Question 3 of 5: pH Significance and Alkalinity from a Double Titration

Question 3: pH Significance and Alkalinity from a Double Titration (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.

(i) pH: Definition and Significance

pH is the negative base-10 logarithm of the hydrogen-ion activity, $pH = -\log_{10}[H^+]$, giving a 0–14 scale on which 7 is neutral (pure water at 25 °C), values below 7 acidic and above 7 alkaline. In water treatment pH governs two process-critical equilibria addressed directly by this exam.

Disinfection. Free chlorine in water exists as an equilibrium between hypochlorous acid (HOCl) and the hypochlorite ion (OCl⁻), $HOCl \rightleftharpoons H^+ + OCl^-$, with a pK_a near 7.5. HOCl, the dominant species below pH ≈7.5, is roughly 80–100 times more effective a biocide than OCl⁻ because its neutral charge lets it penetrate microbial cell walls far more readily. Raising pH shifts the equilibrium toward the far weaker OCl⁻, so a plant must either hold pH low during chlorine contact or dose substantially more chlorine (and provide longer contact time, i.e. a larger CT) to achieve the same log-inactivation at higher pH.

Coagulation-flocculation. Hydrolyzing metal coagulants (alum, ferric chloride) only form the insoluble, positively-charged hydroxide precipitates responsible for charge neutralization and sweep-floc capture within a narrow optimum pH band — roughly 5.5–7.5 for alum, a wider 4–11 for ferric salts. Outside that window the metal hydroxide either fails to precipitate (stays soluble, no floc forms) or, at high pH, redissolves as an anionic aluminate/ferrate, so pH control (often with lime, soda ash, or CO₂) directly ahead of rapid mix is essential to reliable turbidity removal.

(ii) Alkalinity from the Double Titration

Given.

Find. The alkalinity indicated by each end point, its numerical value, and any further alkalinity species (hydroxide/carbonate/bicarbonate) recoverable from the two readings.

Approach. Convert each titrant volume to an alkalinity as mg/L CaCO₃, then apply the standard phenolphthalein/total alkalinity relationships (Sawyer & McCarty) to split the total into its hydroxide, carbonate and bicarbonate components.

1. Name the two end points. The phenolphthalein end point (pH 8.3) measures phenolphthalein alkalinity, $P$ — the alkalinity neutralized down to the point where all hydroxide and half of any carbonate has been converted. The Bromocresol Green end point (pH 4.5) measures total alkalinity, $T$ — all hydroxide, carbonate and bicarbonate alkalinity present, since 4.5 is the carbonic-acid/bicarbonate equivalence point.
2. Convert titrant volumes to alkalinity. $Alkalinity\,(\text{mg/L as }CaCO_3) = \dfrac{V_{acid}(\text{mL}) \times N(\text{eq/L}) \times 50{,}000}{V_s(\text{mL})}$. For $P$: $\dfrac{5 \times 0.02 \times 50{,}000}{50} = \boxed{100 \text{ mg/L as } CaCO_3}$. For $T$: $\dfrac{8 \times 0.02 \times 50{,}000}{50} = \boxed{160 \text{ mg/L as } CaCO_3}$.
3. Identify the governing case. Comparing $P$ to $T/2 = 80$: since $P = 100 > T/2$, the sample’s alkalinity is composed of hydroxide and carbonate only — no bicarbonate is present (Sawyer & McCarty table, case $P > T/2$).
4. Compute the third (other) alkalinity values. Hydroxide: $OH^- = 2P - T = 2(100) - 160 = \boxed{40 \text{ mg/L as } CaCO_3}$. Carbonate: $CO_3^{2-} = 2(T-P) = 2(160-100) = \boxed{120 \text{ mg/L as } CaCO_3}$. Bicarbonate: $HCO_3^- = 0$. Check: $40 + 120 + 0 = 160 = T$, reconstituting the measured total exactly.