---
title: "Chlorine dose and contact time at pH 8 (0.5 mg/l)"
subject: "Water Supply Engineering"
chapter: 6
key: "semester_5/water_supply_engineering:chlorine-dose-contact-c72"
source: https://www.bctengineer.com/q/semester_5/water_supply_engineering/chlorine-dose-contact-c72
site: BCT Engineering
---

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

Water Supply Engineering · Chapter 6: Water Treatment

- Asked in: 2072 Chaitra, 8 marks
- 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-c72
- Whole chapter: https://www.bctengineer.com/md/semester_5/water_supply_engineering/chapter-6

## Question

What dose will be necessary at pH = 8 if 0.5 mg/l of total chlorine is required for disinfection of water at pH = 7.0? Find the contact time required at pH = 8.0, if it is given that initially 10 minutes contact time is required at pH = 7.0. Take n = 1.5 in the equation $c^n t$ = constant, $K_i = 2.7\times10^{-8}$ mol/lit.

## Short answer

HOCl fraction α = 1/(1+Ki/[H+]), α7=0.7874, α8=0.2703. To keep HOCl same, dose at pH 8 = 0.5 × α7/α8 = 1.457 mg/L. With 0.5 mg/L, contact time t8 = 10 × (α7/α8)^1.5 = 49.7 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 = $0.5\times0.7874 = 0.3937$ 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).**

## Check yourself

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

1. What is kept equal at pH 7 and pH 8 to get the same killing power?
   A. Total chlorine dose
   B. HOCl concentration
   C. OCl- concentration
   D. Contact time
   Answer: B. HOCl is the effective disinfectant, so its concentration is held the same. (Key point: Active disinfectant)

2. What is the HOCl fraction at pH 8 with Ki = 2.7x10^-8?
   A. 0.7874
   B. 0.5
   C. 0.2703
   D. 0.027
   Answer: C. alpha = 1/(1 + Ki/[H+]) = 1/(1 + 2.7) = 0.2703. (Key point: HOCl fraction)

3. What total chlorine dose is needed at pH 8?
   A. 0.5 mg/l
   B. 1.457 mg/l
   C. 0.3937 mg/l
   D. 2.91 mg/l
   Answer: B. C8 = 0.5 x 0.7874 / 0.2703 = 1.457 mg/l. (Key point: Dose at pH 8)

4. Which formula gives the new contact time?
   A. t2 = t1 (C1/C2)^n
   B. t2 = t1 (C2/C1)^n
   C. t2 = t1 x C1/C2
   D. t2 = t1 + n(C2 - C1)
   Answer: A. From C^n t = constant, t2 = t1 (C1/C2)^n, giving 2.01 min here. (Key point: Contact time from C^n t)

## Close variants set in other exams

- [Chlorine dose and contact time at pH 8 (0.7 mg/l)](https://www.bctengineer.com/question.md?key=semester_5%2Fwater_supply_engineering%3Achlorine-dose-contact-as67) — Water Supply Engineering, chapter 6, asked 2×