L}}$$
2. Apply the ion product of water. In any aqueous solution at 25 °C the two ion concentrations are locked together by $$K_w = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0 \times 10^{-14}$$ so that $$[\mathrm{OH^-}] = \frac{K_w}{[\mathrm{H_3O^+}]} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-3}}$$ $$\boxed{[\mathrm{OH^-}] = 1.0 \times 10^{-11}\ \text{mol/L}}$$
3. Check with the pH–pOH identity. Taking negative logarithms, $\mathrm{pOH} = 11.0$ and $\mathrm{pH} + \mathrm{pOH} = 3.0 + 11.0 = 14.0$, as required.

These numbers explain why the leachate is described as corrosive and why pH adjustment heads the treatment train. At pH 3 the hydronium concentration is ten thousand times the neutral value of $10^{-7}$ mol/L, which is enough to attack concrete, to corrode carbon steel pipework and pumps, and — most importantly for the environment — to hold heavy metals such as lead, zinc, cadmium and nickel in dissolved, mobile and bioavailable form. Neutralisation is therefore both a materials-protection measure and the step that converts dissolved metals into settleable hydroxide precipitates. The alkali requirement follows directly from the computed acidity: neutralising $1.0 \times 10^{-3}$ equivalents per litre of free acid needs about 0.037 kg of hydrated lime, Ca(OH)₂, per cubic metre of leachate, before any allowance for the additional acidity released as the metals themselves hydrolyse.