Unit 10: Solutions. Student Name: Class Period: Page 1 of 57 Lecture notes. Unit 10: Solutions-Lecture Regents Chemistry Mr.

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1 Unit 10: Solutions Student Name: Class Period: Page 1 of 57

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3 Unit 10 Vocabulary: 1. Aqueous: A solution in which the solvent is water. 2. Colligative Property: A property of a solution that is dependent on concentration. Examples include boiling point, freezing point, and vapor pressure. 3. Molality: The concentration of a solution measured in moles of solute per kilogram of solvent. 4. Molarity: The concentration of a solution measured in moles of solute per liter of solution. 5. Parts per Million: The concentration of a solution measured in mass of solute per mass of solution multiplied by one million. 6. Percent by Mass: The concentration of a solution measured in mass of solute per mass of solution multiplied by one hundred. 7. Percent by Volume: The concentration of a solution measured in volume of solute per volume of solution multiplied by one hundred. 8. Saturated: Any solution that has the maximum concentration of a dissolved solute possible in a given quantity of solvent at a given temperature. A saturated solution is a solution at equilibrium. 9. Solubility: The maximum quantity of a solute that may be dissolved in a given quantity of solvent at a given temperature to make a saturated solution. 10. Solute: Any substance that is broken apart by a solvent and kept separate by the solvent particles. 11. Solution: A homogenous mixture formed when a solute dissolves into a solvent. 12. Solvent: Any substance that attaches to solute particles, breaks the solute apart, and then keeps the separated particles apart in solution. Page 3 of 57

4 13. Supersaturated: A solution that has an excess of solute beyond the solubility point for a given temperature. Excess solute will either precipitate out, or remain in an unstable dissolved state until the supersaturated solution is disturbed. This causes the excess solute to precipitate from solution, leaving the solution saturated. 14. Unsaturated: A solution in which there are solvent particles that have no attached solute particles, and therefore has the capacity to take more solute into solution. Page 4 of 57

5 Unit 10 Homework Assignments: Assignment: Date: Due: Page 5 of 57

6 Notes page: Page 6 of 57

7 Topic: Mixtures Objective: What is the result of combining two dissimilar substances? Mixtures: A mixture is made by physically combining two (or more) substances without any type of chemical reaction occurring. Mixing ionic compounds with water forms aqueous solutions composed of dissolved ions. The polar water molecules attach to the ions, separating them from other ions. The polar water molecules keep the ions separate, holding the ions apart. This property is called molecule-ion attraction. Page 7 of 57

8 Topic: Molecule-Ion Attraction Objective: How are ions situated within a solution? A solid crystal of NaCl is placed into a beaker of water. Immediately the polar water molecules attract the charged ions. The partially positive ends of the water molecules (hydrogen atoms) attract the negatively charged chloride ions. The partially negative ends of the water molecules (oxygen atoms) attract the positively charged sodium ions. When enough water molecules attach to an ion, the natural kinetic motion of the liquid water molecules will tear the ions from each other. This is the molecule-ion attraction. Page 8 of 57

9 The hydrated (aqueous) ions are kept separated by the motion of the water molecules that are attracted to them. In this example, four water molecules around each ion are enough to keep the ions separated. There are still free water molecules in the solution which could tear apart, attract, and keep separate additional solute ions. This is an example of an unsaturated solution. If more NaCl (s) were added until all water molecules were attached to ions, the solution would then become saturated. At the saturation point, any additional NaCl (s) added would simply sink to bottom of the beaker without going into solution, or drive other Na +1 or Cl -1 ions out of solution as a precipitate. The number of water molecules required to keep ions apart depends on water temperature. The higher the temperature, the greater the average kinetic energy, and the faster the water molecules move, so fewer water molecules would be required to keep the ions apart. Page 9 of 57

10 Topic: Solubility Objective: How may we know the amount that enters into solution? Solubility: The quantity of a solute that may be added to a given quantity of solvent at a given temperature and pressure is known as solubility. Saturated: A saturated solution holds as many dissolved particles as it has the capacity to hold. Capacity is the maximum spaces available for anything. The chemistry room has 23 desks; this gives Room 218 the capacity for 23 students unless more desks are brought in, or if desks are taken out. In a saturated aqueous solution, all the solvent water molecules (desks) have an ion (student butt) attached to them, so no more ions (students) could come into the solution (Room 218) and find a place to stay (empty desk). Any additional ions (students) would need to pass through the solution (room). Unsaturated: An unsaturated solution holds fewer dissolved particles as it has the capacity to hold. Room 218 has 23 desks, and the largest chemistry class (for ) has 21 students, so there is capacity for two more students. In an unsaturated aqueous solution, there are solvent water molecules (desks) without an ion (student butt) attached to them, so more ions (students) could come into the solution (Room 218) and find a place to stay (empty desk). Page 10 of 57

11 Supersaturated: A supersaturated solution is very rare solution in where the solution holds more dissolved solute than is theoretically possible. A supersaturated solution is an unstable solution that will cause the excess solute to leave (precipitate) if disturbed. The solution (Room 218) has 23 desks (water molecules) that each can attach to one ion (student butt). If there are more than 23 students (ions) sitting, whereas some desks have more than one student (ion) sitting in them, this is over capacity, or supersaturated. The solution can easily be upset if a disturbance (Murdoch) comes in contact with the system, and forces the extra ions (students) out of the solution (Room 218). Factors affecting solubility: Three factors affect solubility: 1. Temperature 2. Pressure 3. Nature of solute and solvent Watch Crash Course Chemistry Solutions - YouTube - 8:20 Page 11 of 57

12 Topic: Temperature and Solubility Objective: How does temperature affect the solubility of a solution? Temperature: 1. For solid and liquid solutes, solubility in water increases as water temperature increases. 2. For gaseous solutes, solubility in water decreases as water temperature increases. Effect of temperature on solubility of a solid solute in water: An example of the effect of temperature on aqueous solubility of solids is the sugar water needed to form rock candy. A saturated solution is formed at high temperature. As the saturated sugar solution cools, the sugar solute becomes less soluble, causing some water molecules to jump off a sugar molecule and assist other water molecules in holding other sugar molecules apart. This allows the now free sugar molecules to come out of solution, and as more water molecules are pulled off sugar molecules, more and more sugar will precipitate from the solution as sugar rock candy. Page 12 of 57

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14 Effect of temperature on solubility of a gaseous solute in water: An example of the effect of temperature on aqueous solubility of gases is as easy to find as your favorite carbonated beverage. Take two identical bottles of the same carbonated beverage, place one in the refrigerator overnight, and leave the other on the kitchen counter. When you open both bottles side by side you will see that the colder beverage will release much less carbonation (bubbles) than will the warmer beverage. The colder solution allowed more gaseous CO 2 to remain dissolved than did the warmer solution. If you keep observing both bottles, as the colder beverage warms, it will continue to lose dissolved CO 2 (bubbles) during the warming process, while the warmer bottle will form fewer bubbles during the same elapsed time. Page 14 of 57

15 Topic: Pressure and Solubility Objective: How does pressure affect solubility of solutes? Pressure: For solids and liquids, pressure has either no effect or a negligible effect on solubility. For gaseous solutes, solubility increases as pressure increases. i. In commercial soda machines, as water travels through a tube, flavored syrup is added to the water along with carbon dioxide gas. Gases don t have an affinity to form aqueous solutions and require pressure to force the CO 2 gas into the water and syrup solution. When the aqueous syrup and CO 2 mixture is bottled, the gas is trapped within the volume of the container, and equilibrium between dissolved CO 2 molecules and free CO 2 molecules quickly forms. When the container is opened, the pressure decreases, allowing the CO 2 to escape as bubbles. CO 2 gas is soluble at high pressures (sealed container) but nearly insoluble at low pressure (open container). Page 15 of 57

16 Topic: Polar Nature of Solvent Objective: Does the polarity of a solute or solute apply in solutions? Nature of Solute and Solvent (Like dissolves Like): Polar solutes dissolve in polar solvents. o Water is a polar molecule and therefore has partially charged ends that can attract other polar or ionic solutes. This is why ionic solutes such as NaCl and polar molecular solutes such as glucose (sugar) may be dissolved in water. o Water s polar structure has little attraction for nonpolar molecular solutes like oil (remember Hank squishing butter?) Oils will not mix (are immiscible) with water, and being less dense, will float on top of the water. This is why water is NOT used to extinguish oil fires; the water will go under the fire, and float the fire to a new location. Nonpolar gases such as CO 2 and O 2 are even harder to force into aqueous solution because of this. Nonpolar solutes dissolve in nonpolar solvents. o Oils will not dissolve in polar water solvent, but oils will easily dissolve in the nonpolar organic solvent benzene (C 6 H 6 ). Whereas water is a great polar solvent, benzene is a good nonpolar solvent. Acetone is a nonpolar solvent used to remove nonpolar mixtures such as nail polish. Page 16 of 57

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18 Solubility Regents Practice Problems (ungraded): 1. At room temperature, the solubility of which solute listed below would the most affected by a change in pressure? a) Sugar (s) c) Sodium nitrate (aq) b) Methanol (l) d) Carbon dioxide (g) 2. As the pressure on a gas confined above a liquid increases, the solubility of the gas in the liquid a) Increases b) Decreases c) Is unchanged 3. Sublimated carbon dioxide is most soluble in water under the conditions of a) Low pressure and low temperature b) High pressure and low temperature c) Low pressure and high temperature d) High pressure and high temperature 4. At which temperature given below could water contain the most dissolved gas at a pressure of kpa? a) 183 C b) 283 K c) 383 K d) 483 C Page 18 of 57

19 Topic: Solubility Curves Objective: How may we use solubility curves to determine solubility? Reference Table G: Solubility Curves at Standard Pressure: 1. For Table G, the SLOPE of the curve tells you about the solute! a. IONIC salts solutes have UPWARDS slopes; b. GAS solutes (SO 2 and NH 3 ) have DOWNWARDS slopes. 2. Interpreting Reference Table G: a. The Table G x-axis is the temperature in C. If you are given a temperature in K, you ll need to convert (K = C). Note that the numbers are significant to TWO places, as all numbers have a decimal (.) at the end. Note also that the interval is to the tens place (10. C), so precision may be interpolated to the ones place and will still be significant to TWO digits. b. The Table G y-axis is the solubility in grams of solute per 100. grams of water. Again, the numbers are significant to TWO places, as all numbers have a decimal (.) at the end. Note also that the interval is to the tens place (10. grams), so precision may be interpolated to the ones place and will still be significant to TWO digits. c. Also note the segment of the curve for potassium iodide WAY up in the extreme upper left corner. Students have missed the KI curve; be aware of its presence BEFORE you need it! Page 19 of 57

20 3. Table G is only useable at standard pressure (1 atm or kpa); if you have other than standard pressure, you ll need to infer. Solubility of a solute in 100. grams of water at various temperatures: 1. Each line on Table G represents a saturated aqueous solution of the given solute for a given temperature. The farther up the y-axis the saturation line is at a given temperature, the more soluble the solute is. To find the saturation point at a given temperature, simply find the given temperature and then go straight vertically up until you intersect the solubility line you need, and then traverse level left and read the solubility on the y-axis. The table is set for grams of solute/100. grams of H 2 O; if you have other than 100. g of water you will need to find the correct proportion. Page 20 of 57

21 Topic: Degree of Saturation Determining Objective: How Molecular do we use Polarity: solubility curves to determine saturation? Degree of saturation according to Table G: 1. Unsaturated: 2. Saturated: Page 21 of 57

22 3. Supersaturated: Page 22 of 57

23 Topic: Solubility in other than 100. g Objective: Will Table G determine solubility not in 100. g of water? Solubility for other than 100. g of water solvent: Reference Table G may only be directly used in the mass of the solute is dissolved in 100. grams of water as a solvent. With simple math, by finding the proportionality of the given amount of solvent to 100. g of water as listed on Table G, we can easily calculate the solubility at the given mass of water. If given a question with OTHER than 100. grams of water: 1. For the stated temperature in the question, find the saturation solubility in grams of solute per 100. grams of water. 2. Place the question s stated amount of water as a numerator and 100. g of water as the denominator. This will give you a multiplier. i. If you had MORE than 100. grams of water in your question, the multiplier factor will be greater than 1. ii. If you had LESS than 100. grams of water in your question, the multiplier factor will be less than Multiply the saturation solubility from Table G by the multiplier factor from Step #2, and you will have your saturation solubility for other than 100. grams of water. Page 23 of 57

24 I. How many grams of KClO 3 are required to make a saturated aqueous solution in 100. grams of water at 30. C? a. Find 30. C on the Table G x-axis, then move straight up to the KClO 3 curve. Move straight over to the y-axis, and read the scale (12 g). This means that 12 grams of KClO 3 are soluble in 100. grams of water at 30. C. Page 24 of 57

25 II. How many grams of KClO 3 are required to make a saturated aqueous solution in 50. grams of water at 30. C? a. We are now asked for the solubility of KClO 3 in 50. grams of water. Place 50. g of water in the numerator, and 100. g of water in the denominator, and solve. 50. g of water 100. g of water = 0.50 multiplier b. We used Table G to determine that 12 g of KClO 3 are soluble in 100. g of water. We can use the multiplier factor with the 12 g of KClO 3 and determine the mass of KClO 3 soluble in 50. g of water. c. 12 g of KClO 3 x 0.50 = 6.0 g of KClO 3 soluble in 50. g of water at 30. C III. How many grams of KClO 3 are required to make a saturated aqueous solution in 200. grams of water at 30. C? a. We are now asked for the solubility of KClO 3 in 200. grams of water. Place 200. g of water in the numerator, and 100. g of water in the denominator, and solve g of water 100. g of water = 2.00 multiplier b. We used Table G to determine that 12 g of KClO 3 are soluble in 100. g of water. We can use the multiplier factor with the 12 g of KClO 3 and determine the mass of KClO 3 soluble in 200. g of water. c. 12 g of KClO 3 x 2.00 = 24 g of KClO 3 soluble in 200. g of water at 30. C Page 25 of 57

26 IV. Let s say you have to make a saturated NaCl solution using tap water (10 C) but you only have an empty (355 ml) soda can. Of course you know that water has a density of very close to 1.0 gram/ml near 10 C, so you can use the 355 ml volume of your soda can to measure out 355 grams of water. How much NaCl will be needed to form the saturated solution with the 10 C water? a. First, start at 10 C on Table G and go up until you find the solubility curve for NaCl (blue dashed line), and then move left until you reach the y-axis scale (green dotted line). I ll call it 37. However, that 37 g of NaCl is for 100 grams of water! Since you have 355 grams of water, 355 g needs to go in the numerator, with the Table G standard of 100. grams of water as the denominator. 355 g of water 100. g of water = 3.55 multiplier b. We can now take the interpolated 37 grams of NaCl per 100. grams of water and multiply that by 3.55 to give us the saturation point of NaCl in 355 grams of water. 37g Page 26 of 57

27 Table G example questions: 1. What is the solubility of HCl in 100. g of water at 70 C? a. Find 70 C on the x-axis, move straight up to the HCl curve, and then transit level left to the y-axis. b. 52 g of HCl are soluble in 100. g of water at 70 C 2. What type of solution saturation do you have if 20. g of KClO 3 are dissolved in 100. g of water at 40. C? a. Find 40. C on the x-axis, move straight up to the KClO 3 curve, and then transit level left to the y-axis. b. 16 g of KClO 3 are soluble in 100. g of water at 40. C. As you were asked what 20. g of KClO 3 in solution would be, since 20 is greater than 16, the solution would be supersaturated. 3. For the question above, how can we make the existing solution saturated? a. As calculated using Table G, at 40 C only 16 g of KClO 3 could be dissolved to make the solution of aqueous KClO 3 saturated. With the stated 20. g of KClO 3, it was supersaturated. We can make the solution saturated if we increase the temperature to around 48 C, the point that the curve crosses the 20. g solute/100. g H 2 O level. b. What happens in the example if we increase the temperature of the aqueous solution to 75 C without adding more solute? When we find 75 C on the x-axis and go straight up to the curve, we see that 75 C crosses the curve at 40. g solute/100. g H 2 O level. That would mean that with a capacity of 40. g, and a dissolved solute amount of 20. g, the aqueous solution of 20. g per 100. g of water would only be about 50% saturated at 75 C, or about 50% unsaturated. Page 27 of 57

28 Solubility Regents Practice Problems (ungraded): 1. Solubility data for four different salts aqueous solution at 60 C are given in the data table below. Which given salt is the most soluble at 60C? a) Salt A b) Salt B c) Salt C d) Salt D 2. According to Reference Table G, which of these substances is most soluble at 60 C? a) Sodium chloride c) Potassium chlorate b) Potassium chloride d) Ammonium chloride 3. According to Reference Table G, how many grams of potassium nitrate would be needed to saturate 200 grams of water at 70 C? a) 43 g c) 134 g b) 86 g d) 268 g 4. According to Reference Table G, how does decreasing the average kinetic energy affect the solubility of NH 3 and KCl? a) The solubility of NH 3 increases, and the solubility of KCl increases. b) The solubility of NH 3 increases, and the solubility of KCl decreases. c) The solubility of NH 3 decreases, and the solubility of KCl increases. d) The solubility of NH 3 decreases, and the solubility of KCl decreases. Page 28 of 57

29 Student name: Class Period: Please carefully remove this page from your packet to hand in. Solutions Homework Complete the following the chart: (1 pt. ea.) For the solutes below, in aqueous solution: If temperature of the solvent is increased, the solubility of this solute will: If the temperature of the solvent is decreased, the solubility of this solute will: If the surface area of this solute is increased, the solubility of this solute will: NH 3(g) KCl (s) Complete the following chart: (1 pt. ea.) Sketch the structures of the compounds. Is the solute polar, nonpolar, or ionic? Will this solute dissolve in water or benzene? CH 4 KBr H 2 S Explain why, in terms of molecular polarity, why oils will not easily dissolve in water. (1 pt.) Cont d next page Page 29 of 57

30 Use Reference Table G to answer the following: (1 pt. ea.) 1. As the temperature of water decreases, the solubility of gases will? 2. As temperature of water decreases, the solubility of solids will? 3. Which salt on Table G is the least soluble at 90. C? 4. Which gas on Table G is the most soluble at 60. C? 5. Which salt on Table G shows the least change in solubility between 30. C and 70. C? For #6 through #8, determine whether the following solutions are Unsaturated, Saturated, or SuperSaturated in 100. grams of water g of KNO 3 at 60. C g of NH 4 Cl at 30. C g of KCl at 60. C 9. What is the solubility of NaNO 3 in 100. grams of water at 40. C? 10. How much KClO 3 may be dissolved in g of water at 30. C? 11. What is the solubility of NaCl in 50. g of water at 70. C? 12. How many grams of NH 4 Cl may be added to a solution containing 30. g of NH 4 Cl to make a saturated solution in 90. C H 2 O? 13. A solution has 80. g of KNO 3 dissolved in 80. C H 2 O. At what temperature would the solution become saturated? 14. A saturated solution of NaNO 3 cools from 343 K to 293 K. If the solution stays saturated, how many grams of NaNO 3 precipitate? 15. A saturated solution made in 50. g of water at 303 K of NaNO 3 is left out and fully evaporates. What is the mass of NaNO 3 left? 16. At what temperature will KCl and HCl have the same solubility? Page 30 of 57

31 37 Topic: g of NaCl x 3.55 = , Concentration or about 130 grams of NaCl are soluble in 355 grams of water at 10 C Objective: How do we express the amount of solute per unit volume? Concentration: The concentration of a solution is a measure of the amount of substance per unit of volume. a. When discussing solutions, there are several ways to express concentration. 1. Grams of solute/100 g of solvent (water): This is how Reference Table G compares solubility. 2. Molarity (M): Molarity is the number of moles of solute per liter of solvent. Molarity is the most common unit of concentration used in the laboratory. Page 31 of 57

32 Topic: Determining Molarity Objective: How do we work with a single element and compound? Determining Molarity Experimentally: We can experimentally determine the molarity of a solution by 1. Measuring the volume of the liquid solution you have 2. Determine the number of moles of solute in the solution 3. Evaporate all solvent leaving only solute 4. Mass the recovered solute 5. Divide the measured mass of recovered solute by the gram formula mass of the solute Page 32 of 57

33 Molarity examples: 1. What is the molarity of a solution containing 2.0 moles of KNO 3 dissolved in 4.0 L of water? a. M = moles of solute/l of solution 2.0 moles KNO3 KNO3 4.0 L water = 0.5 M KNO 3 2. What is the molarity of a solution containing 2.0 moles of HCl dissolved in 250. ml of water? a. As molarity is in units of liters (L) of solution, first convert ml to L. 1 L b. 250 ml x = L 1000 ml c. M = moles of solute/l of solution 2.0 moles HCl L water = 8.0 M HCl 3. What is the molarity of a solution containing 20. grams of NaOH dissolved in 2.0 L of water? a. As molarity is in units of moles of solute, first convert grams of NaOH to moles of NaOH after finding the gram formula mass for NaOH. b. NaOH = ( ) = 40.0 g / mole NaOH c. Then convert 20. grams of NaOH to moles of NaOH. d. 20. g NaOH x 1 mole NaOH 40.0 g NaOH = 0.5 mole NaOH e. M = moles of solute/l of solution 0.5 moles NaOH 2.0 L water = 0.25 M NaOH Page 33 of 57

34 4. What is the molarity of a solution containing 60. grams of NaOH dissolved in 400. ml of water? a. As molarity is in units of moles of solute, first convert grams of NaOH to moles of NaOH. b. 60. g NaOH x 1 mole NaOH 40.0 g NaOH = 1.5 mole NaOH c. As molarity is in units of liters (L) of solution, first convert ml to L. 1 L d. 400 ml x = L 1000 ml e. M = moles of solute/l of solution NaOH 1.5 moles NaOH L water = 3.75 or 3.8 M Page 34 of 57

35 Topic: Calculating Molarity in Lab Objective: How do we make a solution with a desired molarity? The basic equation for molarity of molarity = moles of solute/liter of solution (Reference Table T) may be rearranged as needed. i. To make a specific volume of a solution with a desired molarity: 1. If you are given a volume and a molarity, you need to solve for moles of solute. 2. Moles of solute = M (molarity) x L (volume of solution) ii. Steps to make your solution: 1. Determine the desired concentration (what is the molarity of the solution you are trying to make? This is usually told to you in the lab handout.) 2. Determine how much solution you want to make (how many L or partial L of solution you need? This, again, is usually told to you in the lab handout.) 3. Multiply the first two step s answers. This will give you your 4. amount of solute in moles. Moles of solute L of solution x L of solution = moles of solute 5. Calculate the gram formula mass for your required solute, to the nearest tenth of a gram/mole. Page 35 of 57

36 6. Multiply the number of moles of solute needed from step 3 by the gram formula mass of your solute. 7. # of moles calculated x Formula mass in g 1 mole = grams of solute needed 8. Measure your calculated grams of solute. 9. Add your solute to a half-full volumetric flask that matches your needed volume. 10. When the solute is dissolved, fill the flask to the needed volume. Wow. Seems like a LOT of work, doesn t it? Why bother? Who would want to do all that? Well, I (or any other chemistry teacher, lab technician, nurse, doctor, veterinarian, etc.) have to do ALL of those steps frequently. Creating a solution to a specific molarity is one of the most commonly done tasks in chemistry. Once you understand the basics, you will be able to do it for life. Page 36 of 57

37 Examples for calculating molarity: 1. How many grams of NaOH are needed to make 4.0 L of 0.50 M aqueous NaOH? a. Determine the number of moles of NaOH needed: Moles = M x L (0.50 moles/l) x (4.0 L) = 2.0 moles of NaOH b. Convert moles of solute to grams of solute: Grams = moles of solute x gram formula mass of solute = (2.0 moles NaOH) x (40.0 g NaOH/mole NaOH) = 80. g of NaOH c. You can now measure and then add 80. grams of NaOH to a partially filled 4.0 L volumetric flask and add distilled water to the 4.0 L level. 2. How many grams of KCl are needed to make 500. ml of M aqueous KCl? a. Molarity is in units of liters (L) of solution, first convert ml to L. 1 L b ml x = L 1000 ml c. Determine the number of moles of KCl needed: Moles = M x L (0.100 moles/l) x (0.500 L) = moles of KCl d. Calculate the gram-formula mass of KCl: KCl = ( ) = 74.6 g KCl/mole e. Convert moles of solute to grams of solute: Grams = moles of solute x gram formula mass of solute = ( moles KCl) x (74.6 g KCl/mole KCl) = 3.73 g of KCl f. You can now measure and then add 3.73 grams of KCl to a partially filled 500. ml volumetric flask and add distilled water to the 500. ml level. Page 37 of 57

38 Topic: Parts-per-million (ppm): Parts-per-Million Objective: How would we calculate the concentration in mass? Parts-per-million is a mass-to-mass concentration for determining impurities or pollutants. Parts-per-million is used to determine trace amounts of ions in drinking water. PPM may also be used to measure atmospheric or industrial air pollutants. For some very toxic materials concentrations may be expressed in parts-per-billion. i. What is the ppm concentration of lead ions when g of lead is dissolved in 100. grams of water? Ppm = ( g of solute g of solution = 0.45 ppm Pb ) x 1,000,000 = ( g of Pb 100. g of solution ) x 1,000,000 Page 38 of 57

39 Ppm examples: 1. A well drilled on property near a landfill was tested before a housing development was constructed. A kg sample of the well water was found to have g of cadmium, a toxic heavy metal found in rechargeable batteries. a. What is the ppm concentration of cadmium in the well water? Ppm is measured in grams of solution, convert kg to grams: kg x g kg = grams g of solute g of Cd Ppm = x 1,000,000 x 1,000,000 = g of solution g of water 3.30 ppm of cadmium 2. What would be ppm concentration of a solution containing 5.00 g of solute dissolved in 250. g of water? For this example you are given the components of the solution, not the total mass of the solution. We need to add both components (solute and solvent) together g of solute g of water = 255 g of solution Ppm = g of solute g of solution x 1,000, ppm 19,600 ppm 5.00 g of solute 255 g of solution x 1,000,000 = Page 39 of 57

40 Topic: Percent by Mass & Volume Objective: Can we calculate concentration as a function or percent? Percent by Mass: Percent by Mass = (grams of solute/grams of solution) x 100 i. A 25.0 g sample of aqueous NaCl solution is evaporated and 2.0 g of NaCl crystals are recovered. What is the percent by mass of the NaCl in the original solution? ii. Percent by Mass = 8.0% NaCl by mass Percent by Volume: g of solute g of solution x 100 = 2.0 g of NaCl 25.0 g of solution Percent by Volume = (ml of solute/ml of solution) x 100 x 100 = i. Percent by volume is often used to describe the concentration of alcohol. Whiskeys are supposed to be 50% ethanol (EtOH) or more; 50% EtOH (aq) is the point where it remains flammable. A customer thinking it was too watered down might ask for proof it wasn t watered-down, and if it didn t burn, it was not strong enough. This is the basis for the term proof in comparing distilled spirits. ii. A 50.0 ml sample of an aqueous ethanol solution is distilled to yield 33.2 ml of EtOH. What percent by volume of EtOH was the original solution? ml of solute 33.2 ml of EtOh Percent by volume = x 100 x 100 ml of solution 50.0 of solution = 66.4% ethanol by volume Page 40 of 57

41 Concentration Regents Practice Problems (ungraded): 1. What is the molarity of a solution containing 20. grams of NaOH in 500 ml of water? a) 1 M c) 0.5 M b) 2 M d) 0.04 M 2. How many moles of solute are in 200 ml of a 1 M solution? a) 1 c) 0.8 b) 0.2 d) What is the concentration of a solution containing 10. moles of copper(ii) nitrate in ml of water? a) 1.0 M c) 5.0 M b) 2.0 M d) 0.50M 4. What is the total number of grams of NaI(s) needed to make 1.0 liter of a M aqueous solution? a) 15 c) 0.15 b) 1.5 d) Using your reference tables, which compound listed would form the most concentrated aqueous solution? a) AgBr c) AgNO 3 b) AgCl d) Ag 2 CO 3 6. What is the concentration of a solution, in parts per million, if 0.02 gram of Na 3 PO 4 is dissolved in grams of water? a) 2 ppm c) 0.2 ppm b) 20 ppm d) 0.02 ppm Page 41 of 57

42 Notes page: Page 42 of 57

43 Student name: Class Period: Please carefully remove this page from your packet to hand in. Concentrations Homework You must show ALL work, including numerical setup, units, and correctly rounded answers. (4 pt. ea.) 1. Calculate the molarities of the following solutions: a) 2.0 moles of NaCl (s) in 4.0 liters of solution b) 34 grams of CaSO 4(s) in ml of solution c) 28 grams of KOH (s) in 100. ml of solution 2. Calculate the mass of solute needed to make the following solutions: a) 1.0 L of 2.0 M NaOH (aq) b) 500. ml of 10.0 M HCl (aq) c) 1.0 L of 5.0 M Ba(NO 3 ) 2(aq) grams of NaOH (s) were recovered after 50.0 ml of NaOH (aq) completely evaporated. What was the molarity of the original solution? 4. Calculate the percent by mass of 35.8 g of Na 2 SO 4(aq) in grams of solution. Page 43 of 57

44 5. Calculate the percent by volume of 4.55 ml of ethanol in 7.83 ml of solution. Parts Per Million: (Show ALL work) 6. A sample is found to contain g of Pb +2 in 10 grams of tap water. Calculate the concentration of the lead in ppm for this sample. 7. The tap water analyzed above came from a site near a landfill. The EPA tested groundwater samples near the landfill, and found the average groundwater sample around the landfill contained 2.7 x 10-4 grams of Pb +2 per kilogram of water. State your conclusion if the lead in the tap water came from the landfill, and back up the answer with your calculations. 8. A ml sample of 8.50 ppm aqueous solution completely evaporates. What will be the mass of the recovered solute? 9. A manufacturing plant has a maximum allowable airborne particulate limit of ppm within the plant clean room. If the clean room contains a total of 450. kg of air, what would be the maximum allowed mass (in grams) of total airborne particulates allowed in the clean room? Page 44 of 57

45 Topic: Colligative Properties Objective: What properties are based on the solution concentration? A property of a solution that is dependent on concentration is known as a Colligative Property. Colligative Properties of Solutions: (Electrolytes) An ionic compound or acid that forms an aqueous solution that conducts electricity is an electrolyte. Electrolytes have free-moving ions in solution that allow the electrons to flow through the solution. Pure (distilled) water will not flow electricity as it contains no dissolved ions (I personally wouldn t work with electricity while standing in a puddle containing distilled water!) Electrolytes will 100% ionize in water in a reaction that resembles a decomposition reaction. The ionization reaction is called dissociation, and is a physical change, not a chemical change. The more ions a solution forms when it dissociates, the higher the boiling point and the lower the freezing point of the electrolytic solution. Page 45 of 57

46 Colligative Properties Examples: (Electrolytes) 1. NaCl (s) Na +1 (aq) + Cl -1 (aq) a. One mole of sodium chloride dissociates into one mole of sodium cations and one mole of chlorine anions (two total moles of ions). 2. CaBr 2(s) Ca +2 (aq) + 2 Br -1 (aq) a. One mole of calcium bromide dissociates into one mole of calcium cations and two moles of chlorine anions (three total moles of ions). 3. Al(NO 3 ) 3(s) Al +3-1 (aq) + 3 NO 3 (aq) a. One mole of aluminum nitrate dissociates into one mole of aluminum cations and three moles of polyatomic nitrate anions (four total moles of ions). Page 46 of 57

47 Nonelectrolytes: Covalently bonded substances (except many acids) do not become electrolytes when forming aqueous solutions. This includes polar molecules that dissolve, but not dissociate, in water. Examples of polar molecules that dissolve but don t dissociate are sugars (C 6 H 12 O 6, C 12 H 22 O 11 ), ethylene glycol (CH 2 OHCH 2 OH), and ethanol (C 2 H 5 OH). These solutes have less of an effect on the freezing and boiling points of solutions than do ionic compounds as they don t dissociate further. Nonelectrolyte examples: 1. C 12 H 22 O 11(s) C 12 H 22 O 11(aq) a. One mole of solid sucrose dissolves in water to form one mole of aqueous sucrose solution. No ions are formed, and no charges exist to flow electrons. 2. CH 2 OHCH 2 OH (l) CH 2 OHCH 2 OH (aq) a. One mole of ethylene glycol dissolves in water to form one mole of aqueous ethylene glycol solution. No ions are formed, and no charges exist to flow electrons. Page 47 of 57

48 Topic: Colligative Properties Objective: What properties are based on the solution concentration? Colligative Properties of Solutions: (Phase change) 1. Freezing Point Depression: The freezing point temperature of water is depressed (decreased) when a solute is added to the water. 2. Boiling Point Elevation: The boiling point temperature of water elevates (increases) when a solute is added to the water. 3. The higher the concentration of a solute in a solution, the lower the freezing point (larger depression) and the higher the boiling point (larger elevation) for the solution. 4. The more particles that a solute dissolves or dissociates into, the lower the freezing point (larger depression) and the higher the boiling point (larger elevation) for the solution. Watch Crash Course Chemistry Water and Solutions video 13:33 Page 48 of 57

49 Colligative Properties Examples: (Phase Change) 1. Which of the following aqueous solutions would boil at the highest temperature? a) 100 g of NaCl in 125 g of water b) 100 g of NaCl in 250 g of water c) 100 g of NaCl in 500 g of water d) 100 g of NaCl in 1000 g of water Concentration = mass of solute mass of solvent i. If the amount of solute (NaCl) remains the same, the less solvent (water) there is would make the solution more concentrated. ii. As concentration increases, the boiling point elevates. The answer would then be choice A. 2. Which of the following aqueous solutions would boil at the lowest temperature? a) 100 g of NaCl in 125 g of water b) 100 g of NaCl in 250 g of water c) 100 g of NaCl in 500 g of water d) 100 g of NaCl in 1000 g of water mass of solute i. Concentration = mass of solvent ii. If the amount of solute (NaCl) remains the same, the more solvent (water) there is would make the solution less concentrated. iii. As concentration decreases, the boiling point decreases. The answer would then be choice D. 3. Which of the following aqueous solutions would freeze at the lowest temperature? a) 100 g of NaCl in 125 g of water b) 100 g of NaCl in 250 g of water c) 100 g of NaCl in 500 g of water d) 100 g of NaCl in 1000 g of water i. Concentration = mass of solute mass of solvent Page 49 of 57

50 ii. If the amount of solute (NaCl) remains the same, the less solvent (water) there is would make the solution more concentrated. iii. As concentration increases, the freezing point depresses. The answer would then be choice A. 4. Which of the following aqueous solutions would freeze at the highest temperature? a) 100 g of NaCl in 125 g of water b) 100 g of NaCl in 250 g of water c) 100 g of NaCl in 500 g of water d) 100 g of NaCl in 1000 g of water mass of solute i. Concentration = mass of solvent ii. If the amount of solute (NaCl) remains the same, the more solvent (water) there is would make the solution less concentrated. iii. As concentration increases, the freezing point increases. The answer would then be choice D. 5. Which of the following aqueous solutions will boil at the highest temperature? a) 1 mole of KBr in 500 g of water b) 1 mole of MgF 2 in 500 g of water c) 1 mole of AlCl 3 in 500 g of water d) 1 mole of C 6 H 12 O 6 in 500 g of water i. The ionic compound AlCl 3 has a total of four atoms, and each dissociates into a separate ion, so a single mole of AlCl 3 makes four moles of ions. The more particles that a solute forms, the higher the boiling point elevates, so the answer is C. Page 50 of 57

51 6. Which of the following aqueous solutions will boil at the lowest temperature? a) 1 mole of KBr in 500 g of water b) 1 mole of MgF 2 in 500 g of water c) 1 mole of AlCl 3 in 500 g of water d) 1 mole of C 6 H 12 O 6 in 500 g of water i. The ionic compound C 6 H 12 O 6 has a total of 1 molecule, so a single mole of C 6 H 12 O 6 makes one mole of particles. The fewer particles that a solute forms, the lower the boiling point elevates, so the answer is D. 7. Which of the following aqueous solutions will freeze at the lowest temperature? a) 1 mole of KBr in 500 g of water b) 1 mole of MgF 2 in 500 g of water c) 1 mole of AlCl 3 in 500 g of water d) 1 mole of C 6 H 12 O 6 in 500 g of water i. The ionic compound AlCl 3 has a total of four atoms, and each dissociates into a separate ion, so a single mole of AlCl 3 makes four moles of ions. The more particles that a solute forms, the lower the freezing point depresses, so the answer is C. 8. Which of the following aqueous solutions will freeze at the highest temperature? a) 1 mole of KBr in 500 g of water b) 1 mole of MgF 2 in 500 g of water c) 1 mole of AlCl 3 in 500 g of water d) 1 mole of C 6 H 12 O 6 in 500 g of water i. The ionic compound C 6 H 12 O 6 has a total of 1 molecule, so a single mole of C 6 H 12 O 6 makes one mole of particles. The fewer particles that a solute forms, the less is the effect on the freezing point, so the answer is D. Page 51 of 57

52 Colligative Properties Regents Practice Problems (ungraded): 1. Which aqueous solution will have the lowest freezing point? a) 1.0 M NaCl c) 1.0 M C 2 H 5 OH b) 1.0 M C 6 H 12 O 6 d) 1.0 M CH 3 COOH 2. As a solute is added to a solvent, what happens to the freezing point and the boiling point of the solution? a) The freezing point decreases and the boiling point decreases. b) The freezing point decreases and the boiling point increases. c) The freezing point increases and the boiling point decreases. d) The freezing point increases and the boiling point increases. 3. Compared to pure water, an aqueous solution of calcium chloride has a a) Lower boiling point and a lower freezing point. b) Lower boiling point and a lower freezing point. c) Higher boiling point and a lower freezing point. d) Lower boiling point and a higher freezing point. 4. A student dissolves 1.0 mole of sucrose (C 12 H 22 O 11 ) in grams of water at 1.0 atmosphere. Compared to the boiling point of pure water, the boiling point of the resulting solution is K K K K 5. Which 1.0 M aqueous solution would have the lowest freezing point? Page 52 of 57

53 The chart above is used to determine the effect that adding ethylene glycol (antifreeze) to your cooling system will have given the concentration. Using the chart above, determine: 6. How many quarts of antifreeze are needed to protect a 16-quart cooling system down to -28F? 7. Based on the bottommost (FREEZE/BOIL PROTECTION CHART) chart, will aqueous ethylene glycol provide a colligative property effect against freezing and/or boiling? Explain your answer. Page 53 of 57

54 Notes page: Page 54 of 57

55 Student name: Class Period: Please carefully remove this page from your packet to hand in. Colligative Properties Homework Ionic compounds (containing BOTH a metal and a nonmetal) are broken apart by water molecules into individual ions. Only ionic bonds may be broken by dissolving, and any covalent bonds within a polyatomic ion (Reference Table E) remain intact. For each of the following compounds determine which ions make up the compound and how many total ions are contained within the compound. 1 pt. ea. Ionic compound # of and symbol of cation # of an symbol of anion Total # of ions in compound KHCO 3 Fe(C 2 H 3 O 2 ) 2 Mg(OH) 2 CaBr 2 Ionic compounds (as described above) form ions, so the resulting aqueous solutions are known as electrolytes. Covalent molecules (containing only nonmetals) do not form ions, and therefore do not act as electrolytes. However, for ionic compounds and covalent compounds, the greater the number of separate aqueous molecules (concentration) in a solution will have a greater impact on both freezing and boiling point. 1 pt. ea. Compound Ionic or Molecular? Electrolyte of Nonelectrolyte? How many aqueous particles are formed per? Rank the order of freezing/boiling point impact (1-less impact; 4-most impact) BaBr 2 LiF C 2 H 6 O Fe(NO 3 ) 3 Cont d next page Page 55 of 57

56 Multiple choice questions: Circle the correct answer. 1 pt. ea. 1. Which of the following 1 M solutions will have the highest boiling point? a) NaCl (aq) c) K 3 PO 4(aq) b) C 6 H 12 O 6(aq) d) Cu(NO 3 ) 2(aq) 2. Which of the following 1 M solutions will have the lowest boiling point? a) NaCl (aq) c) K 3 PO 4(aq) b) C 6 H 12 O 6(aq) d) Cu(NO 3 ) 2(aq) 3. Which of the following 1 M solutions will have the highest freezing point? a) NaCl (aq) c) K 3 PO 4(aq) b) C 6 H 12 O 6(aq) d) Cu(NO 3 ) 2(aq) 4. Which of the following 1 M solutions will have the lowest freezing point? a) NaCl (aq) c) K 3 PO 4(aq) b) C 6 H 12 O 6(aq) d) Cu(NO 3 ) 2(aq) 5. Which of the following aqueous solutions will have the highest boiling point? a) 500. g of solute in 1.00 kg of solvent b) 500. g of solute in kg of solvent c) g of solute in 1.00 kg of solvent d) g of solute in kg of solvent 6. Which of the following aqueous solutions will have the lowest boiling point? a) 500. g of solute in 1.00 kg of solvent b) 500. g of solute in kg of solvent c) g of solute in 1.00 kg of solvent d) g of solute in kg of solvent 7. Which of the following aqueous solutions will have the highest freezing point? a) 500. g of solute in 1.00 kg of solvent b) 500. g of solute in kg of solvent c) g of solute in 1.00 kg of solvent d) g of solute in kg of solvent 8. Which of the following aqueous solutions will have the lowest freezing point? a) 500. g of solute in 1.00 kg of solvent b) 500. g of solute in kg of solvent c) g of solute in 1.00 kg of solvent d) g of solute in kg of solvent Page 56 of 57

57 Notes page: Page 57 of 57