Question 3 of 5: Stack Plume Types and Carbonate Alkalinity

Problem 3: Stack Plume Types and Carbonate Alkalinity (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.

(a) The Three Stack-Plume Shapes

(i) Looping. Occurs under highly unstable atmospheric conditions, when the environmental (measured) lapse rate is steeper than the dry adiabatic lapse rate (a superadiabatic profile), typically on a sunny afternoon with strong surface heating. Vigorous convective eddies of many different sizes carry the plume alternately far above and far below its mean centreline, producing the large, erratic vertical loops shown in sketch (i); ground-level concentrations directly beneath the plume can be briefly very high whenever a large downward eddy brings it to grade, but only intermittently.

(ii) Fanning. Occurs under a strong temperature inversion — a very stable atmosphere in which the environmental lapse rate is much smaller than (or even opposite in sign to) the dry adiabatic lapse rate, common on a clear, calm night or early morning. Vertical turbulence is essentially suppressed, so the plume spreads only sideways with the wind, in the thin, flat horizontal ribbon shown in sketch (ii), and can travel a long distance without ever touching the ground.

(iii) Coning. Occurs under near-neutral to slightly stable conditions (environmental lapse rate close to the dry adiabatic lapse rate), typical of overcast, moderately windy weather. Vertical and horizontal turbulence are comparable in magnitude, so the plume spreads gradually and symmetrically into the widening cone shown in sketch (iii) — the shape closest to the idealized Gaussian plume model.

(b)(i) Definition of Alkalinity

Alkalinity is a water’s capacity to neutralize (buffer) acid — its acid-neutralizing capacity — arising from the sum of its titratable bases, predominantly the carbonate-system species (bicarbonate and carbonate) together with hydroxide, less any free acidity:

$$\text{Alkalinity (eq/L)} = [\text{HCO}_3^-] + 2[\text{CO}_3^{2-}] + [\text{OH}^-] - [\text{H}^+]$$

and it is conventionally reported in mg/L as CaCO₃ (equivalent weight 50 g/eq), the common reference base for comparing alkalinity across samples.

(b)(ii) Alkalinity ≈ [HCO₃⁻] Below pH 8.5

Each term in the alkalinity expression can be compared to [HCO₃⁻] using the two dissociation constants. The carbonate-to-bicarbonate ratio follows directly from K_a2:

$$\dfrac{[\text{CO}_3^{2-}]}{[\text{HCO}_3^-]} = \dfrac{K_{a2}}{[\text{H}^+]} = 10^{\,\text{pH}-\text{pK}_{a2}}$$

At pH = 8.5, this ratio is 10^8.5−10.3 = 10^−1.8 ≈ 0.016, so [CO₃²⁻] is under 2% of [HCO₃⁻] and the 2[CO₃²⁻] term contributes well under 3% of the total alkalinity — and this contribution shrinks further as pH drops below 8.5. Meanwhile [OH⁻] and [H⁺] are both of order 10⁻⁶ mol/L or smaller anywhere near neutral-to-slightly-basic pH, several orders of magnitude below the 10⁻³–10⁻² mol/L range typical of [HCO₃⁻] in natural and wastewater. With all three of the other terms negligible, Alkalinity ≈ [HCO₃⁻] is an excellent approximation for pH below 8.5.

(b)(iii) CO₂ Concentration at pH 7.6

Given. pH = 7.6; total alkalinity = 320 mg/L as CaCO₃; pK_a1 = 6.4 (25 °C).

Find. The concentration of dissolved CO₂ (as H₂CO₃*) in mg/L.

Approach. Since pH = 7.6 < 8.5, alkalinity ≈ [HCO₃⁻] from part (ii); convert the given alkalinity to a molar HCO₃⁻ concentration, then use the K_a1 equilibrium to back out [H₂CO₃*] at the stated pH.

1. Convert alkalinity to molar HCO₃⁻. Using the CaCO₃ equivalent weight of 50 g/eq (HCO₃⁻ is monovalent, so eq = mol): $$[\text{HCO}_3^-] \approx \dfrac{320\ \text{mg/L}}{50{,}000\ \text{mg/eq}} = 6.40\times10^{-3}\ \text{mol/L}$$
2. Apply the K_a1 equilibrium. With [H⁺] = 10^−7.6 = 2.51×10⁻⁸ mol/L and K_a1 = 10^−6.4 = 3.98×10⁻⁷: $$[\text{H}_2\text{CO}_3^{*}] = \dfrac{[\text{H}^+][\text{HCO}_3^-]}{K_{a1}} = \dfrac{(2.51\times10^{-8})(6.40\times10^{-3})}{3.98\times10^{-7}} = 4.04\times10^{-4}\ \text{mol/L}$$
3. Convert to mg/L as CO₂. With MW_CO2 = 44 g/mol: $$C_{CO_2} = (4.04\times10^{-4}\ \text{mol/L})(44{,}000\ \text{mg/mol}) = \boxed{17.8\ \text{mg/L}}$$