1. When water leaves a water treatment plant and enters the distribution system

ID: 1018684 • Letter: 1

Question

1. When water leaves a water treatment plant and enters the distribution system that transports it to homes and other users the water often contains a small chlorine residual* to ensure that the water does not become contaminated before it reaches the consumer. A chlorine residual concentration of ~1mg/l (as chlorine) is common. Hypochlorous acid is a weak acid with a pKa of 7.53

A. If water leaving a treatment plant has a pH of 7 and a chlorine residual (HOCl and OCl- only) of 1.00 mg/l as chlorine, what fraction of the chlorine is present as

i.HOCl

ii.OCl-

B. If hypochlorous acid (HOCl/OCl-) is the only weak acid in the system how many moles of strong base (NaOH, for instance) would be required to raise the pH from 7 to 8?

C. Compare your answer in part B to the number of moles required to increase the pH of pure water from 7 to 8.

*The residual is present in the form of chlorine gas (Cl2), hypochlorous acid (HOCl), and/or hypochlorite ion (OCl-). For this problem all of the chlorine residual is assumed to be present as either HOCl or OCl-.

Explanation / Answer

pH = pKa + log base/acid

7 = 7.53 + log base/acid

log base/acid = -0.53

base/acid = 10-0.53

base/acid = 0.295

So if base or OCl- is 0.295 then HOCl will be 1.

Fraction as OCl- is 0.227 and fraction as HOCl is 0.772.

chlorine present is 1mg/L or 1/35.5 = 0.0281 mmol/L

pH = pKa + log base/acid

8 = 7.53 + log base/acid

log base/acid = 0.47

base /acid = 100.47

base/acid = 2.95

So base = 2.95 and acid = 1

so total chlorine of 0.0281 mmol is present in this

so base form is 0.0281/3.95 x 2.95 = 0.0210

so base increased from 0.295 to 2.95 to do so we need to add 2.95 - 0.295 = 2.65 fraction so NaOH will be

0.0281/3.95 x 2.65 = 0.0188 mmol of NaOH/L

0.0188 x 10-3 mol NaOH/L of water.

C) To raise the pH of pure water from 7 to 8

we will need

[OH-] for pH 7 = pOH 7

[OH-] = 10-pOH

[OH-] = 10-7 = 1 x 10-7 M

when pH = 8

pOH = 6

[OH-] = 10-pOH

[OH-] = 10-6 = 1 x 10-6M

So increase in [OH-] = 1 x 10-6 - 1x 10-7 = 9 x 10-7 moles/L of NaOH is required for pure water

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1. When water leaves a water treatment plant and enters the distribution system

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