---
title: "Chlorine dose and contact time at pH 8 (0.7 mg/l)"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:chlorine-dose-contact-as67"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/chlorine-dose-contact-as67
site: BCT Engineering
---

# Chlorine dose and contact time at pH 8 (0.7 mg/l)

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2067 Asar (old course)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/water_supply_engineering/chlorine-dose-contact-as67
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

If 0.7 mg/lit of total chlorine is required for satisfactory disinfection of water at pH = 7.0, what dosage will be necessary at pH = 8.0? If it is given that initially 12 minutes contact time is required at pH = 7.0, find the contact time required at pH = 8.0, if n = 1.5 in the equation $c^n \times t$ = constant, take $k_i = 2.7\times10^{-8}$ mol/lit.

## Short answer

HOCl fraction at pH 7 is 0.7874, at pH 8 is 0.2703. Required dose at pH 8 = 0.7 × 0.7874 / 0.2703 = 2.039 mg/L. With original 0.7 mg/L dose, contact time t8 = 12 × (0.5512/0.1892)^1.5 = 59.7 min.

## Answer

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 = $0.7\times0.7874 = 0.5512$ mg/l. To get the same HOCl at pH 8 the total dose must be
$$
C_8 = \frac{0.7\times0.7874}{0.2703} = 2.039\ \text{mg/l}
$$

#### 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 0.7 mg/l is applied at pH 8, the HOCl present is only $0.7\times0.2703 = 0.1892$ mg/l, so
$$
t_8 = t_7\left(\frac{C_7}{C_8}\right)^{n} = 12\left(\frac{0.5512}{0.1892}\right)^{1.5} = 12\times(2.913)^{1.5} = 59.7\ \text{min}
$$

(If instead the increased dose of 2.039 mg/l is applied, the HOCl concentration equals that at pH 7, so the contact time stays at 12 min.)

**Answer: dose at pH 8 = 2.039 mg/l (about 2.91 times the dose at pH 7); contact time at pH 8 with the original 0.7 mg/l dose = 59.7 min (about 60 min).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which species is treated as the effective disinfectant when comparing pH 7 and pH 8?
   A. OCl-
   B. HOCl
   C. Cl-
   D. H+
   Answer: B. HOCl is far stronger than OCl-, so equal HOCl gives equal killing power. (Key point: Active disinfectant)

2. What dose is needed at pH 8 if 0.7 mg/l works at pH 7?
   A. 0.7 mg/l
   B. 0.5512 mg/l
   C. 2.039 mg/l
   D. 1.457 mg/l
   Answer: C. C8 = 0.7 x 0.7874 / 0.2703 = 2.039 mg/l. (Key point: Dose at pH 8)

3. Why does the contact time at pH 8 come out shorter (2.41 min) than at pH 7 (12 min)?
   A. Higher pH kills faster
   B. The dose applied is higher, and C^n t is constant
   C. OCl- is the stronger disinfectant
   D. Temperature rises at pH 8
   Answer: B. The dose rises to 2.039 mg/l, so t = 12 (0.7/2.039)^1.5 = 2.41 min. (Key point: Effect on contact time)

## Close variants set in other exams

- [Chlorine dose and contact time at pH 8 (0.5 mg/l)](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Achlorine-dose-contact-c72) — Water Supply Engineering, chapter 6, asked 2×