Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 16

Water pollution and water treatment chemistry

PROBLEMS/SOLUTIONS
1.

A city’s drinking water supply has a pH of 7.2 after filtration and chlorine is to be added as the agent
for disinfection. Calculate the ratio of HOCl to OCl- in the solution after chlorine is added. What can
be done to improve conditions for disinfection?

Solution
HOCl ↔ OCl- + H+

Ka = 3 x 10-8 (Appendix B.4)
pKa = 7.52

Using the Henderson-Hasselbalch form of the weak acid equilibrium expression:
pH = pKa + log [OCl-] / [HOCl]
7.2 = 7.52 + log [OCl-] / [HOCl]
– 0.32 = log [OCl-] / [HOCl]
[OCl-] / [HOCl] = 0.4786
1 / 0.4786 = 2.09 / 1
The ratio of HOCl to OCl- is: 2.1 to 1.
To improve conditions for disinfection, since the deprotonated species is less efficient as a disinfectant than
hypocholorous acid (HOCl), the pH of the water should be lowered to be slightly less than 7. For example a
pH of 6.9 would double the ratio to 4.2 to 1. Note, typical water quality guidelines do not allow the pH of
drinking water to fall below 6.5.

2.

Estimate the total organic carbon (TOC) concentration of waste water whose chemical oxygen demand
(COD) is 500 mg L–1 (O2). What fraction of the total (dissolved and particulate) solids content of 720
mg L–1 is then made up of organic material? Assume the organic fraction can be represented as
{CH2O}. Of what might the remaining solids consist?

Solution

(see Chapter 15, Example 15.5 for a similar calculation)

Use Equation 15.8

{CH2O} + O2 → CO2 + H2O

Given: COD = 500 mg L-1 (as O2)
the molar equivalent of {CH2O} is
500 mg L-1 ÷ 32 g mol-1 = 15.6 mmol L-1

(of O2 = {CH2O})

TOC is then equal to:
15.6 mmol L-1 {CH2O} = 15.6 mmol L-1 (as C)

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 16

Water pollution and water treatment chemistry


15.6 mmol L-1 x 12 g mol-1 = 188 mg L-1

(or 188 ppm)

Using 40% carbon in organic matter, TOC = 188/0.40 = 470 ppm
TS = 720 ppm

% organic = 470 ÷ 720 x 100% = 65 %

The remaining solids (35%) might consist of a variety of Earth materials – clay minerals, Fe(OH)3 and/or
Al(OH)3 plus other inorganic species (dissolved or particulate) including phosphates, nitrates, carbonates,
calcium, magnesium and sodium. If the water were partly from an industrial source, the dissolved and
particulate solids would have composition reflecting the nature of the industry.
3.

An influent wastewater stream contains 330 mg L–1 organic matter (both suspended and soluble) and
27 mg L–1 ammonium ion (as N). Calculate the total BOD. What assumptions is it necessary to make?

Solution
Calculate BOD given 330 mg L-1 OM and 27 mg L-1 ammonium ion as N.
Assume that the organic matter has the generic form of {CH 2O}
{CH2O} + O2 (g) → CO2 (g) + H2O

(Chapter 15, Equation 15.8)

Consider a 1 L sample.
M.W. of {CH2O} = 30.0 g mol-1
0.330 g ÷ 30.0 g mol-1 = 0.011 mol
This requires 0.011 mol of O2 (g) which is:

0.011 mol x 32.0 g mol-1 x 1000 mg g-1 = 352 mg of O2
The BOD from the organic matter in the 1 L sample is: 352 mg L-1 O2
Ammonium can be removed from wastewater through the reaction 15.1:
nitrosomonas
NH4+ (aq) + 2O2(aq) + H2O → NO3- (aq) + 2H3O+ (aq)
nitrobacter
Given 27 mg L-1 N (14 g mol-1) gives: 1.93 mmol of N (and NH4+) which requires 3.86 mmol of O2
yielding 123.4 mg of O2 per 1 L of solution.
The total BOD for this influent stream is
352 mg L-1 + 123 mg L-1 = 475 mg L-1
The assumptions that were necessary to make were: 1) that all of the organic matter was in the simple form of
{CH2O} and 2) that all of the nitrogen measured was in the form of NH4+ and that both degraded completely.

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 16

4.

Water pollution and water treatment chemistry

A wastewater contains 7.2 mg L–1 phosphorus and is treated with 15 mg L–1 aluminium in the form of
an alum solution. Assume that phosphorus is precipitated out by Reaction 16.10 and that any excess
added aluminium forms Al(OH)3. Calculate the mass of inorganic sludge produced in one day in a
plant treating 20 000 m3 of wastewater.

Solution
Wastewater with 7.2 mg L-1 P is treated with 15 mg L-1 aluminium (in the form of alum (Al3+)). Assume the

precipitation process occurs via Reaction 16.10.
Al3+(aq) + PO43-(aq) → AlPO4(s)
Any excess aluminium is further precipitated as Al(OH) 3 (Reactions 16.4-16.6).
Al3+ (aq) + 3 H2O → Al(OH)3 + 3 H+ (aq)
20 000 m3 x 1000 L m-3 = 2.0 x 107 L (of wastewater treated per day)
Total amount of P is
7.2 mg L-1 x 2.0 x 107 L = 1.44 x 108 mg = 144 kg
144 kg P ÷ 0.03097 kg mol-1 = 4.65 x 103 mol of P (= moles of PO43-) which react with Al3+ in a 1:1
ratio forming an equivalent amount of AlPO4(s).
M.M. of AlPO4(s) = 122 g mol-1
4.65 x 103 mol x 122 g mol-1 ÷ 1000 g kg-1 = 567 kg of AlPO4(s)
Excess aluminium?
15 mg L-1 x 2.0 x 107 L ÷ 106 mg kg-1 = 300 kg of aluminium
300 kg ÷ 0.027 kg mol-1 = 1.11 x 104 mol of aluminium
1.11 x 104 – 4.65x103 = 6450 mol of aluminium in excess
Al3+ and Al(OH)3 are equivalent thus giving 6450 mol of Al(OH)3.
M.M. of Al(OH)3 (s) = 78 g mol-1
6.450 x 103 mol x 78.0 g mol-1 ÷ 1000 g kg-1 = 503 kg of Al(OH)3 (s)
The total amount of inorganic sludge involving aluminium in these two simple forms from one day is
503 kg AlPO4(s) + 567 kg Al(OH)3 = 1070 kg
The mass of inorganic sludge produced in one day in a plant treating 20 000 m3 of wastewater is 1070 kg.

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 16

5.


Water pollution and water treatment chemistry

After wastewater treatment in the activated sludge process, nitrogen is mainly in the form of ammonia
and ammonium ion. Plot the fraction of nitrogen that is in the ammonia form (and therefore strippable
by air purging) as a function of pH (at 25ºC) over the pH range 6 to 10.

Solution
NH4+

NH3 + H+

Ka = 5.6 x 10-10 (Appendix B.4)

[NH3][H+]
Ka = --------------- = 5.6 x 10-10
[NH4+]
[NH3]
5.6 x 10-10
------------ = -----------[NH4+]
[H+]
[NH3] + [NH4+] = 1.0
When [H+] = 1 x 10-6

(pH =6)

[NH3]
---------------- = 5.6 x 10-4
1.0 – [NH3]

[NH4+] = 1.0 – [NH3]


or

[NH3] / [NH4+] = 5.6 x 10-4
rearrange and simplify in terms of [NH3]

[NH3] = 5.6 x 10-4 ÷ 1.00056 = 5.5969 x 10-4
[NH4+] = 0.99944
Fraction of NH3 at

(fraction of NH3)

(fraction of NH 4+) - Similarly the other points are calculated and plotted.
pH 6
pH 7
pH 8
pH 9
pH 10

=
=
=
=
=

NH4+

0.00056
0.0056
0.053

0.36
0.85

=
=
=
=
=

0.99944
0.9944
0.947
0.64
0.15

Plot of fraction of NH3 vs. pH

1
Fraction of
NH3

0.8
0.6
0.4
0.2
0

pH

6


7

8

9

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 16

6.

Water pollution and water treatment chemistry

A settling tank treating 5.5 × 106 L of water per day has dimensions as follows: length, 12.2 m; width,
7.0 m; depth, 3.5 m. Calculate the detention time for water in the tank. Calculate the minimum particle
size (expressed as diameter of spheres) that could settle in the tank.

Solution
Volume of tank = 12.2 m x 7.0 m x 3.5 m = 298.9 m3 (or 298 900 L)
Volume of water treated per day is 5.5 x 10 6 L day-1
5.5 x 106 L day-1 ÷ 24 h day-1 = 2.29 x 105 L h-1

1 day = 24 h

The detention time in the tank for the water is the volume of tank divided by the amount of water treated
per hour.

298 900 L ÷ 2.29 x 105 L h-1 = 1.3 h
The detention time for the treated water is 1.3 h.
Minimum particle size for settling: (see Example 16.1 for a similar calculation)
Assume the density of the particle is 2.65 x 103 kg m-3 (as in Example 16.1).
1.3 h = 78 min = 4680 s
The minimum rate of setting (terminal velocity) for a particle in this tank is
3.5 m ÷ 4680 s = 7.48 x 10-4 m s-1
Any slower and the particle will not have time to settle in the tank described.
Using Stoke’s relation: (see equation 16.3 for explanation of terms and symbols)

vt
-4

-1

7.48x10 m s



( p   a ) g d p

2

18 

(2650 kg m-3 – 1000 kg m-3) x 9.8 m s-2 x dp2
= --------------------------------------------------------18 x 1.0x10-3 kg m-1 s-1
dp = 2.9x10-5 m (or 29 µm)

The minimum particle size (expressed as diameter of spheres) that can settle in the tank described above is

29 µm.

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 16

7.

Water pollution and water treatment chemistry

Use Reaction 16.12 to calculate the daily volume of Ca(OH)2 solution of concentration 6.4 g L–1 Ca,
required to treat pH-adjusted wastewater containing 6.1 mg L–1 phosphorus. The treatment plant
processes 27 000 m3 of water each day. Assume a safety factor (fractional excess) of 2.

Solution (See Reaction 16.12)
5 Ca(OH)2(aq) + 3 HPO42- (aq) → Ca5OH(PO4)3 (s) + 6 OH-(aq) + 3 H2O
The amount of water treated in one day is 27 000 m3 = 2.7 x 107 L
The amount of P treated per day is
6.1 mg L-1 x 2.7 x 107 L x 10-6 kg mg -1 = 164.7 kg
1.647 x 105 g ÷ 31 g mol-1 = 5.31 x 103 mol P

(or mol of HPO42-)

stiochiometric factor for reaction with Ca(OH)2 is 1.667

(5/3)

The amount of Ca(OH)2 required is therefore

5.31 x 103 mol HPO42- x 5/3 = 8.85 x 103 mol Ca(OH)2
6.4 g Ca L-1 ÷ 40.0 g mol-1 = 0.16 mol L-1

(1 mol Ca = 1 mol Ca(OH)2)

8.85 x 103 mol ÷ 0.16 mol L-1 = 55 312 L of calcium hydroxide solution
Using a safety factor of 2 means that 1.1 x 10 5 L of 6.4 g L-1 calcium hydroxide solution are required for
the treatment of 27 000 m3 of wastewater with a P concentration of 6.1 mg L-1 per day.

8.

Nitrification of ammonium ion is one of the steps during biological nitrogen removal processes. In
waste water whose pH and alkalinity are 7.2 and 156 mg L–1 (as CaCO3), respectively, a concentration
of 7.8 mg L–1 (as N) ammonium ion is present before the process begins. Calculate the pH and
alkalinity after nitrification has gone to completion, assuming this to be the only reaction that affects
the pH.

Solution
Consider only the nitrification process.
The initial conditions given are: pH = 7.2, alkalinity = 156 mg L-1 (as CaCO3), and there is 7.8 mg L-1
NH4+ (as N).
What is the pH and alkalinity at the end of the reaction?
See Chapter 15, Reaction 15.1
nitrosomonas
NH4+ (aq) + 2O2(aq) + H2O → NO3- (aq) + 2H3O+ (aq)
nitrobacter

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 16

Water pollution and water treatment chemistry

7.8 mg L-1 NH4+ (as N) = 5.57 x 10-4 mol L-1 (NH4+)
Use a 1 L sample volume,
Therefore from the 5.57 x 10-4 mol of NH4+, 1.11 x 10-3 mol of H3O+ will be produced (1:2 ratio).
The hydronium ion will react with the alkalinity, decreasing it along with the pH.
156 mg L-1 (as CaCO3, M.W. = 100.0 g mol-1) = 3.12 x 10-3 mol L-1 (as HCO3-)
therefore, remaining [HCO3-] = 3.12 x 10-3 – 1.11 x 10-3 = 2.01 x 10-3 mol L-1
2.01 x 10-3 mol L-1 ÷ 2 x 100.0 g mol-1 x 1000 mg g-1 = 100 mg L-1
The new alkalinity is 100 mg L-1 (as CaCO3).
at pH 7.2
Ka1

[H+][HCO3-]
= ----------------- = 4.5 x 10-7
[H2CO3]

[H+] = 6.3 x 10-8, [HCO3-] = 0.00312, and [H2CO3] = 4.37 x 10-4
Total carbonate species is 0.00312 + 4.37 x 10 -4 = 3.56 x 10-3 mol L-1
Assume no change in total carbonate species in the closed system and calculate the new pH.

Ka1

[H+][HCO3-]
= ----------------- = 4.5 x 10-7
[H2CO3]


Ka1

[H+][0.00201]
= ----------------- = 4.5 x 10-7
[0.00156]

[H+] = 3.49 x 10-7

therefore the pH = 6.5

The final pH of the solution is 6.5 with an alkalinity of 100 mg L -1 (as CaCO3).

9.

A wastewater treatment plant produces sludge containing 1800 kg of dry organic solids each day.
Assuming the generic formula {CH2O} for the solids and complete anaerobic digestion by reaction
16.9, calculate the fuel value of the generated methane in joules, barrels of oil, and kilowatt hours.

Solution
Consider anaerobic digestion via reaction 16.9.
2 {CH2O} → CH4 (g) + CO2 (g)

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Solutions: Environmental Chemistry - a global perspective 4th Edition
Chapter 16

Water pollution and water treatment chemistry


What is the fuel value obtained from 1800 kg of dry organic solids per day?
Assume the M.M. of the organic matter = 30 g mol-1
1.8 x 106 g ÷ 30 g mol-1 = 6.0 x 104 mol of {CH2O}
This will produce 3.0 x 104 mol of methane (a fuel which can be burned to produce energy).
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l)

ΔHºrxn = ???

ΔHºrxn can be obtained through calculations using ΔHfº data from thermodynamic tables (Appendix B.2) or
directly from a reference. ΔHºrxn = -888.7 kJ mol-1
– 888.7 kJ mol-1 x 3.0 x 104 mol = – 2.67x107 kJ
The amount of energy in joules produced is

2.67 x 10 10 J

In Chapter 8, Table 8.6 provides units of energy and their equivalents in Joules.
The amount of energy in barrels (bbl) (petroleum) is:
1 bbl = 5.8x109 J therefore, 2.67x1010 J ÷ 5.8x109 J bbl-1 = 4.6 bbl (petroleum)
The amount of energy in kilowatt hours (kWh) is:
1 kWh = 3.6x106 J

therefore, 2.67x1010 J ÷ 3.6x106 J kwh-1 = 7419 kWh

In Summary, the fuel value of the generated methane in joules, barrels of oil, and kilowatt hours is 2.67 x
1010 J, 4.6 bbl (petroleum), and 7419 kWh, respectively.

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