Activity Units of Concentration

Transcription

1 KEY Activity Units of Concentration Directions: This Guided Learning Activity (GLA) focuses on chemical calculations related to solution concentration. Part A gives the definitions of the most common concentration units: molarity, % mass (m/m), and ppm. Part B uses these concentration units in mass and mole conversion problems. Part C discusses solution dilution, and the application of C 1 V 1 C 2 V 2 equation. The worksheet is accompanied by instructional videos. See for additional materials. Part A Basic Concentration Units A solution is a homogenous mixture of two or more pure substances. A solvent is defined as the substance present in greatest amount, while the solutes are present in smaller amounts. The concentration of a solute is the amount of solute dissolve in the solution, and can be defined both qualitatively and quantitatively. Qualitatively, a solution can be dilute, containing a small amount of solute, or concentrated, containing a large amount of solute. Three quantitative measures of solution concentration will be discussed here. They are: MMMMMMMM (M) % mmmm ppp mmmmm mmmsmm llllll mmmsmmms mmmm mmmsmm x 111% mmmm mmmsmmms mmmm mmmsmm x 116 mmmm mmmsmmms Mass Percent & Parts Per Million: Recall that the solution is made up of both solute and solvent. The mass of the solution is therefore mass of the solute plus mass of the solvent. Both the mass percent (% mass) and the parts per million (ppm) concentrations can be found directly from the masses of the solute and solution. Generally, ppm is used only for very dilute solutions. Activity Page 1 of 6

2 Example #1. Calculate the % mass and ppm concentrations for each of the following solutions. a g sucrose (C 12 H 22 O 11 ) in 90.0 g of water. % sssssss ppp sssssss mmmm sssssss 2. x 111% mmmm ssssssss mmmm sssssss mmmm ssssssss x 106 b mg sucrose in 2.00 kg of solution. 33 g sssssss x 111% 2. 55% (2. 33 g g)ssssssss g sssssss (2. 35 g g)ssssssss x ppp 1 g sssssss (2. 33 mm sssssss) x 1111 mm 11 3 g sssssss; x 11 3 g sss n 1111 g sss n (2. 00 kk sss n) 1 kg sss n % sssssss x 11 3 g sssssss x 11 3 x 111% % g sss n ppp sssssss x 11 3 g sssssss x 11 3 x ppp g sss n Although we limit this GLA to discussing % mass and ppm on a mass basis, keep in mind that both mass % and ppm can be defined on a mass solute/mass solution (m/m), mass solute/volume solution (m/v) or volume solute/volume solution (v/v) basis. Molar Concentrations: The molar concentration, or molarity, of a solution is used extensively in stoichiometric calculations. The molarity calculation requires that the mass of solute be converted into moles. Example #2. Calculate molar concentration (M) of each of the following solutions. a g of sodium chloride (NaCl) in 2.00 L of solution. 1 mmmm NNNN MMMMM NNNN g NNNN mmmm NNNN g NNNN MMMMMMMM mmmmm NNNN llllll ssssssss 777 mmmm NNNN M L ssssssss b g of urea (CH 4 N 2 O) in ml of solution. (66. 8 g CH 4 N 2 O) 1 mmm CH 4N 2 O mmm uuuu g CH 4 N 2 O MMMMMMMM mmmmm ssssss LLLLLL sss n mmm CH 4N 2 O M L Activity Page 2 of 6

3 Part B Calculations using Solution Concentration Solution concentration directly relates to the amount of solute present, and many calculations in solution chemistry relate the amount of solute to the volume of the solution. To solve these problems, it s important to be familiar with the concentration definitions. Practice: How many moles of magnesium bromide (MgBr 2 ) are found in 6.48 ml of a M aqueous solution? We know: MMMMMMMM mmmmm MMMr 2 llllll ssssssss Plugging in for molarity and volume (6. 44 mm Solving for x: M 1 L 1111 mm x L L): x mmm L ( L) x 11 4 mmm MMMr 2 Alternatively, the concentration can be used as a conversion factor. We know: MMMMMMMM mmmmm MMMr mmm MMMr 2 llllll ssssssss 1 L Solving by dimensional analysis: 1 L sss n mm sss n 1111 mm sss n mmm MMMr x 11 4 mmm MMMr 1 L sss n 2 Example #3. How much ammonium acetate (NH 4 NO 3 ) is needed to prepare a g solution with a concentration of 7.50% NH 4 NO 3? ( g sss n) g NH 4NO g sss n g NH 4NO 3 Activity Page 3 of 6

4 Part C Solution Dilution Calculations In laboratories, often new solutions are prepared from concentrated stock solutions by dilution. When dilutions are performed, the following equation can be used to easily calculate the concentration or volume of the stock or dilute solution. C 1 V 1 C 2 V 2 Where: C 1 concentration of stock solution V 1 volume of stock solution C 2 concentration of diluted solution V 2 volume of diluted solution Keep in mind that this equation is only used when a solution is diluted, and never to perform stoichiometric calculations. Most solution calculations will not use this equation. The concentration can be given in molarity, % mass, or ppm. Likewise, volume can be given in liters, milliliters, or another convenient unit. Practice: What volume in ml of 4.00 M NaOH is needed to prepare 250. ml of dilute M NaOH? C 1 V 1 C 2 V 2 Stock solution: C M NaOH, V 1?? Diluted solution: C M NaOH, V ml V 1 C 2V 2 C 1 ( M)(222. mm) mm (4. 00 M) Example # ml of a M solution of lithium chloride is diluted to ml. What is the final concentration of the solution? C 2 C 1V 1 V 2 ( M LLLL)(5. 00 mm) ( mm) M LLLL Activity Page 4 of 6

5 A serial dilution is a process in which each diluted solution is treated as the stock solution for the next dilution. Serial dilutions are a simple way for scientists to prepare solutions with a wide range of concentrations. Example #5. Solution A is prepared by diluting 1.00 ml of M CaCl 2 to 50.0 ml. Solution B is prepared by diluting 10.0 ml of Solution A to ml. What is the concentration of CaCl 2 in Solution B? The stock solution is solution 1 and solution A is solution 2. C A C sssssv sssss V A ( M)(1. 00 mm) (55. 0 mm) M Solution B is prepared using solution A as the stock, so for this calculation, the stock concentration is M. C B C AV A V B ( M)(11. 0 mm) ( mm) 6. 4 x 10 5 M Activity Page 5 of 6

6 Part D Extra Practice 1. Calculate the grams of solute necessary to prepare ml of the following solutions: 50.0% (m/m) sucrose (C 12 H 22 O 11 ), d1.230 g/ml 1.46 M calcium chloride (CaCl 2 ) 50.0 mm sodium hydrogen phosphate (Na 2 HPO 4 ) 0.36 M glucose (C 6 H 12 O 6 ) 2. Find the mass % concentration of sugar in a g solution containing g of dissolved sugar. 3. How many kilograms of sucrose are in 4.8 kg of a 40.0 % (m/m) sucrose solution? 4. What is the molarity of a solution that contains 32.4 g of potassium bromide in 1.5 L of solution? What is the mass % (d solution 1.02 g/ml)? 5. Which solution contains more moles of sodium ions: ml of M sodium chloride (NaCl), or g of 6.3% (m/m) sodium carbonate (Na 2 CO 3 )? Hint: Recall that soluble ionic compounds dissociate in water. Refer to GLA Complete and Net Ionic Equations for more guidance. 6. Benzene is a known carcinogen that can leak into groundwater. The maximum allowable level of benzene in drinking water is 5 μg/l. The average American consumes 3.9 cups, or 0.92 liters, of water each day. If benzene is present in an individual s drinking water at 5 μg/l, how many grams of benzene will that person consume in one year? 7. A technician wants to prepare a 175 ml of 0.84 M sodium hydroxide solution. What volume of the 5.0 M stock solution of sodium hydroxide does the technician need? 8. A 20.0 ml of a stock solution was diluted to 1.00 L. The concentration of the final solution was 40.0 mm. What was the concentration of the stock solution? 9. What mass of a 22.8 ppm lead (II) nitrate solution will contain mole of lead (II) nitrate? 10. Calculate the number of moles of nitric acid (HNO 3 ) in 500. ml of a solution which is 30.0% HNO 3 by mass. (d solution 1.18 g/ml). Activity Page 6 of 6

7 Key for extra practice GLA For this problem, you need to be familiar with the definition of each concentration unit, and be able to manipulate units to get the correct units in the numerator and denominator of the expression. 50.0% (m/m) sucrose: ( mm sol n) g sss n g sssssss mm sss n 111 g sss n 111 g sssssss For this conversion, recall that the metric prefixes can be applied to any metric unit, including molarity. Therefore, a concentration of 1 M is equivalent to 1000 mm. We will represent 50 mm Na 2 HPO 4 as 55.0 mmmm Na 2HHO 4. 1 L ssssssss 50.0 mm Na 2 HPO 4 : ( mm sss n) g Na 2 HHO 4 1 L sss n 1111 mm sss n 55.0 mmmm Na 2HPO 4 1 L sss n 1 mmm 1111 mmmm g Na 2HPO 4 1 mmm Na 2 HHO 4 For molar concentrations, it s important to remember to convert to moles using molar mass. The XX mmm ssssss XX mmm ssssss molarity can be represented as or as, since 1 L and 1000 ml are 1 L ssssssss 1111 mm ssssssss equivalent volumes M calcium chloride: ( mm sss n) 1.44 mmm CCCl M glucose: ( mm sss n) 0.33 mmm C 6H 11 O mm sss n 1111 mm sss n g CCCl 2 1 mmm CCCl g C 6H 11 O 6 1 mmm C 6 H 11 O 6 mmmm sssss g CCCl g C 6 H 11 O 6 2. For this problem, we need to know % sssss x 111%. The problem gives both the mmmm ssl n mass of the sucrose and the mass of the solution, so we simply plug the two values into the expression. Sometimes, the mass of the solvent is given, and in these cases you ll need to add the mass of the solute and the solvent together to find the mass of the solution. % sssssss g sssss g ssl x 111% % n 3. One of the advantages of concentration units is that they can be used as conversion factors to convert between quantities. In this problem, we recognize that mass percent can be written as a conversion factor. 40.0% means 40.0% of any mass. Because it is a percent measurement, we can use any mass unit we choose. Because the mass of solution is given in kg, we will choose to represent the mass % in kg kk sssssss (4. 8 kk sss n) 111 kk ssl 1. 9 kk sssssss n 4. To calculate the molarity, we need to divide the moles of solute by the liters of solution. It may be convenient to divide the mass by the volume, and then use dimensional analysis to convert the mass of solute to moles of solute. MMMMMMMM mmmmm KKK LLLLLL ssl n 4 g KKK 1 mmm KKK L ssl M KKK n g KKK

8 For the mass percent, we need to remember that the % mass measurement is based on the masses of the solute and solution. The volume of the solution can be converted to mass by using the density as a conversion factor. The density of a solution is most often measured in to distinguish between the mass of the solution and the mass of the solute. ggggg ssssssss mm ssssssss. Take care % KKK mmmm KKK mmmm ssl n x 111%; 33.4 g KKK 1.5 L ssl n 1 L ssl n 1 mm ssl n x 111% 2. 1% KKK 1100 mm ssl n 1.00 g ssl n 5. For this problem we must use the concentration as one of several conversion factors that will allow us to determine the number of sodium ions in each solution. To account for the fact that sodium chloride contains one Na + ion per formula unit while sodium carbonate contains two. This is done by incorporating a mole Na + ions to mole compound conversion factor. ( mm NNNN ssl mmm Naaa 1 mmm Na+ n) 1111 mm ssl n 1 mmm NNNN mmm Na+ iiii ( g Na 2 CO 3 sss n) 6. 3 g Na 2CO g ssl 1 mmm Na 2CO 3 2 mmm Na+ n g Na 2 CO 3 1 mmm Na 2 CO mmm Na + iiii The Na 2 CO 3 contains more sodium ions. 6. This problem requires a number of conversion factors. The question asks how much benzene will be consumed in 1 year, so we will start our dimensional analysis with the quantity 1 year. Another important note is that the benzene concentration is given as 5 μg/l, or 5x10-6 g benzene/l solution. 333 dddd L wwwww 5 μμ bbbbbbb 1 g bbbbbbb (1 yy) 1 yy 1 ddd 1 L wwwww g bbbbbbb μμ bbbbbbb 7. This problem utilizes the dilution equation, C 1 V 1 C 2 V 2. We will rearrange the variables to isolate the variable of interest. Also, notice that the volume and concentrations need not be converted to liters or molarity, respectively. As long as the units cancel by dimensional anlaysis, they do not need to be converted into other units. V sssss C dddddddv ddddddd C sssss (0. 88 M)(111 mm) (5. 0 M) 22 mm sssss 8. Again, this problem employs the dilution equation. The most difficult part of this problem may be associating each value with a variable in the dilution equation. C 1 concentration of stock solution?? C 2 concentration of diluted solution 40.0 mm V 1 vol. of stock solution 20.0 ml V 2 vol. of diluted solution 1.00 L1000 ml Before substituting the values for the variables, we need to convert one of the volume measurement units so these units will cancel. C 1 C 2V 2 V 1 (44. 0 mm)(1111 mm) (22. 0 mm) 2222 mm M

9 9. In this problem, we need to convert moles of lead(ii) nitrate to mass before we can use the ppm measurement to find the mass of the solution containing moles of the compound. Recall that ppm can be written as a conversion factor: 22.8 g PP(NO 3) g ssl n mmm PP(NO 3 ) g PP(NO 3) 2 1 x 106 g sss n 2222 g ssl 1 mmm PP(NO 3 ) g PP(NO 3 ) n As a reminder, mass % is the mass of the solute divided by the mass of the solution, not the volume. To convert to mass, we use the density, which is the mass of the solution over the volume of the solution. (555. mm ssl n) g ssl n 1 mm ssl n g HHO g ssl n 1 mmm HHO g HHO mmm HHO 3