Solution Manual for MWH's Water Treatment: Principles and Design, 3rd Edition

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MWH’S WATER TREATMENT
PRINCIPLES AND DESIGN
3rd Edition
by Crittenden, Trussell, Hand, Howe, and Tchobanoglous
HOMEWORK SOLUTION MANUAL
FOR
Chapter 2
Physical and Chemical
Quality of Water

Homework Solution Manual Page 2 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
PROBLEM 2-1
Problem Statement - Given the following test results, determine the mole fraction of
calcium (Ca 2+
).
Concentration, Concentration,
Cation mg/L Anion mg/L
Ca 2+ 40.0 HCO 3– 91.5
Mg2+ 12.2 SO 42– 72
Na + 15.1 Cl– 22.9
K + 5.1 NO 3– 5.0
Solution
1. By definition, using Eq. 2-2, the mole fraction of Ca 2+ is as follows.
2+
2+
2+ 2+ + + - 2- - -
3 4 3
Ca
Ca
Ca Mg Na K HCO SO Cl NO
n
x n n n n n n n n
= + + + + + + +
2. Substituting in the MW for each of the constituents yields the following:
2+
Ca
40.08
x 40.08 24.31 22.99 39.1 61 96 35.45 62
40.08 0.11
380.93
= + + + + + + +
= =
PROBLEM 2-2
Problem Statement - Determine the mole fraction of magnesium (Mg2+
) for the water
given in Problem 2-1.
Solution
1. By definition, using Eq. 2-2, the mole fraction of Mg2+ is as follows.
2+
2+
2+ 2+ + + - 2- - -
3 4 3
Mg
Mg
Ca Mg Na K HCO SO Cl NO
n
x n n n n n n n n
= + + + + + + +
2. Substituting in the MW for each of the constituents yields the following:

Homework Solution Manual Page 3 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
2+
Mg
24.31
x 40.08 24.31 22.99 39.1 61 96 35.45 62
24.31 0.06
380.93
= + + + + + + +
= =
PROBLEM 2-3
Problem Statement - Determine the mole fraction of sulfate (SO 42–
) for the water given
in Problem 2-1.
Solution
1. By definition, using Eq. 2-2, the mole fraction of 2-
4SO is as follows:
2-
4
2-
4 2+ 2+ + + - - -
3 3
SO
SO
Ca Mg Na K HCO Cl NO
n
x n n n n n n n
= + + + + + +
2. Substituting in the MW for each of the constituents yields the following:
2-
4SO
96
x 40.08 24.31 22.99 39.1 61 96 35.45 62
96 0.25
280.93
= + + + + + + +
= =
PROBLEM 2-4
Problem Statement - Commercial-grade sulfuric acid is about 95 percent H2SO 4 by
mass. If the specific gravity is 1.85, determine the molarity, mole fraction, and normality
of the sulfuric acid. Use Eq. 2-4 to determine molarity.
( )( )
mass of solute, g
M, mole/L molecular weight of solute, g/mole volume of solution, L
=
( )( )
( )( )
0.95 1.85
M, mole/L 0.0179 mole/L
98 1
= =
1. Use Eq. 2-2 to determine the mole fraction.

Homework Solution Manual Page 4 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
2 4
2 4
2 2 4
H SO
H SO
H O H SO
n
x n n
98 0.84
18 98
= +
= =
+
2. Use Eq. 2-6 to determine the normality.
( )( )
mass of solute, g
N, eq/L equivalent weight of solute, g/eq volume of solution, L
=
( )( )
( )( )
0.95 1.85
N, eq/L 0.036 eq/L
49 1
= =
PROBLEM 2-5
Problem Statement - If the UV intensity measured at the surface of a water sample is
180 mW/cm 2
, estimate the average intensity in a petri dish with an average depth
of 15 mm (used to study the inactivation of microorganisms after exposure to UV
light, as discussed in Chap. 13). Assume the absorptivity of the water, kA(λ) at λ =
254 nm, is 0.10 cm –1 and that the following form of the Beer–Lambert law applies:
A
0
I
ln 2.303k ( )x
I
  = − λ 
 
Solution
1. The variables for the Beer-Lambert law are the following:
I = unknown
I 0 = 180 mJ/cm2
kA (λ) = 0.10/cm
x = 1.5 cm
2. Rearrange the Beer-Lambert law given in the problem statement to solve for I as
follows:
( )
( )( )
A-2.303kλ x
0
-2.303 0.10 1.5
2
I I e
I 280e
I 174 mJ/cm
=
=
=

Homework Solution Manual Page 5 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
PROBLEM 2-6
Problem Statement - If the average UV intensity in a Petri dish containing water at a
depth of 10 mm is 120 mW/cm 2
, what is the UV intensity at the surface of the water
sample? Assume the absorptivity of the water, k A(λ) at λ = 254 nm, is 0.125 cm –1
and that the equation given in Problem 2-5 applies.
Solution
1. The variables for the Beer-Lambert law are the following:
I = 120 mJ/cm 2
I 0 = unknown
kA (λ) = 0.125/cm
x = 1.0 cm
2. Rearrange the Beer-Lambert law given in the problem statement to solve for I 0 as
follows:
( )
( )( )
A
0 -2.303kλ x
2
0 -2.303 0.125 1.0
I
I e
120
I 160 mJ/cm
e
=
= =
PROBLEM 2-7
Problem Statement - If the transmittance is 92 percent and a photo cell with a 12-mm
path length was used, what is the absorptivity?
Solution
1. Set up Eq. 2-12 to solve for T in terms of the variables which are given in the
problem statement; -k A(λ), and x.
a. From Eq. 2-10, = -A(λ)
b. T, % = 10 -kA(λ)x x 100
c. Variables for this equation are the following:
T, % = 92
-k A(λ) = unknown

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Homework Solution Manual Page 6 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
x = 1.2 cm
2. Rearrange Eq. 2-12 to solve for absorptivity, k A(λ), as follows:
( )
( )
A
-1
A
T
log 100
kλ x
92
log 100 0.030
1.2
kλ 0.030 cm
 
 
 − =
 
 
 = = −
=
PROBLEM 2-8
Problem Statement - Given the following data obtained on two water supply sources,
determine the constants in Eq. 2-16 and estimate the number of particles in the
size range between 2.1 and 5. Also, comment on the nature of the particle size
distributions.
Particle Count
Bin Size, mm Water A Water B
5.1–10 2500 110
10.1–15 850 80
15.1–20 500 55
20.1–30 250 36
30.1–40 80 25
40.1–50 60 20
50.1–75 28 15
75.1–100 10 10
Solution
1. Calculate the information necessary to graphically represent the data. This
information is shown in the following table.
Bin size, mm
Geometric mean
diameter, (dp)g
Water A particle
number
Water B particle
number

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Homework Solution Manual Page 7 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
5.1 – 10 7.14 a 2500 110
10.1 -15 12.31 850 80
15.1 – 20 17.38 500 55
20.1 - 30 24.56 250 36
30.1 - 40 34.70 80 25
40.1 - 50 44.78 60 20
50.1 - 75 61.30 28 15
75.1 – 100 86.66 10 10
a 7.14 5.1 10= x
2. Prepare a plot of the geometric mean diameter for the bin size, (dp)g, versus the
number of particles in the corresponding bin size.
3. Determine A and β using Eq. 9-4 and the data plot from step 2.
The value of A is determined when dp = 1 mm for the best-fit regression line
through the data, along with β, which is the power law slope coefficient.
a. Water A

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Homework Solution Manual Page 8 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
The equation for the data regression line is shown on the graph in step 2, and
the values for A and β are:
A = 222492 and β = 2.1953
b. Water B
The equation for the data regression line is shown on the graph in step 2, and
the values for A and β are:
A = 864.89 and β = 0.9903
4. Estimate the number of particles in the size range between 2.1 and 5.
a. Calculate the geometric mean size.
2.1 5 3.24× =
b. Estimate the number of particles in water A using the equation for the data
regression line determined in step 2.
( ) 2.1953
n 3.24 222492 3.24 16,842−
= × =
c. Estimate the number of particles in water B using the equation for the data
regression line determined in step 2.
( ) 0.9903
n 3.24 864.89 3.24 270−
= × =
5. Comment on the nature of the particle size distributions.
Parameter A is much greater for water A than for water B. As the value of A
increases, the total number of particles in each size classification increases, thus,
the total number of particles in water A in each size classification is greater than in
water B.
The slope β is a measure of the relative number of particles in each size range. Thus, if
β is less than one 1, the particle size distribution is dominated by large particles, if β is
equal to one all particle sizes are represented equally, and if β is greater the one the
particle size distribution is dominated by small particles. The particle size distribution is
dominated by small particles for water A because β (2.1953) is greater than one. All
particle sizes are represented nearly equally for water B because β (0.9903) is close to
one.

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Homework Solution Manual Page 9 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
PROBLEM 2-9
Problem Statement - The following particle size data were data obtained for the
influent and effluent from a granular medium filter. Determine the constants in Eq.
2-16 and assess the effect of the filter in removing particles.
Particle Count
Bin Size, mm Influent Effluent
2.51–5 20000 101
5.1–10 8000 32
10.1–20 2000 6
20.1–40 800 3.2
40.1–80 400 1.2
80.1–160 85 0.34
160.1–320 40 0.12
Solution
1. Calculate the information necessary to graphically represent the data. This
information is shown in the following table.
Bin size, mm
Geometric mean
diameter, (dp)g
Influent particle
number
Effluent particle
number
2.51 - 5 3.54 a 20000 101
5.1 – 10 7.14 8000 32
10.1 – 20 14.21 2000 6
20.1 - 40 28.35 800 3.2
40.1 – 80 56.64 400 1.2
80.1 - 160 113.21 85 0.34
160.1 - 320 226.34 40 0.12
a 3.54 2.51 5= ×
2. Prepare a plot of the geometric mean diameter for the bin size, (dp)g, versus the
number of particles in the corresponding bin size.

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Homework Solution Manual Page 10 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
3. Determine A and β:
a. Influent
The equation for the data regression line is shown on the graph in step 2,
and the values for A and β are:
A = 136189 and β = 1.514
b. Effluent
The equation for the data regression line is shown on the graph in step 2,
and the values for A and β are:
A = 652.15 and β = 1.595
4. Assess the effect of the filter in removing particles.
a. Parameter A is much greater for the influent than it is for the effluent.
Therefore, the effluent contains fewer particles than the influent, meaning the
filter successfully removed many particles.
b. Both the influent and the effluent are dominated by small particles
because β (1.514 for influent and 1.595 for effluent) is greater than one. The
filter is effective in removing all sizes of particles because β for the effluent is
close to the value for the influent.

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Homework Solution Manual Page 11 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
PROBLEM 2-10
Problem Statement - Determine the alkalinity and hardness in milligrams per liter as
CaCO 3 for the water sample in Problem 2-1.
Solution
1. Determine alkalinity.
a. Use Eq. 2-26 and determine the alkalinity in milliequivalents per liter.
Alk, meq/L = (HCO 3- ) + (CO 32- ) + (OH- ) - (H +)
Alkalinity ions Conc., mg/L mg/meqa meq/L
HCO3- 91.5 61.02 1.50
a Molecular weight/Z (see Eq. 2-7)
b. Adding the conversion from meq/L to mg/L as CaCO 3 (see page xxx of the text)
with Eq. 2-26 yields the following:
3 3Alkalinity as CaCO meq/L of substance x meq/L mass of CaCO , 50 mg/meq=
( )( ) 3Alk 1.50 50 75 mg / L as CaCO= =
2. Determine hardness.
a. Use Eq. 2-23 and determine the hardness in milliequivalents per liter.
Hardness, meq/l = (Ca +2) + (Mg+2)
Hardness ions Conc., mg/L mg/meqa meq/L
Ca2+ 40 20.04b 2.00
Mg2+ 12.2 12.15 1.00
∑ hardness ions 3.00
a Molecular weight/Z (see Eq. 2-8)
b. Adding the conversion from meq/L to mg/L as CaCO 3 (see page xxx of the
text) with Eq. 2-23 yields the following:
3
3
Hardness meq/L of substance x meq/L mass of CaCO , 50 mg/meq
as CaCO =
( )( ) 3Hardness 3 50 150 mg/L as CaCO= =

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Homework Solution Manual Page 12 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
PROBLEM 2-11
Problem Statement - Given the following incomplete water analysis, determine the
unknown values if the alkalinity and noncarbonate hardness are 50 and 150 mg/L
as CaCO 3, respectively:
Ion Concentration,
mg/L
Ca 2+ 42.0
Mg2+ ?
Na + ?
K + 29.5
HCO 3– ?
SO 42– 96.0
Cl– 35.5
NO 3– 4.0
Solution
1. Determine the concentration of HCO 3- using Eq. 2-26.
Alk, meq/L = (HCO 3- ) + (CO 32- ) + (OH- ) - (H +)
a. Convert alkalinity from mg/L as CaCO 3 to meq/L.
3
3
3
3
mg/L alkalinity as CaCO :
meq/L of alkalinity ions x meq/L mass of CaCO , 50 mg/meq
mg/L alkalinity as CaCO
meq/L of alkalinity ions meq/L mass of CaCO
50 mg/L
meq/L of alkalinity ions 1 meq/L
50 mg/meq
=
=
= =
b. Use Eq. 2-26 to determine the alkalinity ions that are present in the water
analysis. The only alkalinity ion listed in the water analysis is -
3HCO , which is
unknown.
c. Convert the concentration of -
3HCO in meq/L to mg/L as follows:
Alkalinity ions Conc., mg/L mg/meqa meq/L

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Homework Solution Manual Page 13 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
HCO3- 61.02 61.02 1.00
a Molecular weight/Z (see Eq. 2-7)
2. Determine the value of Mg2+ using Eq. 2-23. (Non-carbonate hardness is the
concentration of Ca 2+ and Mg2+ associated with nonalkalinity anions such as Cl-
and SO 42-
. In this problem, 1.0 meq/L of Ca 2+ are associated with HCO 3-
an
alkalinity anion. The remaining concentration of Ca 2+
, 1.1 meq/L and all of the
Mg2+ are associated with SO 42- and Cl-
, both nonalkalinity anions.)
a. Convert hardness from mg/L as CaCO 3 to meq/L.
3
3
3
3
mg/L hardness meq/L of hardness ions x meq/L mass of CaCO , 50 mg/meq
as CaCO
mg/L hardness as CaCO
meq/L of hardness ions meq/L mass of CaCO
150 mg/L
meq/L of hardness ions 3 meq/L
50 mg/meq
=
=
= =
b. Use Eq. 2-23 and determine the hardness ions that are present in the water
analysis.
Hardness, meq/l = (Ca 2+) + (Mg2+)
The ions listed in the water analysis are Ca 2+ and Mg2+; the concentration of
Mg2+ unknown.
c. Determine the concentration of Mg2+ and convert the concentration of Mg2+ in
meq/L to mg/L as follows:
Hardness ions Conc., mg/L mg/meqa meq/L
Ca2+ 42.0 20.04b 2.1
Mg2+ 10.94 12.15 0.9
∑ hardness ions 3.0
a Molecular weight/Z (see Eq. 2-8)
3. Determine the concentration of Na + from an anion, cation balance.
a. Prepare a cation-anion balance.
Cation
Conc.,
mg/L mg/meqa meq/L Anion
Conc.,
mg/L mg/meqa meq/L

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Homework Solution Manual Page 14 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
Ca2+ 42.0 20.04b 2.1 HCO3- 61.02 61.02 1.00
Mg2+ 10.94 12.15 0.9 SO42- 96.0 48.03 2.00
Na+ 23 Cl- 35.5 35.45 1.00
K+ 29.5 39.1 0.75 NO3- 4.0 62.01 0.06
∑ cations 3.75 ∑ anions 4.06
a Molecular weight/Z (see Eq. 2-7)
b For calcium, equivalent weight = 40.08/2 = 20.04 g/eq or 20.04 mg/meq
b. Calculate the concentration of Na+ from the difference between the sum of
the cations and the sum of the anions.
+
+
Na meq/L anions cations
Na meq/L 4.06 3.75 0.31
= −
= − =
∑ ∑
( )( )
+ + +
+
Na mg/L Na mg/meq x Na meq/L
Na mg/L 23 0.31 7.13 mg/L
=
= =
PROBLEM 2-12
Instructors Note: See Chap. 19 for a more thorough discussion of hardness, where
total, carbonate, and non-carbonate hardness are explained.
Problem Statement - Given the following incomplete water analysis measured at 25ºC,
determine the unknown values if the alkalinity and noncarbonate hardness are 40
and 180 mg/L as CaCO 3:
Ion Concentration, mg/L
Ca 2+ 55.0
Mg2+ ?
Na + 23.0
K + ?
HCO 3– ?
SO 42– 48.0
Cl– ?
CO 2 4.0

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Homework Solution Manual Page 15 of 18
MWH’s Water Treatment: Principles and Design, 3rd ed. Version 1
Chapter 2 - Physical and Chemical Quality of Water
Solution
1. Determine the concentration of -
3HCO using Eq. 2-26.
a. Convert alkalinity from mg/L as CaCO 3 to meq/L.
3
3
3
3
mg/L alkalinity =meq/L of alkalinity ions x meq/L mass of CaCO , 50 mg/meq
as CaCO
mg/L alkalinity as CaCO
meq/L of alkalinity ions meq/L mass of CaCO
40 mg/L
meq/L of alkalinity ions 0.8 meq/L
50 mg/meq
=
= =
b. Use Eq. 2-26 and determine the alkalinity ions that are present in the water
analysis.
- 2- - +
3 3Alk, meq/L = (HCO ) + (CO ) + (OH ) - (H )
The only alkalinity ion listed in the water analysis is -
3HCO , which is unknown.
c. Convert the concentration of -
3HCO in meq/L to mg/L as follows:
Alkalinity ions Conc., mg/L mg/meqa meq/L
HCO3- 48.82 61.02 0.80
a Molecular weight/Z (see Eq. 2-8)
2. Determine the value of Mg2+ using Eq. 2-23, which is for non-carbonate hardness.
a. Convert hardness from mg/L as CaCO 3 to meq/L.
3
3
3
3
mg/L hardness meq/L of hardness ions x meq/L mass of CaCO , 50 mg/meq
as CaCO
mg/L hardness as CaCO
meq/L of hardness ions meq/L mass of CaCO
180 mg/L
meq/L of hardness ions 3.6 meq/L
50 mg/meq
=
=
= =
b. Use Eq. 2-23 and determine the hardness ions that are present in the water
analysis.
Hardness, meq/l = (Ca 2+) + (Mg2+)
The hardness ions listed in the water analysis are Ca 2+ and Mg2+, and the
concentration of Mg2+ is unknown.

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