Question 5 of 7: Contaminant Partitioning, Equilibrium Chemistry and Reactor Balances

Question 5: Contaminant Partitioning, Equilibrium Chemistry and Reactor Balances (20 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) The Terms in $K_d = \alpha \cdot f_{OC} \cdot K_{OW}$

$\alpha$ is an empirical proportionality (calibration) coefficient that relates the organic-carbon-normalized partition coefficient to the compound's octanol-water partition coefficient; it captures the fact that natural organic carbon is not identical to octanol, so a correction factor (commonly of order 0.4–0.6, e.g. the widely used Karickhoff value $\approx 0.63$ for $K_{OC}$ vs $K_{OW}$) is needed to convert one to the other.

$f_{OC}$ is the fraction of organic carbon in the soil or sediment solids — a property of the sorbent, dimensionless (mass organic carbon / mass dry solids). It sets how much sorption capacity the solid phase actually has: a mineral soil with $f_{OC} \approx 0.001$ sorbs far less of a hydrophobic contaminant than an organic-rich sediment with $f_{OC} \approx 0.05$.

$K_{OW}$ is the octanol-water partition coefficient of the chemical itself — a property of the contaminant, measuring its hydrophobicity (how strongly it prefers a non-polar organic phase over water). A high-$K_{OW}$ compound (e.g., many chlorinated pesticides) partitions strongly to organic carbon wherever it is present.

Together, a high $f_{OC}$ solid and a high-$K_{OW}$ chemical produce a high $K_d$: strong sorption to soil/sediment solids, low aqueous mobility (slow transport in groundwater), but correspondingly higher persistence in sediment and bioaccumulation potential in the food chain.

(ii) NH₃/NH₄⁺ Equilibrium in the Polishing Lagoon

Given. TAN $= 20\ \text{mg/L}$, pH $= 9$, $T = 25\,{}^{\circ}\text{C}$, $K_b = 2\times10^{-5}$ for $\text{NH}_3+\text{H}_2\text{O}\rightleftharpoons\text{NH}_4^{+}+\text{OH}^{-}$.

Find. The percentage of TAN present as $\text{NH}_3\text{-N}$ and as $\text{NH}_4^{+}\text{-N}$.

Approach. Convert pH to $[\text{OH}^-]$, use $K_b$ to find the $[\text{NH}_4^+]/[\text{NH}_3]$ ratio, then convert that ratio to percentages of the fixed TAN pool.

1. Hydroxide concentration from pH. $\text{pOH} = 14 - \text{pH} = 14 - 9 = 5$, so $[\text{OH}^-] = 10^{-5}\ \text{M}$.
2. Species ratio from $K_b$. $K_b = \dfrac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}$, so $$\frac{[\text{NH}_4^+]}{[\text{NH}_3]} = \frac{K_b}{[\text{OH}^-]} = \frac{2\times10^{-5}}{1\times10^{-5}} = \boxed{2.0}.$$
3. Fractions of TAN. With $\text{NH}_4^+ = 2\,\text{NH}_3$ and $\text{NH}_3+\text{NH}_4^+=\text{TAN}$, $$\begin{aligned} f_{\text{NH}_3} &= \frac{1}{1+2} = 33.3\% \\ f_{\text{NH}_4^+} &= \frac{2}{1+2} = \boxed{66.7\%} \end{aligned}$$
4. Concentrations. Applying the fractions to the 20 mg/L TAN pool, $$\begin{aligned} \text{NH}_3\text{-N} &= 0.333(20) = \boxed{6.67\ \text{mg/L}} \\ \text{NH}_4^+\text{-N} &= 0.667(20) = \boxed{13.33\ \text{mg/L}} \end{aligned}$$

(iii) CMFR Residence Time for 99% Nitrite Removal

Given. First-order decay, rate constant $k$; target $C_{out}/C_{in} = 0.01$ (99% removal).

Find. (a) mean residence time $\tau$ for one CMFR; (b) total mean residence time for four equal-volume CMFRs in series, same overall removal.

Approach. Apply the steady-state CMFR design equation for a first-order reaction, once for a single tank and once for $N$ equal tanks in series (each with residence time $\tau/N$).

1. (a) Single CMFR. The steady-state mass balance on a CMFR with first-order decay gives $C_{out}/C_{in} = 1/(1+k\tau)$, so $$k\tau = \frac{C_{in}}{C_{out}} - 1 = \frac{1}{0.01}-1 = \boxed{99} \quad\Rightarrow\quad \tau = \frac{99}{k}.$$
2. (b) Four equal CMFRs in series. For $N$ tanks of equal residence time $\tau/N$ in series, $C_{out}/C_{in} = 1/(1+k\tau/N)^{N}$. With $N=4$ and the same target ratio, $$\left(1+\frac{k\tau}{4}\right)^{4} = 100 \;\Rightarrow\; 1+\frac{k\tau}{4} = 100^{1/4} = 3.162 \;\Rightarrow\; \frac{k\tau}{4}=2.162,$$ $$k\tau = 4(2.162) = \boxed{8.65} \quad\Rightarrow\quad \tau = \frac{8.65}{k}.$$