POGIL EXERCISE 16 Concentration Terms and There Use

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1 RUN TIME = 80 MIN POGIL 16 Page 1 of 10 POGIL EXERCISE 16 Concentration Terms and There Use Each member should locate his/her role in Table 1 and assume his or her role at this time. The new manager takes charge of the POGIL folder and hands out the GRF and RRF to the appropriate members. The new recorder should record the names of the group members on the new GRF and the Reflector should record the start time on her/his form. Table 1. Group Member Role Assignments GROUP TYPE -> GROUPS OF THREE GROUPS OF FOUR MEMBER NO. -> Manager + + Reporter + + Recorder + + Reflector + + Technician Encourager + + SFUC * + + * OBSERVATION I: The most important skill set related to chemistry demanded by employers and other academic programs is the ability to understand concentration terms and to properly prepare solutions. Concentration terms are used to indicate a quantity of a solute per unit of measure - usually volume (e.g. g/ml) but sometimes mass (e.g. g/g) of a solution. In our world there are two types of concentration terms that differ in the type of quantity being delivered (Table 2). Type I is a set of concentration terms that deliver a specific number of solute formula units. This type of concentration is used primarily in the scientific community particular in the chemical based sciences. Type II is a set of concentration terms that delivers a certain amount in terms of mass or volume. This type of unit is the most widely used in our culture from medicinal doses to amounts of nutrients in our food. Table 2. Classification of Concentration Terms Studied in Our Course I. Delivers an Exact Number of Solute Units Per Unit Solvent Molarity (M; n/l) (Covered in Text: P ) Molality (O; n /Kg solvent) (Covered in Text: ) NOTE: n in Type I (Column 1) stands for moles II. Delivers an Exact Amount of Solute Per Unit Solvent Percent Mass/Volume (%(m/v)) (Covered in Text: P459) Percent Mass/Mass (%(m/m)) (Called Mass Percent in Text: P ) Percent Volume/Volume (%(v/v)) (Covered in Text: P ) + 5 MIN

2 RUN TIME = 80 MIN POGIL 16 Page 2 of 10 OBSERVATION I -- Continued: Let s try to understand the difference between the two types of concentration terms by doing some simple calculations based on principles we already know. First of all we know that mass and numbers are related by the mole concept as seen in Equations 1 and 2 below: EQ1: Numbers of formula units (FU) = moles (n); 6.02 x FU EQ2: n = mass of formula units Formula Mass (FM) Suppose we have on our shelf (Solution A) that is labeled 20 %(m/v) NaOH (FM=40). This is a Type II concentration term. This means that there are 20 grams of NaOH per 100 ml of solution or 0.20 g/ml. Another container on the shelf is labeled 6.00 M NaOH (A Type I term). This means that this solution contains six moles of NaOH per liter of solution or n/ml. Since as chemists we need to know numbers of formula units to make any kind of quantitative decisions let s find the answer to the following question: Which solution has the largest number of NaOH units per 100 ml of solution? 1. Since 20% %(m/v) looks larger than 6 M let s start with this term: We know that 100 ml of this solution contains exactly 20g (20g/100 ml x 100 ml = 20 g). Now calculate how many FU of NaOH are in 20 g using Equations 1 and 2. (Remember: g n #.) 2. From Observation I we know that there are 6.00 moles of NaOH per liter (1000 ml) of solution. Calculate how many moles are in a 100 ml. 3. Now calculate how many FU there are in the number of moles from Item What observation can you make comparing your answers for Items 1 and 3? + 15 MIN

3 RUN TIME = 80 MIN POGIL 16 Page 3 of 10 OBSERVATION II: Why do chemists use Type I concentration terms? We know from our study of stoichiometry and chemical equations that chemical reactions are maximized by having the number of reactant molecules in the proper proportion. Chemists absolutely require concentration terms that deliver the same number of molecules to quantitatively predict the outcome; i.e. the same concentration of two different compounds deliver the same number of formula units of each compound. Molarity was developed for this function and this is why molarity is the most important concentration term for chemists. Molarity (M) is defined as the number of moles (n) per liter (L) of solution. Equation 3 defines molarity. EQ3: M = n/l A rearrangement of Equation 1 gives a new definition for a mole as can be seen in Equation 4. EQ4: M*L = n From previous studies we know that a mole can also be defined by the ratio of the mass of the sample to its formula mass (FM) as seen in Equation 2. Using these rules we are able to measure the exact number of FU we need by measuring masses of the solutes to be added to the solvent. Now let s practice using the concepts in Observation II. 5. How many moles are present in 50 ml of a 2.00 M of HCl? 6. How many moles are present in 100 ml of 1.00 M NaOH? 7. How many grams of NaOH are required to prepare 100 ml of 1.00 M NaOH? 8. If you added 20 g of NaOH to water and made the total volume of the solution 250 ml, what would be the molar concentration of the solution? Present your responses to Items 5-8 to the instructor for validation MIN

4 RUN TIME = 80 MIN POGIL 16 Page 4 of 10 OBSERVATION III: Suppose a chemist wanted to make 100 ml of a 2.00 M solution of NaCl (FW=58.443) in our lab. What mass of the compound would she have to measure? Using Equation 3 we can calculate how many moles will be required to prepare the solution. Since molar concentrations are defined with the volume in liters, we must convert 100 ml to liters before we use the equation ml x 1.00 L = L ; M = _n_ ; n = M x L = (2.00 n/l)(0.100 L) = n 1000 ml L 1 Next, we use a rearrangement of Equation 2 to calculate the number of grams NaCl we need. n = g ; g = n x FM = (.2000 n)( g/n) = g FM She would weigh g of NaCl. (NOTE: All masses for our lab should have three decimal points). Using the information in Observation III and the above example, please complete Table 3. SHOW YOUR CALCULATIONS ON BACK OF SHEET: Table 3. What Mass of Solute Is Required to Prepare the Indicated Solutions? Mass Required Volume M Solute (Name It Here) NaOH g 100 ml M ml 4.00 M 500 ml M *Formula masses were calculated to the nearest whole number Submit your table to the instructor for validation. FM* (sodium hydroxide) KH 2 PO 4 ( ) C 6 H 12 O 6 ( ) MIN

5 RUN TIME = 80 MIN POGIL 16 Page 5 of 10 OBSERVATION IV: PREPARATION OF MOLAR SOLUTIONS. The first thing to remember when preparing any solution is to be as accurate in your measurements as you need to be. For molar solutions you need to be using the tools and techniques that give you the highest degree of accuracy. This means that mass measurements should be made to the nearest milligram (0.001 g). You need to use the most accurate volumetric devices you have in the lab. In most labs this requirement is met by using volumetric flasks. Figure 1 pictures the most common volumetric flasks used today. Each flask has a volume mark etched into the neck of the flask. When the bottom of the meniscus touches this line, the proper volume is in the flask. The procedure used to prepare a solution is to pick the flask that matches exactly the volume required. If the volume required does not match a flask the calculations are adjusted so the next largest Figure 1. Common Volumetric Flasks flask can be used. The general procedure for the preparation of a 100 ml of M NaOH (Exact FM = g) solution is as follows: STEP 1: Locate the proper volumetric flask (i.e. 100 ml volumetric flask) and add approximately 50 ml to the flask. STEP 2: Calculate the mass of NaOH required using Equations 4 and 2 (Calculations shown in Equation 5) and weigh the mass of the solute (in an appropriate EQ5: 100 ml x L = L x 1.00 n = n NaOH x g NaOH = g NaOH 1000 ml L n NaOH 1 weighing dish) required for the preparation. STEP 3: Quantitatively transfer the NaOH to the flask and mix by swirling until dissolved. STEP 4: Add more distilled water in approximately 10 ml portions until the meniscus is just below the volume line. Mix by inverting the flask 3 or 4 times be sure to place the stopper on the top of the flask after each addition.) + 30 MIN

6 RUN TIME = 80 MIN POGIL 16 Page 6 of 10 STEP 5: Bring the volume of the solution to the volume line using a water bottle filled with distilled water. Mix by inversion 3-4 times. STEP 6: Transfer solution to appropriate storage bottle. The bottle should be labeled immediately as indicated in the lab s standard operating procedure in our case: (1) the concentration and formula of reagent, the date prepared and group number of the group that prepared the solution. STEP 7: Store as directed by instructor. Clean up. Process the following items using the concepts in Observation IV and the information in Table 3. ITEMS 11-14: These items concern the preparation of 250 ml of a M NaOH solution. Circle the letter of the option that best answers the question. Show calculations on the back of the sheet on which the problem appears. 11. Which of the following measuring devices should be used to make this solution? a. 250 ml graduated cylinder c. 250 ml beaker d. 250 ml graduated flask b. 250 ml volumetric flask e. 500 ml volumetric flask 12. Approximately how much distilled water should you add to the container before you added the NaOH? a. 250 ml c. 200 ml d. 125 ml b. 100 ml volumetric flask e. none 13. What mass of NaOH would you measure for the preparation? a. 10 g c g d g b g e g 13. A general rule of thumb to follow in the lab is that solutes are added to solvents rather than solvents added to solutes. a. True b. False 14. The label for the storage container of the solution should contain (mark all correct answers). a. FM of solute c. Formula of solvent d. Group number b. Formula (name) of solute e. concentration + 40 MIN

7 RUN TIME = 80 MIN POGIL 16 Page 7 of On the back of this sheet, write a step by step procedure for the preparation of this solution. Recorders should submit responses to Items to instructor for validation. MOLALITY (O) OBSERVATION V. There are some processes that depend on the total number of units in solution. In biology the process of osmosis is a process that depends on the total number of units in solution. A common example in daily life is the freezing point of water. The normal freezing point is 0 C. If you add enough glucose to the water it would lower the freezing point of water significantly. This is why it takes a lower temperature to freeze ice cream than to freeze ice cubes. Glucose does not break apart when dissolved in water so if you added one mole of glucose to a volume of water you would have one mole units in the water. However, if you added the same molar concentration of NaCl to the water the freezing point will be lowered more than the glucose because this salt breaks up into sodium and chloride ions (Equation 4). EQ4: NaCl Na + + Cl - (2 units produced by ionization) Thus, if you added one mole of NaCl to a volume of water, there would be two moles of ions. This is used by the highway department in the winter when they salt the roads to prevent them from icing or when salt is added to ice and water in the homemade ice cream. But if you were to add the same concentration of ammonium sulfate as NaCl to water, the freezing point is lowered even further (Equation 5). And if we used the same concentration of aluminum chloride, the freezing point would be lowered even more (Equation 6). EQ5: (NH 4 ) 2 SO 4 2 NH SO 4-2 (3 units produced by ionization) EQ6: AlCl 3 Al Cl - (4 units produced by ionization) Molality is defined by the number of moles dissolved in a kilogram of solvent as seen in Equation 7 below. Mass is used as a base because the volume of a solution changes EQ7: O = n Kg Solvent with temperature and if the volume changes the concentration changes. Mass does not vary with temperature and thus concentrations based on mass do not change. This is required in some very precise processes fortunately none of these are in our lab. Please respond to Items (next page) using the information in Observation IV. Show all calculations on the back of the next page MIN

8 RUN TIME = 80 MIN POGIL 16 Page 8 of Which of the following solutions would lower the freezing point of water the most assuming that they were the same molality? a. Sucrose (table sugar) d. NaCl b. Ammonium sulfate e. Aluminum nitrate c. Sulfuric acid f. Sodium sulfide 17. What mass of NaOH (FM = 40) is required to prepare 2.00 kg solution of a 2.00 O solution? 18. What mass of water is required to prepare 2.00 kg solution of a 2.00 O NaOH solution? (HINT: Total mass of solution = mass of solute + mass of solvent.) Recorders should submit responses to Items to instructor for validation. OBSERVATION VI - PERCENT CONCENTRATION TERMS (TYPE II): Type II concentration terms are the most common concentration terms seen by most people including health professionals. These concentration terms are used to assure the delivery of a certain quantity of (usually mass) of solute to the system. All of these terms are based on parts per hundred i.e. a percent basis. The base is always 100; the units are either in volume (v) or mass (g). The three most common in use are: percent mass of solute/volume of solution (%m/v) defined by Equation 8, percent volume of solute/volume of solution EQ8: % m/v = mass of solute x ml solution (%v/v) defined by Equation 9, and percent mass of solute per mass of solution (called EQ9: % m/v = ml of solute x ml solution mass percent in the text; %m/m) defined by Equation 10. EQ10: % m/v = mass of solute x g solution + 50 MIN

9 RUN TIME = 80 MIN POGIL 16 Page 9 of 10 OBSERVATION VI continued: As an example to work any percent concentration problem let s use this example: What mass is required to prepare 300 ml of a 20% (m/v) solution of sodium hydroxide? By definition we understand that a 20% (m/v) solution would contain 20 g of NaOH per 100 ml solution (Equation 11). In a similar manner we know that 300 ml of a EQ11: 20 %(m/v) = 20 g NaOH. 100 ml solution 20% solution is defined by Equation 12. Since both equations are equal to 20% (m/v), EQ12: 20 %(m/v) = x g NaOH. 300 ml solution they are equal to each other (Equation 13). Resolving Equation 13 as seen in EQ13: 20 g NaOH = X g NaOH. 100 ml solution 300 ml solution Equation 14 we get the correct answer that the preparation would require 60 g NaOH. EQ13: 20 g NaOH = X g NaOH ; X g NaOH = (20 g NaOH)(3) = 60 g 100 ml solution 300 ml solution 1 3 This solution would best be prepared in by locating 300 ml volumetric flask and adding about 150 ml of distilled water to the flask. Measure g of NaOH and quantitatively transfer to the volumetric flask. Mix by swirling until all the NaOH is dissolved. Water is added in approximately ml amounts with mixing by inversion in between. Finally, water is added with water bottle until meniscus is touching the volume line. The solution is mixed by inversion and transferred to a storage bottle. The bottle is labeled 20% NaOH, a designation as to who prepared the solution and date of preparation. Using information in Observation VI process Items on the next page MIN

10 RUN TIME = 80 MIN POGIL 16 Page 10 of 10 INSTRUCTIONS: For the following problems record the calculations on the back of this sheet; place the answers in the blanks below. 19. What mass of NaOH would be required to prepare 250 ml of a %(m/v) solution? 20. How would you prepare the solution mentioned in Item 19? 21. What would be the percent m/v concentration of a solution if you placed 5 g of NaCl in a 250 ml volumetric flask and filled it up to the volume mark? 21. How many ml of pure isopropyl alcohol would be required to make commercial 70% (v/v) solution? 23. What mass of ammonium sulfate would be required to prepare kg of a 10 %(m/m) solution? 24. What mass of water would be required to prepare the solution in Item 23?. Recorders should submit responses to Items to instructor for validation. EXERCISE END. The Recorder and Reflector should complete their forms and give them to the manager. The manager should attach the forms together and place them in t he left pocket of the group folder. NOTIFY INSTRUCTOR WHEN FINISHED. WAIT FOR INSTRUCTIONS. EXERCISE END. 80 MIN