Alkalinity speciation from a two-end-point titration; UV disinfection: 16-Civ-B5 Water Supply and Wastewater Treatment, December 2018, Q2

Question 2: Alkalinity speciation from a two-end-point titration; UV disinfection (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) Alkalinity types and their values (15 marks)

Given. A 20 mL water sample is titrated with standardised 0.02 N sulfuric acid to two successive indicator end-points.

Given data — titration of the water sample
Quantity	Symbol	Value
Sample volume	$V_s$	20 mL
Titrant normality	$N$	0.02 N $\text{H}_2\text{SO}_4$
Acid to the phenolphthalein end-point (pH 8.3)	$V_P$	6 mL
Acid to the bromocresol-green end-point (pH 4.5)	$V_T$	8 mL (cumulative from the start)

Find. The name and value of the alkalinity measured at each end-point, and the further alkalinity species that the pair of readings allows to be calculated.

The two indicator end-points cut the titration into segments. Phenolphthalein (pH 8.3) is reached when all hydroxide and exactly half the carbonate have been neutralised; bromocresol green (pH 4.5) is reached when the remaining bicarbonate has been converted to carbonic acid.

Approach. Convert each titrant volume to an alkalinity in mg/L as $\text{CaCO}_3$ with the Standard Methods 2320 B expression, then use the relative size of the two results to identify which carbonate species can coexist and to split the total into hydroxide, carbonate and bicarbonate alkalinity.

Name the two end-points. The pH 8.3 end-point, developed with phenolphthalein, defines the phenolphthalein alkalinity $P$; the pH 4.5 end-point, developed here with bromocresol green, defines the total alkalinity $T$ (also called methyl-orange alkalinity from the traditional indicator). The second reading is cumulative — the titration is continued in the same beaker — so $V_T = 8\ \text{mL}$ is the acid added from the very start, not an additional 8 mL.
Convert each volume to alkalinity as calcium carbonate. Standard Methods 2320 B expresses alkalinity as the equivalent mass of $\text{CaCO}_3$, whose equivalent weight is $100.09/2 = 50.04 \approx 50\ \text{mg/meq}$: $$\text{Alkalinity}\ \left(\text{mg/L as CaCO}_3\right) = \frac{V_{\text{acid}} \times N \times 50\,000}{V_s}$$
Phenolphthalein alkalinity. Substituting the first end-point, $$P = \frac{6 \times 0.02 \times 50\,000}{20} = \boxed{300\ \text{mg/L as CaCO}_3}$$ Equivalently, the acid delivered is $6 \times 0.02 = 0.12\ \text{meq}$ into 20 mL, i.e. 6.0 meq/L, and $6.0 \times 50.04 = 300\ \text{mg/L}$.
Total alkalinity. Substituting the second end-point in the same way, $$T = \frac{8 \times 0.02 \times 50\,000}{20} = 400\ \text{mg/L as CaCO}_3$$ so the water has a total acid-neutralising capacity of $\boxed{400\ \text{mg/L as CaCO}_3}$, of which three-quarters is already spent by the time the phenolphthalein end-point is reached.
Identify which species can coexist. Hydroxide and bicarbonate cannot exist together in the same water — they react to form carbonate — so at most two of the three species are present, and the relative size of $P$ and $T$ decides which. Here $$P = 300 > \tfrac{T}{2} = 200\ \text{mg/L}$$ which is the hydroxide-plus-carbonate case: all of the hydroxide and half of the carbonate have been neutralised at pH 8.3, and no bicarbonate is present.
Split the total into its parts. Because the acid consumed to pH 8.3 is $\text{OH}^- + \tfrac12\,\text{CO}_3^{2-}$ and the acid consumed to pH 4.5 is $\text{OH}^- + \text{CO}_3^{2-}$, solving the two expressions together gives $$\text{Hydroxide (caustic) alkalinity} = 2P - T = 2(300) - 400 = \boxed{200\ \text{mg/L as CaCO}_3}$$ $$\text{Carbonate alkalinity} = 2(T - P) = 2(400 - 300) = 200\ \text{mg/L as CaCO}_3, \qquad \text{Bicarbonate alkalinity} = 0$$ The three parts sum to $200 + 200 + 0 = 400\ \text{mg/L}$, which recovers $T$ exactly and is the arithmetic check worth writing down.
Sanity-check the answer against the chemistry. Converting the hydroxide alkalinity back to its own units, $200/50.04 = 4.00\ \text{meq/L}$, i.e. $[\text{OH}^-] = 4.0 \times 10^{-3}\ \text{M}$ or 68 mg/L as $\text{OH}^-$, which corresponds to $\text{pOH} = 2.40$ and $\text{pH} = 11.6$ at $25\,{}^{\circ}\text{C}$; the carbonate alkalinity of 200 mg/L is 2.0 mmol/L, or 120 mg/L as $\text{CO}_3^{2-}$. A pH near 11.6 is not a natural surface or ground water — it is what excess-lime softening produces — and that consistency is what confirms the hydroxide branch was chosen correctly.

Final results — Q2(a)
Quantity	Basis	Value
Phenolphthalein alkalinity, $P$ (pH 8.3 end-point)	mg/L as $\text{CaCO}_3$	300
Total alkalinity, $T$ (pH 4.5 end-point)	mg/L as $\text{CaCO}_3$	400
Hydroxide (caustic) alkalinity, $2P-T$	mg/L as $\text{CaCO}_3$	200
Carbonate alkalinity, $2(T-P)$	mg/L as $\text{CaCO}_3$	200
Bicarbonate alkalinity	mg/L as $\text{CaCO}_3$	0
Implied hydroxide concentration and pH	mg/L as $\text{OH}^-$ / –	68.0 / 11.6

Check: the second reading is taken as cumulative. Standard Methods runs the two end-points on one aliquot, so 8 mL is the total acid added from the start and $T = 400\ \text{mg/L}$. If the 8 mL were instead read as a further 8 mL after the first end-point, the total would be 14 mL, $T = 700\ \text{mg/L}$, and $P = 300 < T/2 = 350$ would put the water on the carbonate-plus-bicarbonate branch with $\text{CO}_3^{2-} = 600$ and $\text{HCO}_3^- = 100\ \text{mg/L}$ — a completely different answer. The cumulative reading is the correct one and is confirmed by the resulting pH of 11.6 being a realistic softening-plant value.

(b) Principle and working of UV disinfection (10 marks)

Ultraviolet disinfection inactivates micro-organisms by photochemical damage to their nucleic acid, not by chemical oxidation. Germicidal UV in the UV-C band, with the low-pressure mercury lamp emitting essentially monochromatically at 253.7 nm, is absorbed by the pyrimidine bases of DNA and RNA; adjacent thymine bases on the same strand dimerise, the covalent cross-link blocks the replication fork, and the organism — although still metabolically alive and still countable as a cell — can no longer reproduce and therefore cannot cause infection. The absorption maximum of DNA at about 260 nm is what makes 254 nm so efficient.

A UV reactor is accordingly a hydraulic device, not a contact tank. Water passes through a closed vessel or an open channel containing banks of lamps sheathed in quartz sleeves; the delivered dose (properly, the UV fluence) is $$D = I_{\text{avg}} \times t$$ in $\text{mJ/cm}^2$, where $I_{\text{avg}}$ is the average germicidal irradiance in the reactor and $t$ the residence time. A typical validated drinking-water design dose of 40 $\text{mJ/cm}^2$ is delivered by an average irradiance of 10 $\text{mW/cm}^2$ in about four seconds — three orders of magnitude faster than a chlorine contact tank. Because both terms in the product vary across the reactor cross section, dose is not calculated from first principles for a real unit; it is established by bioassay validation, and the reactor is then operated within its validated envelope of flow, UV transmittance and lamp power.

Water quality enters through the UV transmittance. Attenuation follows the Beer–Lambert law, so a water of 65 per cent UVT per centimetre has an absorption coefficient of $-\log_{10}(0.65) = 0.187\ \text{cm}^{-1}$ and passes only 11.6 per cent of the incident intensity across a 5 cm lamp spacing, against 77.4 per cent for a 95 per cent-UVT water — a factor of 6.7 in delivered dose for the same lamps. Organic colour, iron and turbidity all depress UVT, and particles additionally shield organisms bodily, so UV is placed downstream of filtration and never treats a poorly clarified water.

Advantages over chlorination. UV forms no halogenated disinfection by-products, since nothing is added to the water; it is extremely effective against the chlorine-resistant protozoa, needing only about 12 $\text{mJ/cm}^2$ for 3-log Cryptosporidium inactivation where chlorine is essentially useless; contact times are seconds rather than tens of minutes, so the footprint is small; there is no taste, odour or pH effect, no chemical delivery, storage or gas-hazard management, and no risk of overdose. For wastewater effluent it also avoids the toxicity of chlorine residual to the receiving water and the dechlorination step that would otherwise be required.

Disadvantages over chlorination. The decisive one is that UV leaves no residual: it cannot protect the distribution system against regrowth or post-treatment contamination, so a chlorine or chloramine residual must still be added and UV supplements rather than replaces it. Performance is invisible — there is no simple field test equivalent to a residual measurement, only lamp-intensity monitoring — and it is strongly degraded by low UVT, fouling of the quartz sleeves (requiring mechanical or chemical wipers) and lamp ageing, so lamps must be replaced on a schedule regardless of apparent output. Certain viruses are notably resistant: 4-log adenovirus inactivation requires about 186 $\text{mJ/cm}^2$, more than fifteen times the Cryptosporidium dose, which drives the validated dose upward wherever full virus credit is sought. Some organisms can also photoreactivate or repair damage in the dark if the dose is marginal. Finally the power demand is continuous and the mercury lamps are a hazardous waste at end of life.

Topic. Carbonate-system alkalinity by two-end-point titration, and disinfection by an agent that leaves no residual.

Key relations. $\text{Alk} = V N \times 50\,000/V_s$ in mg/L as $\text{CaCO}_3$; the $P$–$T$ decision table — $P = 0$: bicarbonate only; $P < T/2$: carbonate $= 2P$ and bicarbonate $= T - 2P$; $P = T/2$: carbonate only; $P > T/2$: hydroxide $= 2P - T$ and carbonate $= 2(T-P)$; $P = T$: hydroxide only. UV dose $D = I_{\text{avg}}t$; Beer–Lambert attenuation $I/I_0 = 10^{-a d}$ with $a = -\log_{10}(\text{UVT})$.

Why this works. The titration works because the carbonate system has two well-separated dissociation steps, $\text{p}K_{a1} = 6.35$ and $\text{p}K_{a2} = 10.33$. The pH 8.3 end-point sits close to the mid-point between them, where essentially all carbonate has become bicarbonate and essentially none has gone further; the pH 4.5 end-point sits below $\text{p}K_{a1}$, where all bicarbonate has become carbonic acid. Each end-point therefore counts a defined and different set of proton acceptors, and two independent readings resolve a three-species system because the incompatibility of hydroxide with bicarbonate removes one unknown. The UV analogue is equally simple: the germicidal effect depends on total absorbed photons, so intensity and time trade off directly, and everything that competes for those photons — organics, iron, particles — reduces the dose that reaches the organism.

Common pitfall. Reading the second titration volume as an increment rather than as the cumulative total, which moves the water onto the wrong branch of the $P$–$T$ table and changes every subsequent number. A close second is quoting hydroxide and bicarbonate alkalinity in the same water, which is chemically impossible. On the UV side, the recurring error is claiming UV as a replacement for chlorine: with no residual it cannot satisfy a distribution-system requirement.

Source. APHA/AWWA/WEF, Standard Methods, Method 2320 B; Davis & Cornwell, Introduction to Environmental Engineering, 5th ed., Ch. 4 (alkalinity and softening); Crittenden et al., MWH's Water Treatment, 3rd ed., Ch. 13–14 (UV disinfection, validated dose, UVT).

Question 2 of 5: Alkalinity speciation from a two-end-point titration; UV disinfection

Question 2: Alkalinity speciation from a two-end-point titration; UV disinfection (25 marks)

(a) Alkalinity types and their values (15 marks)

(b) Principle and working of UV disinfection (10 marks)