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a sample chapter from WSO Water
Treatment, Grade 2, get your full copy
from AWWA.
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© 2016 American Water Works Association
Contents
Chapter 1 Basic Microbiology and Chemistry
Chemical Formulas and Equations
Moles and Molarity
Equivalent Weights and Normality
Dilution Calculations
Standard Solutions
1
1
6
7
8
12
15
Chapter 2 Operator Math
Volume Measurements
15
Conversions22
Average Daily Flow
30
Surface Overflow Rate
33
Weir Overflow Rate
36
Filter Loading Rate
38
Filter Backwash Rate
41
Mudball Calculation
43
Detention Time
44
Pressure47
Flow Rate Problems
52
Chemical Dosage Problems
55
Chapter 3 USEPA Water Regulations
Types of Water Systems
Disinfection B
­ y-­product and Microbial Regulations
71
Chapter 4 Coagulation and Flocculation Process Operation
Operation of the Processes
Dosage Control
Safety Precautions
Record Keeping
89
71
72
89
93
93
94
97
Chapter 5 Sedimentation and Clarifiers
Process Description
97
Sedimentation Facilities
98
Other Clarification Processes
104
Regulations107
Operation of the Process
107
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iv WSO Water Treatment Grade 2
Chapter 6 Filtration115
Equipment Associated With Gravity Filters
115
Operation of Gravity Filters
123
Pressure Filtration
135
Regulations138
Safety Precautions
138
Record Keeping
139
Chapter 7 Chlorine Disinfection
Gas Chlorination Facilities
Hypochlorination Facilities
Operation of the Chlorination Process
Chlorination Operating Problems
Safety Precautions
Record Keeping
143
143
155
157
162
165
170
173
Chapter 8 Iron and Manganese Treatment
Excessive Iron and Manganese
173
Control Processes
175
Control Facilities
178
Regulations182
Manganese Greensand Filter Operation
182
Process Monitoring
188
Operating Problems
188
Record Keeping
188
Chapter 9 Fluoridation Process Operation
Operation of the Fluoridation Process
Fluoridation Operating Problems
Control Tests
Safety Precautions
Record Keeping
191
Chapter 10 Water Quality Testing
Testing and Laboratory Procedures
Physical and Aggregate Properties of Water
199
Chapter 11 Corrosion Control
Purposes of Corrosion and Scaling Control
Water System Corrosion
Scale Formation
Corrosion and Scaling Control Methods
Corrosion and Scaling Control Facilities
Chemical Feed Equipment
221
191
193
194
195
196
199
209
221
222
226
227
230
233
237
Chapter 12 Lime Softening
Lime Softening Chemical Reactions
237
Lime Softening Facilities
239
Regulations244
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Contents v
Chapter 13 Ion Exchange
247
Ion Exchange Softening
247
Facilities247
Operation of Ion Exchange Processes
252
Operating Problems
254
Ion Exchange for Removal of Arsenic, Barium, Radium, Nitrate,
TOC, and Uranium
255
Activated Alumina Fluoride Removal Process
257
Adsorptive Media
257
261
Chapter 14 Activated Carbon Adsorption
The Principle of Adsorption
261
Adsorption Facilities
262
Powdered Activated Carbon
263
Granular Activated Carbon
265
Regulations268
Operating Procedures for Adsorption
268
Operating Problems
272
Control Tests
273
Record Keeping
276
Chapter 15 Aeration279
Water-Into-Air Aerators
279
Air-Into-Water Aerators
284
Combination Aerators
285
Chapter 16 Membrane Treatment
Microfiltration Facilities
Pleated Membrane Facilities
Reverse Osmosis Facilities
291
Chapter 17 Plant Waste Treatment and Disposal
Removal of Sludge from Conventional Sedimentation Processes
Softening Sludge Handling, Dewatering,
and Disposal
Solids Separation Technologies
303
291
294
297
303
305
306
313
Chapter 18 Instrumentation and Control Systems
Flow, Pressure, and Level Measurement
313
Other Operational Control Instruments
316
Automation317
Computerization318
Chapter 19 Centrifugal Pumps
Operation of Centrifugal Pumps
Centrifugal Pump Maintenance
Record Keeping
Pump Safety
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321
321
325
336
336
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Chapter 20 Treatment Plant Safety and Security Practices
Treatment Plant Safety Review
Plant Security
339
339
348
355
Chapter 21 Administration, Records, and Reporting Procedures
Process Records
355
Reporting359
Plant Performance Reports
360
Public Relations
360
Chapter 22 Additional Level 2 Study Questions
363
Study Question Answers 369
Glossary 379
Index 387
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Chapter
Chapter 11
Basic Microbiology and
Chemistry
Chemical Formulas and Equations
A group of chemically bonded atoms forms a particle called a molecule. The simplest molecules contain only one type of atom, such as when two atoms of oxygen
combine (O2) or when two atoms of chlorine combine (Cl2). Molecules of compounds are made up of the atoms of at least two different elements; for example,
one oxygen atom and two hydrogen atoms form a molecule of the compound
water (H2O). “H2O” is called the chemical formula of water. The formula is
a shorthand way of writing what elements are present in a molecule of a compound, and how many atoms of each element are present in each molecule.
Reading Chemical Formulas
The following are examples of chemical formulas and what they indicate.
Example 1
The chemical formula for calcium carbonate is
CaCO3
According to the formula, what is the chemical makeup of the compound?
First, the letter symbols given in the formula indicate the three elements
that make up the calcium carbonate compound:
Ca = calcium
C = carbon
O = oxygen
Second, the subscripts (the small numbers at the lower right corners of
the letter symbols) in the formula indicate how many atoms of each element
are present in a single molecule of the compound. There is no number just to
the right of the Ca or C symbols; this indicates that only one atom of each
element is present in the molecule. The subscript 3 to the right of the O symbolizing oxygen indicates that there are three oxygen atoms in each molecule.
CaCO3
1 atom
3 atoms
1 atom
chemical formula
Using the chemical
symbols for each
element, a shorthand
way of writing what
elements are present in a
molecule and how many
atoms of each element
are present in each of the
molecules.
1
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Determining Percent by Weight of Elements in a
Compound
If 100 lb of sodium chloride (NaCl) were separated into the elements that make
up the compound, there would be 39.3 lb of pure sodium (Na) and 60.7 lb of
pure chlorine (Cl). We say that sodium chloride is 39.3 percent sodium by weight
and that it is 60.7 percent chlorine by weight. The percent by weight of each
element in a compound can be calculated using the compound’s chemical formula
and atomic weights from the periodic table.
The first step in calculating percent by weight of an element in a compound
is to determine the molecular weight (sometimes called formula weight) of the
compound. The molecular weight of a compound is defined as the sum of the
atomic weights of all the atoms in the compound.
For example, to determine the molecular weight of sodium chloride, first
count how many atoms of each element a single molecule contains:
NaCl
1 atom
1 atom
Next, find the atomic weight of each atom, using the periodic table:
atomic weight of Na = 22.99
atomic weight of Cl = 35.45
Finally, multiply each atomic weight by the number of atoms of that element in
the molecule, and total the weights:
Number of
Atoms
Atomic
Weight
Total
Weight
sodium (Na)
1
×
22.99
=
22.99
chlorine (Cl)
1
×
35.45
=
35.45
=
58.44
molecular weight of NaCl
Once the molecular weight of a compound is determined, the percent by
weight of each element in the compound can be found with the following formula:
percent element by weight =
percent by weight
The proportion,
calculated as a
percentage, of each
element in a compound.
molecular weight
The sum of the atomic
weights of all the atoms
in the compound. Also
called formula weight.
000200010272023365_ch01_p001-014.indd 2
weight of element in compound
molecular weight of compound
× 100
Using the formula, first calculate the percent by weight of sodium in the
compound:
percent Na by weight =
=
weight of Na in compound
molecular weight of compound
× 100
22.99
× 100
58.44
= 0.393 × 100
= 39.3% sodium by weight
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Basic Microbiology and Chemistry 3
Then, calculate percent by weight of chlorine in the compound:
percent Cl by weight =
=
weight of Cl in compound
molecular weight of compound
× 100
35.45
× 100
58.44
= 0.607 × 100
= 60.7% chlorine by weight
To check the calculations, add the percentages. The total should be 100:
39.3%
+ 60.7%
100.0%
Na
Cl
NaCl
Chemical Equations
A chemical equation is a shorthand way, through the use of chemical formulas,
to write the reaction that takes place when certain chemicals are brought together.
As shown in the following example, the left side of the equation indicates the
reactants, or chemicals that will be brought together; the arrow indicates which
direction the reaction occurs; and the right side of the equation indicates the products, or results, of the chemical reaction.
calcium
bicarbonate
plus
calcium
hydroxide
Ca(HCO3)2
+
Ca(OH)2
Reactants
react to
form
calcium
carbonate
plus
water
2CaCO3
+
2H2O
Products
The 2 in front of CaCO3 is called a coefficient. A coefficient indicates the
relative number of molecules of the compound that are involved in the chemical
reaction. If no coefficient is shown, then only one molecule of the compound is
involved. For example, in the preceding equation, one molecule of calcium bicarbonate reacts with one molecule of calcium hydroxide to form two molecules
of calcium carbonate and two molecules of water. Without the coefficients, the
equation could be written
Ca(HCO3)2 + Ca(OH)2
CaCO3 + CaCO3 + H2O + H2O
If you count the atoms of calcium (Ca) on the left side of the equation and
then count the ones on the right side, you will find that the numbers are the same.
In fact, for each element in the equation, as many atoms are shown on the left
side as on the right. An equation for which this is true is said to be balanced. A
balanced equation accurately represents what really happens in a chemical reaction: because matter is neither created nor destroyed, the number of atoms of
each element going into the reaction must be the same as the number coming out.
Coefficients allow balanced equations to be written compactly.
000200010272023365_ch01_p001-014.indd 3
chemical equation
A shorthand way, using
chemical formulas, of
writing the reaction
that takes place when
chemicals are brought
together. The left side of
the equation indicates
the chemicals brought
together (the reactants);
the arrow indicates in
which direction the
reaction occurs; and the
right side of the equation
indicates the results
(the products) of the
chemical reaction.
coefficient
An indication of the
relative number of
molecules of the
compound that are
involved in the chemical
reaction.
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4 WSO Water Treatment Grade 2
Coefficients and subscripts can be used to calculate the molecular weight of
each term in an equation, as illustrated in the following example.
Example 2
Calculate the molecular weights for each of the four terms in the following
equation:
Ca(HCO3)2 + Ca(OH)2
2CaCO3 + 2H2O
First, calculate the molecular weight of Ca(HCO3)2:
Number of
Atoms
Atomic
Weight
Total
Weight
calcium (Ca)
1
×
40.08
=
40.08
hydrogen (H)
2
×
1.01
=
2.02
carbon (C)
2
×
12.01
=
24.02
oxygen (O)
6
×
16.00
=
96.00
=
162.12
molecular weight of Ca(HCO3)2
The molecular weight for Ca(OH)2 is determined as follows:
Number of
Atoms
Atomic
Weight
Total
Weight
calcium (Ca)
1
×
40.08
=
40.08
oxygen (O)
2
×
16.00
=
32.00
hydrogen (H)
2
×
1.01
=
2.02
=
74.10
molecular weight of Ca(OH)2
The coefficient 2 in front of the next term of the equation (2CaCO3) indicates that two molecules of CaCO3 are involved in the reaction. First find the
weight of one molecule, then double that weight to determine the weight of
two molecules:
Number of
Atoms
000200010272023365_ch01_p001-014.indd 4
Atomic
Weight
Total
Weight
calcium (Ca)
1
×
40.08
=
40.08
carbon (C)
1
×
12.01
=
12.01
oxygen (O)
3
×
16.00
=
48.00
weight of one molecule of CaCO3
=
100.09
weight of two molecules of CaCO3
=
(2)(100.09)
=
200.18
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Basic Microbiology and Chemistry 5
The coefficient in front of the fourth term in the equation (2H2O) also
indicates that two molecules are involved in the reaction. As in the last calculation, first determine the weight of one molecule of H2O, then the weight of
two molecules:
Number of
Atoms
Atomic
Weight
Total
Weight
hydrogen (H)
2
×
1.01
=
2.02
oxygen (O)
1
×
16.00
=
16.00
weight of one molecule of H 2O
=
18.02
weight of two molecules of H 2O
=
(2)(18.02)
=
36.04
In summary, the weights that correspond to each term of the equation are
Ca(HCO3)2 + Ca(OH)2
162.12
2CaCO3 + 2H2O
74.10
200.18
36.04
Notice that the total weight on the left side of the equation (236.22) is equal
to the total weight on the right side of the equation (236.22), meaning the
equation is balanced.
The practical importance of the weight of each term of the equation is that
the chemicals shown in the equation will always react in the proportions indicated by their weights. For example, from the calculation above, you know that
Ca(HCO3)2 reacts with Ca(OH)2 in the ratio 162.12:74.10. This means that, given
162.12 lb of Ca(OH)2, you must add 74.10 lb of Ca(HCO3)2 for a complete
reaction. Given twice the amount of Ca(HCO3)2 (that is, 324.24 lb), you must
add twice the amount of Ca(OH)2 (equal to 148.20 lb) to achieve complete reaction. The next two examples illustrate more complicated calculations using the
same principle.
Example 3
If 25 g of Ca(OH)2 were added to some Ca(HCO3)2, how many grams of
Ca(HCO3)2 would react with the Ca(OH)2?
Remember, the molecular weights indicate the weight ratio in which the
two compounds will react. The molecular weight of Ca(HCO3)2 is 162.12,
and the molecular weight of Ca(OH)4)2 is 74.10. Use this information to set
up a proportion in order to determine how many grams of Ca(HCO3)2 will
react with the Ca(OH)2:
known ratio
74.10 g Ca(OH)2
162.12 g Ca(HCO3)2
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desired ratio
=
25 g Ca(OH)2
x g Ca(HCO3)2
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6 WSO Water Treatment Grade 2
Next, solve for the unknown value:
74.10
25
=
162.12
x
(x)(74.10)
= 25
162.12
x=
(25)(162.12)
74.10
x = 54.7 g Ca(HCO3)2
Given the molecular weights and the chemical equation indicating the ratio by which the two chemicals would combine, we were able to calculate that
54.7 g of Ca(HCO3)2 would react with 25 g of Ca(OH)2.
Moles and Molarity
You may sometimes find chemical reactions described in terms of moles of a substance. The measurement mole (an abbreviation for ­gram-­mole) is closely related
to molecular weight. The molecular weight of water, for example, is 18.02—and
1 mol of water is defined to be 18.02 g of water. (The abbreviation for mole is
mol.) The general definition of a mole is as follows:
A mole of a substance is a number of grams of that substance, where the number
equals the substance’s molecular weight.
In example 2, you saw that the following equation and molecular weights
were correct:
Ca(HCO3)2 + Ca(OH)2
162.12
74.10
2CaCO3 + 2H2O
2(100.09) 2(18.02)
If 162.12 g of Ca(HCO3)2 were used in the reaction, then the ratio equations
given in example 3 would show that the weights of each of the substances in the
reaction were as follows:
Ca(HCO3)2 + Ca(OH)2
mole
The quantity of a
compound or element
that has a weight in
grams equal to the
substance’s molecular or
atomic weight. Used in
this text generally as
an abbreviation for
­gram-­mole.
000200010272023365_ch01_p001-014.indd 6
162.12 g
2CaCO3 + 2H2O
74.10 g (2)(100.09) g (2)(18.02) g
Because of the way a mole is defined, this could also be written in a more compact
form:
Ca(HCO3)2 + Ca(OH)2
1 mole
1 mole
2CaCO3 + 2H2O
2 moles 2 moles
Reading this information, a chemist could state, “One mole of Ca(HCO 3)2 is
needed to react with one mole of Ca(OH)2, and the reaction yields two moles of
CaCO3 and two moles of water.”
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Basic Microbiology and Chemistry 7
When measuring chemicals in moles, always remember that the weight of a
mole of a substance depends on what the substance is. One mole of water weighs
18.02 g—but one mole of calcium carbonate weighs 100.09 g.
Example 4
A lab procedure calls for 3.0 mol of sodium bicarbonate (NaHCO3) and
0.10 mol of potassium chromate (K2CrO4). How many grams of each compound are required?
To find the grams required of NaHCO3, first determine the weight of
1 mol of the compound:
Number of
Atoms
Atomic
Weight
Total
Weight
sodium (Na)
1
×
22.99
=
22.99
hydrogen (H)
1
×
1.01
=
1.01
carbon (C)
1
×
12.01
=
12.01
oxygen (O)
3
×
16.00
=
48.00
=
84.01
molecular weight of NaHCO3
Therefore, 1 mol of NaHCO3 weighs 84.01 g.
Next, multiply the weight of 1 mol by the number of moles required.
The required amount of NaHCO3 is 3 mol. The weight of 3 mol NaHCO3 is
(3)(84.01 g) = 252.03 g.
To find grams required of K2CrO4, first determine the weight of 1 mol of
the compound:
Number of
Atoms
Atomic
Weight
Total
Weight
potassium (K)
2
×
39.10
=
78.20
chromium (Cr)
1
×
52.00
=
52.00
oxygen (O)
4
×
16.00
=
64.00
=
194.20
molecular weight of K 2CrO4
Therefore, 1 mol of K2CrO4 weighs 194.20 g.
Next, multiply the weight of 1 mol by the number of moles required.
The required amount of K2CrO4 is 0.10 mol. This amount weighs (0.10)
(194.20 g) = 19.42 g.
Equivalent Weights and Normality
Another method of expressing the concentration of a solution is normality. Normality depends in part on the valence of an element or compound. An element
or compound may have more than one valence, and it is not always clear which
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8 WSO Water Treatment Grade 2
valence (and therefore what concentration) a given normality represents. Because
of this problem, normality is being replaced by molarity as the expression of concentration used for chemicals in the lab.
In the lab, you will often have detailed, s­ tep-­by-step instructions for preparing
a solution of a needed normality. Nonetheless, it is useful to have a basic idea of
what the measurement means. To understand normality, you must first understand equivalent weights.
Equivalent Weights
The equivalent weight of an element or compound is the weight of that element
or compound that, in a given chemical reaction, has the same combining capacity as 8 g of oxygen or as 1 g of hydrogen. The equivalent weight may vary with
the reaction being considered. However, one equivalent weight of a reactant will
always react with one equivalent weight of the other reactant in a given reaction.
Although you are not expected to know how to determine the equivalent
weights of various reactants at this level, it will help if you remember one characteristic: the equivalent weight of a reactant either will be equal to the reactant’s
molecular weight or will be a simple fraction of the molecular weight. For example, if the molecular weight of a compound is 60.00 g, then the equivalent weight
of the compound in a reaction will be 60.00 g or a simple fraction (usually 1∕2, 1∕3,
1∕4, 1∕5, or 1∕6) of 60.00 g.
Normality
Normality is defined as the number of equivalent weights of solute per liter of
solution. Therefore, to determine the normality of a solution, you must first determine how many equivalent weights of solute are contained in the total weight of
dissolved solute. Use the following equation:
number of equivalent weights =
equivalent weight
The weight of a
compound that contains
one equivalent of a
proton (for acids) or
one equivalent of a
hydroxide (for bases).
The equivalent weight
can be calculated by
dividing the molecular
weight of a compound
by the number of H+
or OH – present in the
compound.
normality
The number of
equivalent weights
of solute per liter of
solution.
000200010272023365_ch01_p001-014.indd 8
total weight
equivalent weight
Dilution Calculations
Sometimes a particular strength of solution will be created by diluting a strong
solution with a weak solution of the same chemical. The new solution will have a
concentration somewhere between the weak and the strong solutions.
Although there are several methods available for determining what amounts
of the weak and strong solutions are needed, perhaps the easiest is the rectangle
method (sometimes called the dilution rule), shown in Figure 1-1.
Solution A,
= A%
higher
concentration
Solution B,
= B%
lower
concentration
C – B = Parts of A
required
Solution C,
desired
concentration
= C%
Sum = total
parts in
desired solution
A – C = Parts of B
required
Figure 1-1 Schematic for rectangle method
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Basic Microbiology and Chemistry 9
The following example illustrates how the rectangle method is used.
Example 5
What volumes of a 3 percent solution and an 8 percent solution must be mixed
to make 400 gal of a 5 percent solution?
Use the rectangle method to solve the problem. First write the given concentrations into the proper places in the rectangle:
Higher
concentration
8
Lower
concentration
3
Desired
concentration
5
Now complete the rectangle by subtraction:
8
5–3= 2
Total = 5 Parts
5
3
8–5= 3
The circled numbers on the right side of the rectangle indicate the volume ratios of the solutions to be mixed. The sum of the circled numbers is a total of
five parts to be added (2 parts + 3 parts = 5 parts total).
Two parts out of the five parts (2/5) of the new solution should be made
up of the 8 percent solution. And three parts out of the five parts (3/5) of the
new solution should be made up of the 3 percent solution.
Using the ratios, determine the number of gallons of each solution to be
mixed:
()
()
2
(400 gal) = 160 gal of 8% solution
5
3
(400 gal) = 240 gal of 3% solution
5
Mixing these amounts will result in 400 gal of a 5 percent solution.
Solutions
A solution consists of two parts: a solvent and a solute. These parts are completely and evenly mixed, forming what is referred to as a homogeneous mixture.
The solute part of the solution is dissolved in the solvent (Figure 1-2).
In a true solution, the solute will remain dissolved and will not settle out. Salt
water is a true solution; salt is the solute and water is the solvent. In contrast,
sand mixed into water does not form a solution—the sand will settle out when the
water is left undisturbed.
In water treatment, the most common solvent is water. Before it is dissolved,
the solute may be solid (such as dry alum), liquid (such as sulfuric acid), or gaseous (such as chlorine).
The concentration of a solution is a measure of the amount of solute dissolved in a given amount of solvent. A concentrated (strong) solution is a solution
000200010272023365_ch01_p001-014.indd 9
solution
A liquid containing a
dissolved substance. The
liquid alone is called the
solvent, the dissolved
substance is called the
solute. Together they are
called a solution.
concentration
In chemistry, a
measurement of how
much solute is contained
in a given amount of
solution. Concentrations
are commonly measured
in milligrams per liter
(mg/L).
5/2/16 9:05 AM
10 WSO Water Treatment Grade 2
Solute
Figure 1-2 Solution,
composed of a solute and
a solvent
Solvent
in which a relatively great amount of solute is dissolved in the solvent. A dilute
(or weak) solution is one in which a relatively small amount of solute is dissolved
in the solvent.
There are many ways of expressing the concentration of a solution, including
the following:
77
77
77
77
77
Milligrams per liter
Grains per gallon
Percent strength
Molarity
Normality
Molarity and normality were discussed previously in the chapter. The other expressions of concentration are briefly described next.
Milligrams per Liter and Grains per Gallon
The measurements milligrams per liter (mg/L) and grains per gallon (gpg) each
express the weight of solute dissolved in a given volume of solution. The mathematics involved in using milligrams per liter are covered in Chapter 2.
Percent Strength
The percent strength of a solution can be expressed as percent by weight or percent by volume. The percent-by-weight calculation is used more often in water
treatment. Conversions between milligrams per liter and percentages are discussed
in Chapter 2.
Percent Strength by Weight
The equation used to calculate percent by weight is as follows:
percent strength (by weight) =
000200010272023365_ch01_p001-014.indd 10
weight of solute
weight of solution
× 100
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Basic Microbiology and Chemistry 11
where
weight of solutions = weight of solute + weight of solvents
Use of the equation is illustrated in the following examples.
Example 6
If 25 lb of chemical is added to 400 lb of water, what is the percent strength
of the solution by weight?
Recall the formula:
percent strength (by weight) =
weight of solute
weight of solution
× 100
The weight of the solute is given as 25 lb of chemical, but the weight of the
solution is not given. Instead, the weight of the solvent (400 lb of water) is
given. To determine the weight of the solution, combine the weights of the
solute and the solvent:
weight of solution = weight of solute + weight of solvent
= 25 lb + 400 lb
= 425 lb
Using this information, calculate the percent concentration:
percent strength (by weight) =
=
weight of solute
weight of solution
25 lb chemical
425 lb solution
× 100
× 100
= 0.059 × 100
= 5.9% strength
Example 7
If 40 lb of chemical is added to 120 gal of water, what is the percent strength
of the solution by weight?
First, calculate the weight of the solution. The weight of the solution is
equal to the weight of the solute plus the weight of solvent. To calculate this,
first convert gallons of water to pounds of water using the following formula:
volume of water (in gallons) × 8.34 = weight of water (in pounds)
Therefore:
(120 gal)(8.34 lb/gal) = 1,001 lb water
Then calculate the weight of solution:
weight of solution = weight of solute + weight of solvent
= 40 lb + l,001 lb
= 1,041 lb
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Now calculate the percent strength of the solution:
percent strength (by weight) =
=
weight of solute
weight of solution
40 lb chemical
1,041 lb solution
× 100
× 100
= 0.038 × 100
= 3.8% strength
Standard Solutions
A standard solution is any solution that has an accurately known concentration.
Although there are many uses of standard solutions, they are often used to determine the concentration of substances in other solutions. Standard solutions are
generally made up based on one of three characteristics:
77
77
77
Weight per unit volume
Dilution
Reaction
Weight per Unit Volume
When a standard solution is made up by weight per unit volume, a pure chemical
is accurately weighed and then dissolved in some solvent. By the addition of more
solvent, the amount of solution is increased to a given volume. The concentration
of the standard is then determined in terms of molarity or normality, as discussed
previously.
Dilution
When a given volume of an existing standard solution is diluted with a measured
amount of solvent, the concentration of the resulting (more dilute) solution can be
determined from the following equation:
(
normality of
solution 1
)(
volume of
solution 1
) (
=
normality of
solution 2
)(
volume of
solution 2
)
This equation can be abbreviated as follows:
(N1)(V1) = (N2)(V2)
When this equation is being used, it is important to remember that both
volumes—V1 and V2—must be expressed in the same units. That is, both must
be in liters (L) or both must be in milliliters (mL).
Example 8
You have a standard 1.4N solution of H2SO4. How much water must be added
to 100 mL of the standard solution to produce a 1.2N solution of H2SO4?
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Basic Microbiology and Chemistry 13
First determine the total volume of the new solution by using the relationship between solution concentration and volume:
(N1)(V1) = (N2)(V2)
(1.4N)(100 mL) = (1.2N)(x mL)
Solve for the unknown value:
(1.4)(100)
= x mL
1.2
116.67 = x mL
Therefore, the total volume of the new solution will be 116.67 mL. Since
the volume of the original solution is 100 mL, 16.67 mL of water (that is,
116.67 mL – 100 mL) needs to be added to obtain the 1.2N solution:
100 mL of 1.4N solution + 16.67 mL of water = 116.67 mL of 1.2N solution
Reaction
A similar equation can be used for calculations involving reactions between samples of two solutions, as illustrated in the following example.
Example 9
You know that 32 mL of a 0.1N solution of HCl is required to react with (neutralize) 30 mL of a certain base solution. What is the normality of the base solution?
(N1)(V1) = (N2)(V2)
(0.1N)(32 mL) = (xN)(30 mL)
(0.1)(32)
= x normality
30
0.1 = x normality
WATCH THE VIDEOS
Operator Chemistry 1 and Operator Chemistry 2 (www.awwa.org/
wsovideoclips)
Study Questions
1. Molarity is the number of
a. gram equivalent weights of solute per liter of solution.
b. gram atomic weight per oxidation number.
c. moles of solute per kilogram of solvent.
d. moles of solute per liter of solution.
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2.“CaCO3” is the ______ of calcium carbonate.
a. chemical symbol
b. atomic formula
c. chemical formula
d. chemical equation
3.The first step in calculating percent by weight of an element in a compound is to determine the
a. percent by weight.
b. molecular weight.
c. chemical equation.
d.coefficient.
4. What term refers to the number of equivalent weights of solute per liter
of solution?
a.Normality
b. Equivalent weight
c. Molecular weight
d.Coefficient
5. A solution consists of two parts: a solvent and a
a.diluent.
b.solute.
c.concentrate.
d.suspension.
6.The chemical formula for calcium hydroxide (lime) is Ca(OH)2. Determine the number of atoms of each element in a molecule of the
compound.
7.The equation of the reaction between calcium carbonate (CaCO3) and
carbonic acid (H2CO3) is shown. If 10 lb of H2CO3 is used in the reaction, how many pounds of CaCO3 will react with the H2CO3?
8.You need to prepare 25 gal of a solution with 2.5 percent strength. Assume the solution will have the same density as water: 8.34 lb/gal. How
many pounds of chemical will you need to dissolve in the water?
9.How much water should be added to 80 mL of a 2.5N solution of
H2CO3 to obtain a 1.8N solution of H2CO3?
10. How many atoms of oxygen are present in Ca(HCO3)2?
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