Calculating Concentration PDF - Science 20 - 2006

1.5 Calculating Concentration Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached...

1.5 Calculating Concentration Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page. Figure A1.20: Some insect repellents contain 25% DEET. Health Canada regulates that there should be no more than 0.010 ppm of lead in drinking water. A lab technician determines that a solution of cobalt(II) nitrate has a concentration of 0.200 mol/L. Three Methods for Communicating Concentration You can sometimes determine the relative concentrations of solutions by observing their qualitative properties; but it is often very useful to know specifically how much solute is dissolved in a particular solvent. In science and technology, a variety of ways are used to communicate the concentration of a solution. The table “Expressing Concentration” summarizes three common ways to express concentrations. EXPRESSING CONCENTRATION Symbol Formula Use Used By communicating the Percent by % V/V mL of solute ¥ 100% volume of a liquid solute manufacturers of Volume mL of solution dissolved in the total consumer products volume of a solution communicating levels g of solute of a substance (like a agencies that set health Parts Per Million ppm ¥ 106 ppm g of solution pollutant) in very dilute and safety standards aqueous solutions communicating the amount of moles of Molar mol of solute scientists and lab C a pure substance Concentration L of solution dissolved in the total technicians volume of a solution In this lesson you will explore how and when to use these forms of communicating concentrations. By understanding how concentration is communicated, you should be better able to understand the labels of some of the consumer goods you purchase. You should also be able to make connections to environmental and scientific issues currently being debated. 40 Unit A: Chemical Change Percent By Volume (% V/V) Sometimes, solving problems requires using algebra to Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page. Percent by volume is commonly used for liquids dissolved rearrange the equation. in liquids. This form of concentration is usually used for consumer products like drinks and cleaners. The equation for Example Problem 1.7 calculating percent by volume is as follows: A mosquito repellent says that DEET makes up 45.0% volume of solute dissolved in the of the total volume. If you have a 75-mL sample of this solution (mL) repellent, determine the volume of DEET within the V sample. (% V/V ) = V solute ¥100% solution Solution total volume of List the knowns and the unknown. percent by volume the solution (mL) concentration of the solution (% V/V ) = 45.0% Vsolution = 75 mL Vsolute = ? Example Problem 1.6 Rearrange the percent by volume equation so Vsolute is isolated. A hair product requires you to combine 20.0 mL of V hydrogen peroxide with enough water to produce a (% V/V ) = V solute ¥ 100% solution solution with a total volume of 120.0 mL. Determine the (% V/V ) Vsolute ¥ 1000% percent by volume concentration of the solution. = Vsolution ¨ Divide both 100% 100% sides by 100%. Solution Vsolute (% V/V ) Vsolute = Vsolute = 20.0 mL (% V/V ) = Vsolution ¥ 100% 100% Vsolution Vsolution = 120.0 mL (% V/V ) Vsolute 20.0 mL ¥ Vsolution = ¥ Vsolution ¨ Multiply both = ¥ 100% 100% Vsolution (% V/V ) = ? 120.0 mL sides by Vsolution. = 16.7% (% V/V ) ¥ Vsolution = Vsolute ¨ Vsolute is now isolated. 100% The percentage by volume concentration of the solution is 16.7%. Substitute the values into the equation. Note: The rules of significant digits state that the (% V/V ) Vsolute = ¥ Vsolution final answer must be expressed to the same number 100% of significant digits as the value with the least number 45.0% = ¥ 75 mL of significant digits. Because the volume of the solute 100% consists of only three significant digits, the final answer = 33.75 mL must consist of only three significant digits. For more = 34 mL ¨ two significant digits information, refer to “Calculations with Significant The repellent sample contains 34 mL of DEET. Digits” on pages 534 to 536. When problem-solving throughout this course, there are It is not necessary to show all of the algebra in your key points that you will want to remember: solutions. Most students simply write the basic equation and then write the rearranged version on the next line. Mathematical solutions begin by listing the known and unknown values for this problem. V (% V/V ) = V solute ¥100% The list of knowns and unknowns is used to choose solution the proper equation. The data, including units, is substituted into the equation. (% V/V ) The final answer is expressed with the correct Vsolute = ¥ Vsolution 100% number of significant digits and with the proper The important thing to remember is that the second units. equation is simply a rearranged version of the basic equation; it is not a new equation. Chapter 1: Aqueous Solutions 41 Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page. Example Problem 1.8 Practice Insecticidal soap is an environmentally friendly way 28. To make a hand cleaner, a technician mixes 30 mL to control insect pests on plants. A gardener needs of antiseptic with enough liquid soap to make a solution with a percent by volume concentration 70 mL of solution. Determine the percent by volume of insecticidal soap of 5.0%. If the total volume of concentration of the antiseptic in the hand cleaner. the solution was 4000 mL, calculate the volume of 29. A solution of rubbing alcohol is labelled 60% (V/V ). insecticidal soap needed to make this solution. Determine the volume of rubbing alcohol present in a 200-mL sample of the solution. Solution 30. A bottle of insect repellent states that it has a DEET V (% V/V ) = 5.0% (% V/V ) = V solute ¥ 100% percent by volume concentration of 25%. If you just solution bought a 250-mL container of this product, what Vsolution = 4000 mL (% V/V ) volume of DEET have you purchased? Vsolute = ¥ Vsolution Vsolute = ? 100% 5..0% Parts Per Million (ppm) = ¥ 4000 mL 100% Parts per million is a unit of concentration used for very = 200 mL dilute solutions. You will often come across this unit when = 2.0 ¥ 10 2 mL you are investigating situations involving very small amounts To make this solution, 2.0 ¥ 102 mL of insecticidal soap of substances in contaminated water systems or food. The is needed. table “Allowable Toxic Levels in Drinking Water” lists the current guidelines established by Canada Health for some Note: Because the percent by volume has the least toxic elements in your drinking water. number of significant digits (two), the final answer must ALLOWABLE TOXIC LEVELS IN be expressed with only two significant digits. To do this, DRINKING WATER in this case, scientific notation is required. Maximum Acceptable Substance Concentration (ppm) Y ID arsenic 0.025 ? O D U chromium 0.050 If you combined 50 mL of ethanol with 50 mL of fluoride 1.5 water, you get a solution with a volume of 95 mL. How lead 0.010 can this be? The molecules of ethanol and the molecules of water do not directly stack on top of each other. mercury 0.001 Instead, the two liquids intermingle with each other and, uranium 0.02 thus, fill in some of the spaces between each other’s molecules. This causes them to occupy less volume. The term 1 part per million means “one part solute for For this reason, if you are making a solution a every million parts of solution.” The details of the equation specific concentration, you first need to measure the used to calculate parts per million is as follows: volume of your liquid solute; then you need to add a sufficient amount of solvent to produce a final total mass of solute (g) volume for your solution. msolute parts per million = ¥ 106 ppm msolution parts per million concentration of mass of solution (g) the solution Remember that 1.000 mL of water has a mass of 1.000 g. 42 Unit A: Chemical Change  Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page. Example Problem 1.9 A 200-g sample from a bottle of water contains 5.4 ¥ 10-3 g of mercury. a. Calculate the concentration of mercury in the sample in parts per million. b. Use the information in the table “Allowable Toxic Levels in Drinking Water” to determine if this water is safe to drink. Solution msolute a. msolute = 5.4 ¥ 10 - 3 g parts per million = ¥ 106 ppm msolution msolution = 200 g 5.4 ¥ 10 - 3 g = ¥ 106 ppm parts per million = ? 200 g = 27 ppm The concentration of mercury in this sample is 27 ppm. Calculator Tip: You will be required to perform the calculations that involve entering values in scientific notation into your calculator. For most calculators, this requires using a special key labelled EXP or EE. Pressing these keys includes the “ ¥ 10 ” part of the scientific notation. So, you only need to enter the exponent. Value Keystrokes on Calculator 5.4 ¥ 10- 3 5. 4 EE (-) 3 106 = 1 ¥ 106 1 EE 6 Consult the user’s guide that came with your calculator for more information. b. The mercury concentration in the water is well above the maximum acceptable concentration of 0.001 ppm. This water is not safe to drink. Example Problem 1.10 Carbon monoxide, CO(g), is a deadly gas that takes the place of oxygen molecules and binds to hemoglobin in blood. If you are smoking, the concentration of carbon monoxide that reaches your lungs is approximately 200 ppm. Determine the mass of CO(g) that would be present in a sample of air having a mass of 9.6 g (approximately one breath). Express your answer in scientific notation. Solution msolute parts per million = 200 ppm parts per million = ¥ 106 ppm msolution msolution = 9.6 g parts per million msolute = ¥ msolution msolute = ? 106 ppm 200 ppm = ¥ 9.6 g 106 ppm = 0.0019 g = 1.9 ¥ 10 - 3 g There would be 1.9 ¥ 10-3 g of CO(g) present. Chapter 1: Aqueous Solutions 43  Note: In Example Problem 1.10, the answer displayed on Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page. the calculator is not in scientific notation. To write the Practice answer in scientific notation, follow these steps: 31. A 250-g sample of water contains 8.30 ¥ 10-3 g of step 1: Move the decimal to the right until it is on the lead. right side of the first non-zero digit. a. Calculate the concentration of lead in the 0.00192 sample of water in parts per million. step 2: Round the number to the specific number of b. Determine if this water is safe to drink. significant digits. In this case, round the value to 32. Canadian law prevents the sale of fish for two significant digits. consumption containing more that 2.00 ppm of PCBs. A 227-g sample of fish is tested and found 1.9 to contain the maximum allowable concentration of step 3: Insert the power of ten the value should be PCBs. Determine the mass of PCBs in this sample multiplied by. Because the decimal moved three of fish. places to the right, the exponent in the power of 33. It is considered unsafe to have more than 50.0 ppm 10 is - 3. of arsenic in drinking water. If you have a bottle of 1.9 ¥ 10-3 water containing 250 g of water with this level of arsenic, what mass of arsenic would you ingest? Alternatively, many calculators can be set to automatically display answers in scientific notation. This is a very useful feature because it reduces errors. Molar Concentration, C (mol/L) On many calculators, using the MODE key Earlier in this chapter you saw how diagrams of the atomic automatically causes the answers to be displayed in structure of a water molecule can communicate that two scientific notation. Check the user’s guide that came with hydrogen atoms covalently bond to one oxygen atom. If your calculator to learn how to set this up. you were to break up this molecule, you would expect to get twice as many molecules of hydrogen gas, H2(g), as you would oxygen gas, O2(g). The difficulty is that any Y sample of water contains so many atoms of hydrogen and ID ? O oxygen that it is impossible to count them individually. D U Technologists and scientists, who need to know the amount If you put 4 drops of ink into a rain barrel that of the substance, have solved this problem by counting holds 210 L of water, the concentration of the ink would particles in a very large group called a mole. be approximately 1 ppm. mole: a specific amount of a substance that consists of 6.022 ¥ 1023 particles molar concentration (molarity): the amount of solute, in moles, per litre of solution The mole is a very useful quantity because it enables technologists to combine the precise amounts of substances so that all of the reactants are completely converted into products in a chemical reaction. Since many reactions occur in solutions, concentrations within the scientific community are most often communicated using molar concentration or molarity. The molar concentration of a solution can be calculated using the following relationship: number of moles of solute dissolved in n C= the solution (mol) V total volume of concentration of the solution (L) the solution (mol/L) This technique is very useful because it communicates the number of molecules or ions of solute dissolved or dissociated in a specified volume of a solution. 44 Unit A: Chemical Change  Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page. Example Problem 1.11 Example Problem 1.12 A sample of water taken from a nearby lake is found to You dissolve 30.0 g of sodium sulfate, Na2SO4(s), into have 0.0035 mol of salt in a 100-mL solution. Determine 300 mL of water. the concentration of the salt in the lake. a. Determine the number of moles of sodium sulfate in this solution. Solution n b. Calculate the molar concentration of this sodium n = 0.0035 mol C= V sulfate solution. 1L 0.0035 mol V = 100 mL ¥ = 1000 mL 0.100 L Solution = 0.100 L Note: Because the volume = 0.035 mol/L a. First, determine the molar mass, M, of Na2SO4(s). of the solution must be C =? = 3.5 ¥ 10 - 2 mol/L in litres (L), convert the original volume using a 11 Na 16 S 8 O conversion factor. sodium sulfur oxygen 22.99 32.06 16.00 The concentration of the salt in the lake is 3.5 ¥ 10-2 mol/L. Na+ S2– O2– sodium sulfide oxide Throughout this course, when you record measurements M = 2 ( M of Na ) + ( M of S ) + 4 ( M of O ) or do calculations, you should always include the units. In = 2 ( 22.99 g/mol ) + (32.06 g/mol ) + 4 (16.00 g/mol ) Example Problem 1.11, the volume of the solution was given = 142.04 g/mol in millilitres (mL) instead of litres (L). It is essential that you communicate, in a clear and concise way, how to convert one Now, calculate the number of moles. unit into another. One of the best ways to do this is by using m m = 30.0 g n= a conversion factor. M M = 142.04 g 30.0 g conversion factor: a fraction used to = n=? 142.04 g/mol convert one set of units into another = 0.211 208 110 4 Conversion factors offer a concise and consistent way = 0.211 mol to sort out units. Conversion factors not only keep your solutions organized, they help keep your thinking clear by The solution contains 0.211 mol of sodium sulfate. giving you a reliable way to handle units in all situations. For more information, refer to “Conversion Factors” on b. n = 0.211 208 110 4 mol page 532. 1L V = 300 mL ¥ 1000 mL Molar Mass = 0.300 L Sometimes it is necessary to first calculate the number of moles using the mass of the sample and information from the C =? periodic table to determine the molar mass. The equation for Note: Although the number of moles was rounded to three significant digits in part a., you should use the unrounded calculating the number of moles in a substance is as follows: value in this calculation to get a more accurate answer to part b. Remember, the number of moles is still considered to have three mass of the molar mass: the mass significant digits in this calculation. substance (g) of 1 mol of a substance n m C= n= V M 0.211 208 110 4 mol number of moles of = a substance (mol) molar mass of the 0.300 L substance (g/mol) = 0.704 mol/L The process of finding the number of moles and then The molar concentration of the sodium sulfate calculating the molar concentration is illustrated in Example solution is 0.704 mol/L. Problem 1.12. Chapter 1: Aqueous Solutions 45 Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page. Practice 34. Determine the molar concentration for each solution. a. 0.435 mol of sodium chloride, NaCl(s), dissolves in 200 mL of water. b. 800 mL of water contains 0.674 mol of sodium hydroxide, NaOH(aq). 35. 30.0 g of NaCl(s) is added to water to make 800 mL of salt solution. a. Use the periodic table on pages 554 and 555 to determine the number of moles in 30.0 g of NaCl(s). b. Calculate the molar concentration of this salt solution. 36. In a science lab, 5.00 g of NaOH(s) is dissolved in 300 mL of water. What is the molar concentration of the resulting solution? Standard Solutions Technologists in a number of industries often need to make solutions with a specific concentration. In the field of health care, prescriptions can involve creams or liquids that require an exact amount of medicinal solute to be combined with the appropriate solvent. During manufacturing, specific solutions may have to be prepared. Each solution is an essential component, allowing for the precise control of a chemical reaction. Why don’t these facilities simply order the necessary solutions already mixed in the proper concentrations? As is the case with consumer goods, it is less expensive to ship highly concentrated solutions. Another reason is that sometimes the solutions need to be modified. This provides greater flexibility when it comes to mixing the specific solution needed for specific situations. Clearly, in many industries, technologists follow careful procedures to prepare standard solutions. Making a standard solution requires practice and skill because the slightest error can cause the concentration of your standard solution to be less accurate. In the next investigation you will have an opportunity to produce a standard solution and a dilute solution. standard solution: a solution having a precisely known concentration Figure A1.21: In a number of industries, solutions need to be very specific. Investigation Developing Technological Skills with Solutions Purpose Science Skills You will practise the skills for making a standard solution and for making a dilute solution. Performing and Recording Analyzing and Interpreting Materials 50-mL beaker 2, 100-mL volumetric flasks anhydrous copper(II) sulfate, CuSO4(s) distilled water Use gloves, safety glasses, and eyedropper a lab apron for this investigation. 10-mL volumetric pipette Part A: Making a Standard Solution In this part of the investigation you will make 100 mL of a 0.200 mol/L solution of copper(II) sulfate from anhydrous copper(II) sulfate. 46 Unit A: Chemical Change  Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page. Pre-Lab Analysis 1. Complete the following calculation to determine the number of moles of solute you will need for your solution: 1L n V = 100 mL ¥ C= V 1000 mL n = CV = 0.100 L =( )( ) C = 0.200 mol/L = n=? 2. Complete the following calculation to determine the mass of the copper(II) sulfate, CuSO4, you need. m n= n= M M = 1( M of Cu) + 1( M of S ) + 4 ( M of O ) m = nM =( )+( )+4( ) =( )( ) = = m=? Procedure step 1: Measure the appropriate mass of solute, CuSO4(s), and place it into a small beaker. step 2: Dissolve the solute in the beaker with as little distilled water as possible. step 3: Carefully transfer the dissolved solute into a volumetric flask. Be sure to rinse your beaker with a little water, and add this rinse to the volumetric flask as well. step 4: Carefully fill the volumetric flask until the top of the meniscus reaches the 100-mL line. step 5: Use an eyedropper to add water until the bottom of the meniscus touches the 100-mL line. This is the key step. Do not overshoot the line. If you do overshoot, discard the solution and start over. step 6: Stopper the volumetric flask. Firmly hold the stopper in place, and invert the flask 15 times to mix the solution. step 7: Show the solution to your teacher to check the accuracy of your work. Keep your solution. It will be used in Part B of this investigation. Part B: Diluting a Standard Solution In this part of the investigation you will make 100 mL of a new solution that is a 10% dilution of your standard solution from Part A. Figure A1.22: The correct measurement of 100 mL occurs when the bottom of the meniscus is even with the line on the volumetric flask. If the bottom of Procedure the meniscus is above the line on the volumetric flask, you have more than step 1: Have your teacher demonstrate the proper 100 mL of solution. technique for using a pipette. step 2: Pipette 10 mL of your standard solution into another 100-mL volumetric flask. step 3: Carefully fill the volumetric flask with water so the meniscus is just at the bottom of the 100-mL line. step 4: Stopper the volumetric flask, and mix the contents of the flask by inverting and carefully shaking it. Chapter 1: Aqueous Solutions 47 Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page. step 5: Transfer the dilute solution to an appropriate storage container. Calculate the concentration of your dilute solution. Label the container with the name, chemical formula, and concentration. Also, Do not pour the remainder of your indicate which WHMIS symbols should be placed on the label. standard solution down the drain. Consult your teacher regarding the Analysis proper disposal of your solution. 3. List the potential problems you may have encountered with each step of the procedure. Identify steps that could affect the accuracy of the concentration of your standard solution. 4. Are you confident that your standard solution is exactly 0.200 mol/L? Give a reason for your answer. 5. Do you think it would be easier or more difficult to make a standard solution that is colourless? Support your answer. 6. Explain the importance of knowing the exact concentration of a standard solution. 7. Explain how volumetric flasks and pipettes help you measure volumes with greater precision. Evaluation In this investigation you were instructed not to pour your standard solution down the drain. One reason for doing this is that once the water has evaporated, the crystals can be reconstituted and used by other classes. Another reason for not pouring the solutions down the drain has to do with the harmful effects of copper substances on the environment. 8. Use the Internet and other sources of information to determine the harmful effects of copper compounds released into the environment. Diluting Solutions Some chemical solutions are transported in concentrated form to save costs. It may be necessary to dilute such a solution to obtain a solution with a specifically desired concentration. There are also times when water is removed from a solution to make it more concentrated. In each case, the amount of solute is unchanged, but the amount of solvent changes. Adding Solvent to a Solution When solvent is added to a solution, the number of moles of solute, n, is unchanged. initial solution final solution n n Ci = Cf = Vi Vf n = Ci Vi number of moles of n = Cf Vf solute after dilution number of moles of solute before dilution Since the number of moles of solute is constant, Note: The final volume is the total volume of the solution, not initial concentration CV i i = Cf Vf final volume the amount of solvent added. initial volume final concentration 48 Unit A: Chemical Change  Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page. Example Problem 1.13 You have 65.0 mL of a 0.759-mol/L solution of sodium chloride, NaCl(aq). a. Calculate the final concentration of the solution if it is diluted to a final volume of 100.0 mL. b. Calculate the final concentration of a solution prepared by adding 100.0 mL of water to the original solution. c. How much water do you need to add to the original solution to obtain a solution with a concentration of 0.200 mol/L? d. How much water needs to evaporate from the original solution to obtain a solution with a concentration of 0.890 mol/L? Solution a. Vi = 65.0 mL Ci Vi = Cf Vf Ci = 0.759 mol/L Ci Vi Cf = Vf Vf = 100.0 mL (0.759 mol/L ) (65.0 mL ) Cf = ? = ¨ Note: There is no need to convert millilitres to 100.0 mL litres because they cancel each other out. = 0.493 mol/L The final concentration of the solution is 0.493 mol/L. b. Vi = 65.0 mL Ci Vi = Cf Vf Ci Vi Ci = 0.759 mol/L Cf = Vf Vf = 65.0 mL + 100.0 mL (0.759 mol/L ) (65.0 mL ) = 165.0 mL = 165.0 mL Cf = ? = 0.299 mol/L The final concentration of the solution is 0.299 mol/L. c. Vi = 65.0 mL Ci Vi = Cf Vf Ci Vi Ci = 0.759 mol/L Vf = Cf Cf = 0.200 mol/L = (0.759 mol/L ) (65.0 mL ) Vf = ? 0.200 mol/L Vadded = ? = 246.675 mL The amount added can be found by subtracting the initial volume from the final volume. Vadded = Vf - Vi = 246.675 mL - 65.0 mL =181.675 mL continued on next page =182 mL You need to add 182 mL of water. Chapter 1: Aqueous Solutions 49 Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page. d. Vi = 65.0 mL Practice Ci = 0.759 mol/L Cf = 0.890 mol/L Vf = ? Vevaporated = ? Ci Vi = Cf Vf Ci Vi Vf = Cf = (0.759 mol/L ) (65.0 mL) 0.890 mol/L = 55.432 584 27 mL The amount evaporated can be found by subtracting the final volume from the initial volume. Vevaporated = Vi - Vf = 65.0 mL - 55.432 584 27 mL = 9.567 415 73 mL Figure A1.23: A battery technician checks the concentration of = 9.6 mL sulfuric acid. 37. The battery is the primary source of electrical energy The volume of water needed to evaporate is 9.6 mL. used in vehicles. Most automotive batteries produce electricity by using the chemical reaction between two different types of lead in a solution of sulfuric Diluting Acids acid. This application is the reason why some In previous science courses you studied compounds that people refer to sulfuric acid as battery acid. In one form solutions that turn blue litmus paper red because they automotive battery, 360 mL of concentrated sulfuric produce hydrogen ions. These compounds are called acids. acid, 17.8 mol/L, is combined with 640 mL of water to form the solution with the proper concentration. acid: a substance that a. Determine the total volume of the solution in the produces hydrogen ions when battery. dissolved in water to form a conducting aqueous solution b. Calculate the molar concentration of the sulfuric acid solution in the battery. Soft drinks, tomato sauce, and pickles are foods that are c. Determine the percent by volume concentration all slightly acidic. Acid content in food can be identified of the sulfuric acid solution in the battery. with a sour taste. While these products are relatively d. Consider your answers to questions 37.b. harmless, concentrated acids can be quite hazardous. and 37.c. Which method of communicating Concentrated acids can cause acid burns on skin, blindness concentration is likely used by scientists if splashed in the eyes, and death if swallowed. researching new designs for batteries, and which method is best for a brochure for customers The most common acid used in industry is sulfuric acid. explaining the features of a particular battery? In fact, more sulfuric acid is produced each year than any 38. Sulfuric acid is used to make chlorine dioxide other industrial chemical. As you’ll see in the next set of for bleaching in the pulp and paper industry. A Practice questions, concentrated sulfuric acid is used in a technician in a pulp mill needs to make 275 L of a wide variety of applications. solution containing 4.25 mol/L of sulfuric acid. Safety Tip: When diluting an acid, always add the acid a. Calculate the volume of concentrated sulfuric to the water in small amounts. The rearrangement of solute acid (17.8 mol/L) that the technician must and solvent when an acid is diluted is exothermic and can measure to make the required solution. be a safety risk. b. Determine the volume of water needed to make the required solution. 50 Unit A: Chemical Change  Science 20 © 2006 Alberta Education (www.education.gov.ab.ca). Third-party copyright credits are listed on the attached copyright credit page. 39. Approximately two-thirds of all the sulfuric acid produced for industry is used in the production of fertilizers. Ammonium sulfate and potassium sulfate are both fertilizers that are manufactured using sulfuric acid. At a large fertilizer manufacturing plant, a technician begins to prepare some tests using a standard solution of sulfuric acid. a. The technician measures out 2.50 L of a standard solution of sulfuric acid with a concentration of 10.0 mol/L. Determine the amount of water that must be added to create a solution with a concentration of 3.75 mol/L. b. A large beaker contains 655 mL of a standard solution of sulfuric acid with a concentration of 10.0 mol/L. This beaker is placed in a fume hood where evaporation can occur. Determine the amount of water that would have to evaporate in order for the solution to have a concentration of 11.0 mol/L. Y ID ? O Figure A1.24: The voltage output of an automotive battery is checked. D U When the engine is running, the vehicle’s recharging Many natural compounds, such as the oils in system prevents this situation from occurring by supplying onions, contain sulfur. When you slice an onion, a gas an electric current to the battery that flows in the opposite is released that rises upwards and combines with the direction. As the electrons are forced into the battery, the water in your eyes. The result is that a dilute sulfuric chemical reactions are reversed, the concentration of sulfuric acid solution forms in your eyes. In response to this acid rises, and the battery is said to be fully charged. irritation, your eyes automatically start to blink and The concentration of sulfuric acid in the battery is a good tear to flush out the sulfuric acid. indicator of the state of charge of the battery. In the next chapter you will learn more about the role of electrons in chemical reactions and how all this relates to the batteries. 1.5 Summary Having quantitative units for concentrations allows you to know specifically how much solute is dissolved in a solution. Percent by volume is a unit of concentration use