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 hydrogen-ion activity, $pH=-\log_{10}[H^+]$, giving the familiar 0–14 scale on which 7 is neutral (pure water at 25 °C), values below 7 acidic and above 7 alkaline. Two process-critical equilibria in water treatment depend directly on it.

Disinfection. Aqueous free chlorine exists in equilibrium between hypochlorous acid and the hypochlorite ion, $HOCl \rightleftharpoons H^+ + OCl^-$, with a pK_a near 7.5. The neutral, uncharged HOCl species — dominant below pH ≈7.5 — diffuses through microbial cell walls roughly 80–100 times faster than the charged OCl⁻ ion and is correspondingly the far more effective biocide, so raising pH weakens chlorine disinfection unless dose and contact time (CT) are increased to compensate.

Coagulation-flocculation. Hydrolyzing metal coagulants only precipitate as the insoluble, positively-charged hydroxide flocs responsible for charge neutralization and sweep coagulation within a narrow optimum pH window — roughly 5.5–7.5 for alum, a somewhat wider 4–11 for ferric salts. Below or above that window the metal either stays in soluble form (no floc) or, at high pH, redissolves as an anionic aluminate/ferrate, so pH adjustment (lime, soda ash, or CO₂/acid) immediately ahead of rapid mix is essential to reliable turbidity removal, exactly as described in Question 2.

(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 cumulative 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 titrated down to the point where all hydroxide and half of any carbonate present has been neutralized. The Bromocresol Green end point (pH 4.5) measures total alkalinity, $T$ — hydroxide, carbonate and bicarbonate alkalinity combined, since pH 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 $P$: $\dfrac{4\times0.02\times50{,}000}{20}=\boxed{200\text{ mg/L as }CaCO_3}$. For $T$: $\dfrac{6\times0.02\times50{,}000}{20}=\boxed{300\text{ mg/L as }CaCO_3}$.
3. Identify the governing case. Comparing $P$ to $T/2=150$: since $P=200>T/2$, the sample’s alkalinity is composed of hydroxide and carbonate only, with no bicarbonate present (Sawyer & McCarty table, case $P>T/2$).
4. Compute the third (other) alkalinity values. Hydroxide: $OH^-=2P-T=2(200)-300=\boxed{100\text{ mg/L as }CaCO_3}$. Carbonate: $CO_3^{2-}=2(T-P)=2(300-200)=\boxed{200\text{ mg/L as }CaCO_3}$. Bicarbonate: $HCO_3^-=0$. Check: $100+200+0=300=T$, reconstituting the measured total exactly.