Chlorine dose and contact time at pH 8 (0.5 mg/l)

Hypochlorous acid (HOCl) is the effective disinfectant; the hypochlorite ion (OCl$^-$) is far weaker and is neglected. The same killing power is obtained at both pH values if the HOCl concentration is kept the same.

$$HOCl \rightleftharpoons H^+ + OCl^-, \qquad K_i=\frac{[H^+][OCl^-]}{[HOCl]}=2.7\times10^{-8}\ \text{mol/l}$$

Fraction of total chlorine present as HOCl:

$$\alpha = \frac{[HOCl]}{[HOCl]+[OCl^-]}=\frac{1}{1+K_i/[H^+]}$$

Step 1: HOCl fraction at the two pH values

• pH 7.0: $[H^+]=10^{-7}$; $K_i/[H^+] = 0.27$; $\alpha_7 = 1/1.27 = 0.7874$
• pH 8.0: $[H^+]=10^{-8}$; $K_i/[H^+] = 2.7$; $\alpha_8 = 1/3.7 = 0.2703$

Step 2: Dose at pH 8.0

HOCl at pH 7 = $0.5\times0.7874 = 0.3937$ mg/l. To get the same HOCl at pH 8 the total dose must be

$$C_8 = \frac{0.5\times0.7874}{0.2703} = 1.457\ \text{mg/l}$$

Step 3: Contact time at pH 8.0

The kill depends on the HOCl concentration, so $C$ in $C^n t=\text{constant}$ is the HOCl concentration. If the same total dose of 0.5 mg/l is applied at pH 8, the HOCl present is only $0.5\times0.2703 = 0.1351$ mg/l, so

$$t_8 = t_7\left(\frac{C_7}{C_8}\right)^{n} = 10\left(\frac{0.3937}{0.1351}\right)^{1.5} = 10\times(2.913)^{1.5} = 49.7\ \text{min}$$

(If instead the increased dose of 1.457 mg/l is applied, the HOCl concentration equals that at pH 7, so the contact time stays at 10 min.)

Answer: dose at pH 8 = 1.457 mg/l (about 2.91 times the dose at pH 7); contact time at pH 8 with the original 0.5 mg/l dose = 49.7 min (about 50 min).