Chapter 7 Solutions 1

Transcription

1 1 Chapter 7 Solutions

2 Solutions: Solute and Solvent Solutions are homogeneous mixtures of two or more substances form when there is sufficient attraction between solute and solvent molecules have two components: the solvent, present in a much larger amount, and the solute, present in a smaller amount 2

3 Solutes Solutes spread evenly throughout the solution cannot be separated by filtration can be separated by evaporation are not visible, but can give a color to the solution A solution of copper(ii) sulfate (CuSO 4 ) forms as particles of solute dissolve and become evenly dispersed among the solvent (water) molecules. 3

4 Types of Solutes and Solvents Solutes and solvents may be solids, liquids, or gases. 4

5 Water as a Solvent Water is one of the most common solvents in nature is a polar molecule due to polar O H bonds molecules form hydrogen bonds important in many biological compounds 5

6 Solutions with Ionic and Polar Solutes NaCl crystals undergo hydration as water molecules surround each ion and pull it into solution. solid NaCl(s) H 2 O(l) separate ions Na + (aq) + Cl - (aq) 6

7 Solutions with Nonpolar Solutes Nonpolar solutes require nonpolar solvents to form a solution. 7

8 Like Dissolves Like Like dissolves like : polarity of solute and solvent must be similar to form a solution. 8

9 Learning Check Identify the solute in each of the following solutions. A. 2 g of sugar and 100 ml of water B ml of ethyl alcohol and 30.0 ml of methyl alcohol C ml of water and 1.50 g of NaCl D. Air: 200 ml of O 2 and 800 ml of N 2 9

10 Solution Identify the solute in each of the following solutions. A. 2 g of sugar and 100 ml of water The solute is sugar. B ml of ethyl alcohol and 30.0 ml of methyl alcohol The solute is methyl alcohol. C ml of water and 1.50 g of NaCl The solute is NaCl. D. Air: 200 ml of O 2 and 800 ml of N 2 The solute is O 2. 10

11 Learning Check Which of the following solutes will dissolve in water? Why? A. Na 2 SO 4 B. gasoline (nonpolar) C. I 2 D. HCl 11

12 Solution Which of the following solutes will dissolve in water? Why? A. Na 2 SO 4 will dissolve, ionic B. gasoline (nonpolar) will not dissolve, nonpolar C. I 2 will not dissolve, nonpolar D. HCl will dissolve, polar Most polar and ionic solutes dissolve in water because water is a polar solvent. Nonpolar solutes will not dissolve in water. 12

13 Solutes and Ionic Charge In water, strong electrolytes produce ions and conduct an electric current weak electrolytes produce a few ions nonelectrolytes do not produce ions 13

14 Strong Electrolytes Strong electrolytes dissociate in water, producing positive and negative ions conduct an electric current in water in equations show the formation of ions in aqueous (aq) solutions 100% ions NaCl(s) H 2 O(l) Na + (aq) + Cl - (aq) 14

15 Weak Electrolytes A weak electrolyte dissociates only slightly in water in water forms a solution of a few ions and mostly undissociated molecules dissociation HF(aq) H + (aq) + F - (aq) recombination 15

16 Nonelectrolytes Nonelectrolytes dissolve as molecules in water do not produce ions in water do not conduct an electric current C 12 H 22 O 11 (s) sucrose H 2 O(l) C 12 H 22 O 11 (aq) solution of sucrose 16

17 Classification of Solutes in Aqueous Solutions 17

18 Equivalents An equivalent (Eq) is the amount of an electrolyte or an ion that provides 1 mole of electrical charge (+ or ). 18

19 Electrolytes in Body Fluids In replacement solutions for body fluids, the electrolytes are given in milliequivalents per liter (meq/l). Ringer s Solution Na meq/l Cl 154 meq/l K + Ca 2+ 4 meq/l 4 meq/l The milliequivalents per liter of cations must equal the milliequivalents per liter of anions. 19

20 Electrolyte Concentrations in IV Solutions 20

21 Electrolytes in Body Fluids 21

22 Learning Check 1. In 1 mole of Fe 3+, there are A. 1 Eq B. 2 Eq C. 3 Eq 2. In 2.5 moles of SO 2 4, there are A. 2.5 Eq B. 5.0 Eq C. 1.0 Eq 3. An IV bottle contains NaCl. If the Na + is 34 meq/l, the Cl is A. 34 meq/l B. 0 meq/l C. 68 meq/l 22

23 Solution 1. In 1 mole of Fe 3+, there are C. 3 Eq 2. In 2.5 moles of SO 4 2, there are B. 5.0 Eq 2.5 mole SO Eq = 5.0 Eq 1 mole SO An IV bottle contains NaCl. If the Na + is 34 meq/l, the Cl is A. 34 meq/l 23

24 Solubility Solubility is the maximum amount of solute that dissolves in a specific amount of solvent temperature sensitive for solutes expressed as grams of solute in 100 grams of solvent, usually water g of solute 100 g water 24

25 Unsaturated Solution Unsaturated solutions contain less than the maximum amount of solute can dissolve more solute 25

26 Saturated Solution Saturated solutions contain the maximum amount of solute that can dissolve have undissolved solute at the bottom of the container contain solute that dissolves as well as solute that recrystallizes in an equilibrium process solute + solvent solute dissolves solute recrystallizes saturated solution 26

27 Effect of Temperature on Solubility Solubility depends on temperature of most solids increases as temperature increases of gases decreases as temperature increases In water, most common solids are more soluble as the temperature increases. 27

28 Solubility and Pressure Henry s law states that the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid at higher pressures, more gas molecules dissolve in the liquid 28

29 Insoluble Soluble Solubility Rules 1) All nitrates (NO 3- ) and acetates (C 2 H 3 O 2- ) are soluble. 2) All sodium (Na + ), Potassium (K + ), and ammonium (NH 4+ ) salts are soluble. [Except Ammonium hydroxide (NH 4 OH)] 3) All chlorides (Cl - ), bromides (Br - ), and iodides (I - ) are soluble except silver (Ag + ), mercury (Hg 2+ ), and lead (Pb 2+ ) salts. 4) Most sulfates (SO 4 2- ) are soluble except Group IIA (Ca 2+, Ba 2+ ) salts and lead (Pb 2+ ) salts. 5) Most oxides (O 2- ) and hydroxides (OH - ) are insoluble except Na +, K +, NH 4 + salts and Group IIA (Ca 2+, Ba 2+ ) salts. 6) Most sulfides (S 2- ), carbonates (CO 3 2- ), phosphates (PO 4 3- ), chromates (CrO 4 2- ) are insoluble except Na +, K +, NH 4 + salts.

30 Soluble vs. Insoluble Salts Not all ionic compounds, salts, are soluble in water. 30

31 Learning Check What ions make each of these compounds insoluble in water? 31

32 Solution Which of the following solutes will dissolve in water? Why? A. CdS(s) S 2 compounds are generally insoluble. B. FeS(s) S 2 compounds are generally insoluble. C. PbI 2 (s) I is soluble unless combined with Pb 2+. D. Ni(OH) 2 (s) OH compounds are generally insoluble. 32

33 Solution Concentration The concentration of a solution is expressed as amount of solute x amount of solution Units of concentration include: Mass percent (m/m) Volume percent (v/v) Mass/volume percent (m/v) Molarity (moles solute/liters solution) 33

34 Mass Percent Mass percent (% m/m) is the concentration by mass of solute in a solution mass percent = g of solute 100 g of solute + g of solvent amount in g of solute in 100 g of solution (conversion factor for mass percent) mass percent = g of solute x 100 g of solution 34

35 Calculating Mass Percent The calculation of mass percent (% m/m) requires the grams of solute (g KCl) and grams of solution (g KCl + g water) or g of KCl g of solvent (water) g g g of KCl solution = g 8.00 g KCl (solute) 100 = 16.0% (m/m) g KCl solution 35

36 Guide to Calculating Solution Concentration 36

37 Learning Check A solution is prepared by mixing 15.0 g of Na 2 CO 3 and 235 g of H 2 O. Calculate the mass percent (% m/m) of the solution. A. 15.0% (m/m) Na 2 CO 3 B. 6.38% (m/m) Na 2 CO 3 C. 6.00% (m/m) Na 2 CO 3 37

38 Solution A solution is prepared by mixing 15.0 g of Na 2 CO 3 and 235 g of H 2 O. Calculate the mass percent (% m/m) of the solution. Step 1 Determine the quantities of solute and solution. mass solute = 15.0 g Na 2 CO 3 mass solution = 15.0 g Na 2 CO g H 2 O = 250. g Step 2 Write the concentration expression. Use g solute/g solution ratio: mass % (m/m) = g solute 100 g solution 38

39 Solution A solution is prepared by mixing 15.0 g of Na 2 CO 3 and 235 g of H 2 O. Calculate the mass percent (% m/m) of the solution. Step 3 Substitute solute and solution quantities into the expression. mass % (m/m) = 15.0 g Na 2 CO = 6.00% Na 2 CO g solution The answer is C., 6.00% Na 2 CO 3. 39

40 Volume Percent The volume percent (% v/v) is percent volume (ml) of solute (liquid) to volume (ml) of solution volume % (v/v) = ml of solute 100 ml of solution solute (ml) in 100 ml of solution (conversion factor for volume percent) volume % (v/v) = ml of solute 100 ml of solution NB: You can t add the volumes of solute and solvent together like you can with the masses! 40

41 Mass/Volume Percent The mass/volume percent (% m/v) is percent mass (g) of solute to volume (ml) of solution mass/volume % (m/v) = g of solute 100 ml of solution solute (g) in 100 ml of solution (conversion factor for mass/volume percent) mass/volume % (m/v) = g of solute 100 ml of solution 41

42 Molarity Molarity (moles of solute/liter of solution) is moles of solute per volume (L) of solution. M = moles solute = 1.0 mole NaCl liter of solution 1 L solution = 1.0 M NaCl solution A 1.0 M NaCl solution is prepared by weighing out 58.5 g of NaCl (1.0 mole) and adding water to make 1.00 liter of solution. 42

43 Molarity Calculations What is the molarity of L of NaOH solution if it contains 6.00 g of NaOH? Step 1 Determine the quantities of solute and solution. Solute: 1 mole of NaOH = 40.0 grams of NaOH 1 mole NaOH and 40.0 g NaOH x 40.0 g NaOH 1 mole NaOH 6.00 g NaOH 1 mole NaOH = mole of NaOH 40.0 g NaOH The volume of the solution is L. 43

44 Molarity Calculations What is the molarity of L of NaOH solution if it contains 6.00 g of NaOH? Step 2 Write the concentration expression. M = moles solute L solution Step 3 Substitute solute and solution quantities into the expression. M = moles solute = mole NaOH = M NaOH L solution L solution 44

45 Learning Check What is the molarity of L of a KNO 3 solution containing 34.8 g of KNO 3? A M B M C M 45

46 Solution What is the molarity of L of a KNO 3 solution containing 34.8 g of KNO 3? Step 1 Determine the quantities of solute and solution. Solute: 1 mole of KNO 3 = grams of KNO 3 1 mole KNO 3 and g KNO g KNO 3 1 mole KNO g KNO 3 1 mole KNO 3 = mole of KNO g KNO 3 The volume of the solution is L. 46

47 Solution Step 2 Write the concentration expression. M = moles solute L solution Step 3 Substitute solute and solution quantities into the expression. M = moles solute = mole of KNO 3 = 1.53 M KNO 3 L solution L solution The answer is B, 1.53 M KNO 3. 47

48 Molarity as a Conversion Factor The units of molarity are used as conversion factors in calculations with solutions. Molarity Equality 3.5 M HCl 1 L = 3.5 moles of HCl Written as Conversion Factors 3.5 moles HCl and 1 L 1 L 3.5 moles HCl Molarity = # of moles Liters of sol n # of moles = Liters of sol n X Molarity 48

49 Conversion Factors from Concentrations 49

50 Dilution In a dilution, water is added volume increases concentration decreases mass of solute remains the same moles of solute remains the same 50

51 Solute Concentration in Initial Diluted Solutions In the initial and diluted solution, the moles of solute are the same the concentrations and volumes are related by the following equations: for percent concentration: C 1 V 1 = C 2 V 2 initial diluted for molarity: M 1 V 1 = M 2 V 2 initial diluted 51

52 Guide to Calculating Dilution Quantities 52

53 Dilution: Molarity What is the final concentration when 0.50 L of 6.0 M HCl solution is diluted to a final volume of 1.0 L? Step 1 Prepare a table of the concentrations and volumes of the solutions. C 1 = 6.0 M C 2 =? M V 1 = 0.50 L V 2 = 1.0 L 53

54 Dilution: Molarity What is the final concentration when 0.50 L of 6.0 M HCl solution is diluted to a final volume of 1.0 L? Step 2 Rearrange the dilution expression to solve for the unknown quantity. M 1 V 1 = M 2 V 2 Step 3 Substitute the known quantities into the dilution expression and solve. 6.0 M 0.50 L = 3.0 M HCl 1.0 L 54

55 Learning Check What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 ml of 14.0% (m/v) HCl solution? 55

56 Solution What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 ml of 14.0% (m/v) HCl solution? Step 1 Prepare a table of the concentrations and volumes of the solutions. C 1 = 14.0% (m/v) V 1 = 25.0 ml C 2 = 2.00% (m/v) V 2 =? Step 2 Rearrange the dilution expression to solve for the unknown quantity. C 1 V 1 = C 2 V 2 V 2 = V 1 C 1 C 2 56

57 Solution What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 ml of 14.0% (m/v) HCl solution? Step 3 Substitute the known quantities into the dilution expression and solve. V 2 = (25.0 ml)(14.0%) = 175 ml of solution 2.00% 57

58 Learning Check What is the molarity (M) of a solution prepared by diluting L of M HNO 3 to L? 58

59 Solution What is the molarity (M) of a solution prepared by diluting L of M HNO 3 to L? Step 1 Prepare a table of the concentrations and volumes of the solutions. M 1 = M V 1 = L M 2 =? V 2 = L Step 2 Rearrange the dilution expression to solve for the unknown quantity. M 1 V 1 = M 2 V 2 M 2 = M 1 V 1 V 2 59

60 Solution What is the molarity (M) of a solution prepared by diluting L of M HNO 3 to L? Step 3 Substitute the known quantities into the dilution expression and solve. M 2 = (0.600 M)(0.180 L) = M solution L 60

61 Solutions, Colloids and Suspensions Solutions are transparent do not separate contain small particles, ions or molecules that cannot be filtered and pass through semipermeable membranes Colloids have medium-size particles cannot be filtered can be separated by semipermeable membranes Suspensions have very large particles settle out can be filtered must be stirred to stay suspended Examples include blood platelets, muddy water, and calamine lotion. 61

62 Solutions, Colloids, and Suspensions Properties of different types of mixtures: (a) suspensions settle out; (b) suspensions are separated by a filter; (c) solution particles go through a semipermeable membrane, but colloids and suspensions do not. 62

63 Question: Learning Check A mixture that has solute particles that do not settle out but are too large to pass through a semipermeable membrane is called a. A. solution B. colloid C. Suspension Solution: B. colloid 63

64 Colligative Properties and Electrolytes Colligative properties depend on the number of solute particles in solution include freezing point lowering include boiling point elevation The solute particles disrupt the formation of solid crystals, lowering the freezing point of the solvent. Ethylene glycol is added to a radiator to form an aqueous solution that has a lower freezing point and a higher boiling point than water. Electrolytes break up into ions in solution, increasing the number of particles include freezing point lowering include boiling point elevation 1 mole of CaCl 2 = 3 moles of ions 64

65 Osmosis In osmosis, water (solvent) flows from the lower solute concentration into the higher solute concentration the level of the solution with the higher solute concentration rises the concentrations of the two solutions become equal with time Water flows into the solution with a higher solute concentration until the flow of water becomes equal in both directions. 65

66 Osmotic Pressure Osmotic pressure is produced by the solute particles dissolved in a solution equal to the pressure that would prevent the flow of additional water into the more concentrated solution greater as the number of dissolved particles in the solution increases Osmotic Pressure in Blood Red blood cells have cell walls that are semipermeable membranes maintain an osmotic pressure that cannot change or damage occurs must maintain an equal flow of water between the red blood cell and its surrounding environment 66

67 Dialysis An isotonic solution exerts the same osmotic pressure as body fluids such as red blood cells, RBCs of 5.0% (m/v) glucose or 0.90% (m/v) NaCl are typical isotonic solutions A hypotonic solution has a lower solute concentration than red blood cells means water flows into cells by osmosis The increase in fluid causes the cells to swell and burst, a condition called hemolysis. A hypertonic solution has a higher solute concentration than RBCs water goes out of the cells by osmosis causes crenation: RBCs shrink in size In dialysis, solvent and small solute particles pass through an artificial membrane large particles are retained inside waste particles such as urea from blood are removed using hemodialysis (artificial kidney) (a) In an isotonic solution, a red blood cell retains its normal volume. (b) Hemolysis: In a hypotonic solution, water flows into a red blood cell, causing it to swell and burst. (c) Crenation: In a hypertonic solution, water leaves the red blood cell, causing it to shrink. 67

68 Learning Check Indicate if each of the following solutions is A. isotonic B. hypotonic C. hypertonic 1. 2% NaCl solution 2. 1% glucose solution % NaCl solution 4. 5% glucose solution 68

69 Solution Indicate if each of the following solutions is A. isotonic B. hypotonic C. hypertonic 1. _C 2% NaCl solution 2. _B 1% glucose solution 3. _B 0.5% NaCl solution 4. _A 5% glucose solution 69

70 Learning Check When placed in each of the following, indicate if a red blood cell will A. not change B. undergo hemolysis C. undergo crenation 1. 5% glucose solution 2. 1% glucose solution % NaCl solution 4. 2% NaCl solution 70

71 Solution When placed in each of the following, indicate if a red blood cell will A. not change B. undergo hemolysis C. undergo crenation 1._A 5% glucose solution 2._B 1% glucose solution 3._B 0.5% NaCl solution 4._C 2% NaCl solution 71

72 Concept Map 72