Moles and Molarity 6 Equivalent Weights and Normality 7 Dilution Calculations 8 Standard Solutions 12

Home , Chemical equation, Equivalent (chemistry), Equivalent weight

This is the table of contents and a sample chapter from WSO Water Treatment, Grade 2, get your full copy from AWWA. awwa.org/wso

Chapter 1 Basic Microbiology and Chemistry 1 Chemical Formulas and Equations 1 Moles and Molarity 6 Equivalent Weights and Normality 7 Dilution Calculations 8 Standard Solutions 12

Chapter 2 Operator Math 15 Volume Measurements 15 Conversions 22 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 Pressure 47 Flow Rate Problems 52 Chemical Dosage Problems 55

Chapter 3 USEPA Water Regulations 71 Types of Water Systems 71 Disinfection By-product and Microbial Regulations 72

Chapter 4 Coagulation and Flocculation Process Operation 89 Operation of the Processes 89 Dosage Control 93 Safety Precautions 93 Record Keeping 94

Chapter 5 Sedimentation and Clarifiers 97 Process Description 97 Sedimentation Facilities 98 Other Clarification Processes 104 Regulations 107 Operation of the Process 107

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Chapter 6 Filtration 115 Equipment Associated With Gravity Filters 115 Operation of Gravity Filters 123 Pressure Filtration 135 Regulations 138 Safety Precautions 138 Record Keeping 139

Chapter 7 Chlorine Disinfection 143 Gas Chlorination Facilities 143 Hypochlorination Facilities 155 Operation of the Chlorination Process 157 Chlorination Operating Problems 162 Safety Precautions 165 Record Keeping 170

Chapter 8 Iron and Manganese Treatment 173 Excessive Iron and Manganese 173 Control Processes 175 Control Facilities 178 Regulations 182 Manganese Greensand Filter Operation 182 Process Monitoring 188 Operating Problems 188 Record Keeping 188

Chapter 9 Fluoridation Process Operation 191 Operation of the Fluoridation Process 191 Fluoridation Operating Problems 193 Control Tests 194 Safety Precautions 195 Record Keeping 196

Chapter 10 Water Quality Testing 199 Testing and Laboratory Procedures 199 Physical and Aggregate Properties of Water 209

Chapter 11 Corrosion Control 221 Purposes of Corrosion and Scaling Control 221 Water System Corrosion 222 Scale Formation 226 Corrosion and Scaling Control Methods 227 Corrosion and Scaling Control Facilities 230 Chemical Feed Equipment 233

Chapter 12 Lime Softening 237 Lime Softening Chemical Reactions 237 Lime Softening Facilities 239 Regulations 244

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Chapter 13 Ion Exchange 247 Ion Exchange Softening 247 Facilities 247 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

Chapter 14 Activated Carbon Adsorption 261 The Principle of Adsorption 261 Adsorption Facilities 262 Powdered Activated Carbon 263 Granular Activated Carbon 265 Regulations 268 Operating Procedures for Adsorption 268 Operating Problems 272 Control Tests 273 Record Keeping 276

Chapter 15 Aeration 279 Water-Into-Air Aerators 279 Air-Into-Water Aerators 284 Combination Aerators 285

Chapter 16 Membrane Treatment 291 Microfiltration Facilities 291 Pleated Membrane Facilities 294 Reverse Osmosis Facilities 297

Chapter 17 Plant Waste Treatment and Disposal 303 Removal of Sludge from Conventional Sedimentation Processes 303 Softening Sludge Handling, Dewatering, and Disposal 305 Solids Separation Technologies 306

Chapter 18 Instrumentation and Control Systems 313 Flow, Pressure, and Level Measurement 313 Other Operational Control Instruments 316 Automation 317 Computerization 318

Chapter 19 Centrifugal Pumps 321 Operation of Centrifugal Pumps 321 Centrifugal Pump Maintenance 325 Record Keeping 336 Pump Safety 336

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Chapter 20 Treatment Plant Safety and Security Practices 339 Treatment Plant Safety Review 339 Plant Security 348

Chapter 21 Administration, Records, and Reporting Procedures 355 Process Records 355 Reporting 359 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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Chemical Formulas and Equations A group of chemically bonded atoms forms a particle called a molecule. The sim- plest 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 chemical formula element is present in the molecule. The subscript 3 to the right of the O sym- Using the chemical bolizing oxygen indicates that there are three oxygen atoms in each molecule. symbols for each element, a shorthand

CaCO3 way of writing what elements are present in a molecule and how many 1 atom 3 atoms atoms of each element are present in each of the 1 atom molecules.

000200010272023365_ch01_p001-014.indd 1 5/2/16 9:05 AM 2 WSO Water Treatment Grade 2 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: Na Cl 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 Atomic Total Atoms Weight Weight sodium (Na) 1 × 22.99 = 22.99 chlorine (Cl) 1 × 35.45 = 35.45 molecular weight of NaCl = 58.44

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:

weight of element in compound percent element by weight = × 100 molecular weight of compound

percent by weight Using the formula, first calculate the percent by weight of sodium in the The proportion, compound: calculated as a weight of Na in compound percentage, of each percent Na by weight = × 100 element in a compound. molecular weight of compound 22.99 molecular weight = × 100 The sum of the atomic 58.44 weights of all the atoms = 0.393 × 100 in the compound. Also called formula weight. = 39.3% sodium by weight

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Then, calculate percent by weight of chlorine in the compound: weight of Cl in compound percent Cl by weight = × 100 molecular weight of compound 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% Na + 60.7% Cl 100.0% 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 calcium react to calcium plus plus water bicarbonate hydroxide form carbonate chemical equation A shorthand way, using Ca(HCO ) + Ca(OH) 2CaCO + 2H O chemical formulas, of 3 2 2 3 2 writing the reaction that takes place when Reactants Products chemicals are brought together. The left side of the equation indicates 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 the chemicals brought reaction. If no coefficient is shown, then only one molecule of the compound is together (the reactants); involved. For example, in the preceding equation, one molecule of calcium bi- the arrow indicates in which direction the carbonate reacts with one molecule of calcium hydroxide to form two molecules reaction occurs; and the of calcium carbonate and two molecules of water. Without the coefficients, the right side of the equation equation could be written indicates the results Ca(HCO ) + Ca(OH) CaCO + CaCO + H O + H O (the products) of the 3 2 2 3 3 2 2 chemical reaction. 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. coefficient In fact, for each element in the equation, as many atoms are shown on the left An indication of the side as on the right. An equation for which this is true is said to be balanced. A relative number of balanced equation accurately represents what really happens in a chemical re- molecules of the action: because matter is neither created nor destroyed, the number of atoms of compound that are each element going into the reaction must be the same as the number coming out. involved in the chemical reaction. Coefficients allow balanced equations to be written compactly.

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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 Atomic Total Atoms Weight 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

molecular weight of Ca(HCO3)2 = 162.12

The molecular weight for Ca(OH)2 is determined as follows:

Number of Atomic Total Atoms Weight Weight calcium (Ca) 1 × 40.08 = 40.08 oxygen (O) 2 × 16.00 = 32.00 hydrogen (H) 2 × 1.01 = 2.02

molecular weight of Ca(OH)2 = 74.10

The coefficient 2 in front of the next term of the equation (2CaCO3) indi-

cates 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 Atomic Total Atoms Weight 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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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 calcu-

lation, first determine the weight of one molecule of H2O, then the weight of two molecules:

Number of Atomic Total Atoms Weight Weight hydrogen (H) 2 × 1.01 = 2.02 oxygen (O) 1 × 16.00 = 16.00

weight of one molecule of H2O = 18.02

weight of two molecules of H2O = (2)(18.02) = 36.04

In summary, the weights that correspond to each term of the equation are

Ca(HCO3)2 + Ca(OH)2 2CaCO3 + 2H2O 162.12 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 indi- cated 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 desired ratio

74.10 g Ca(OH) 25 g Ca(OH) 2 = 2 162.12 g Ca(HCO3)2 x g Ca(HCO3)2

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Next, solve for the unknown value: 74.10 25 = 162.12 x (x)(74.10) = 25 162.12 (25)(162.12) x = 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 2CaCO3 + 2H2O 162.12 74.10 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 2CaCO3 + 2H2O

mole 162.12 g 74.10 g (2)(100.09) g (2)(18.02) g The quantity of a Because of the way a mole is defined, this could also be written in a more compact compound or element form: that has a weight in

grams equal to the Ca(HCO3)2 + Ca(OH)2 2CaCO3 + 2H2O substance’s molecular or atomic weight. Used in 1 mole 1 mole 2 moles 2 moles this text generally as Reading this information, a chemist could state, “One mole of Ca(HCO3)2 is an abbreviation for needed to react with one mole of Ca(OH) , and the reaction yields two moles of gram- mole. 2 CaCO3 and two moles of water.”

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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 Atomic Total Atoms Weight 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

molecular weight of NaHCO3 = 84.01

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 2K CrO4, first determine the weight of 1 mol of the compound:

Number of Atomic Total Atoms Weight Weight potassium (K) 2 × 39.10 = 78.20 chromium (Cr) 1 × 52.00 = 52.00 oxygen (O) 4 × 16.00 = 64.00

molecular weight of K2CrO4 = 194.20

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. Nor- mality 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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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, step- 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 capac- ity 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 char- acteristic: 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:

total weight number of equivalent weights = equivalent weight

equivalent weight The weight of a compound that contains Dilution Calculations one equivalent of a proton (for acids) or Sometimes a particular strength of solution will be created by diluting a strong one equivalent of a solution with a weak solution of the same chemical. The new solution will have a hydroxide (for bases). concentration somewhere between the weak and the strong solutions. The equivalent weight Although there are several methods available for determining what amounts can be calculated by of the weak and strong solutions are needed, perhaps the easiest is the rectangle dividing the molecular method (sometimes called the dilution rule), shown in Figure 1-1. weight of a compound by the number of H+ – Solution A, or OH present in the Parts of A higher = A% C – B = required compound. concentration Solution C, Sum = total normality desired concentration parts in The number of = C% desired solution Solution B, equivalent weights Parts of B lower = B% A – C = required of solute per liter of concentration solution. Figure 1-1 Schematic for rectangle method

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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 8 concentration Desired concentration 5 Lower 3 concentration Now complete the rectangle by subtraction:

8 5 – 3 = 2

5 Total = 5 Parts

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) solution Mixing these amounts will result in 400 gal of a 5 percent solution. A liquid containing a dissolved substance. The liquid alone is called the Solutions solvent, the dissolved substance is called the A solution consists of two parts: a solvent and a solute. These parts are com- solute. Together they are pletely and evenly mixed, forming what is referred to as a homogeneous mixture. called a solution. 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 concentration water is a true solution; salt is the solute and water is the solvent. In contrast, In chemistry, a sand mixed into water does not form a solution—the sand will settle out when the measurement of how water is left undisturbed. much solute is contained In water treatment, the most common solvent is water. Before it is dissolved, in a given amount of the solute may be solid (such as dry alum), liquid (such as sulfuric acid), or gas- solution. Concentrations eous (such as chlorine). are commonly measured The concentration of a solution is a measure of the amount of solute dis- in milligrams per liter solved in a given amount of solvent. A concentrated (strong) solution is a solution (mg/L).

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Solute

Figure 1-2 Solution, Solvent composed of a solute and a 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 Milligrams per liter 77 Grains per gallon 77 Percent strength 77 Molarity 77 Normality Molarity and normality were discussed previously in the chapter. The other ex- pressions of concentration are briefl y 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 mathe- matics 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:

weight of solute percent strength (by weight) = × 100 weight of solution

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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: weight of solute percent strength (by weight) = × 100 weight of solution 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: weight of solute percent strength (by weight) = × 100 weight of solution 25 lb chemical = × 100 425 lb solution

= 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: weight of solute percent strength (by weight) = × 100 weight of solution

40 lb chemical = × 100 1,041 lb solution

= 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 Weight per unit volume 77 Dilution 77 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 volume of normality of volume of = ( solution 1 )(solution 1 ) ( solution 2 )(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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First determine the total volume of the new solution by using the relation- ship 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 sam- ples 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 (neutral- ize) 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

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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. De- termine 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 reac-

tion, 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. As- sume 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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