Chapter 7 Solutions 7.1 Solutions 1 Copyright © 2009 by Pearson Education, Inc.

Solute and Solvent Solutions are homogeneous mixtures of two or more substances. consist of a solvent and one or more solutes. 2

Nature of Solutes in Solutions 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. 3 Copyright © 2009 by Pearson Education, Inc.

Examples of Solutions The solute and solvent in a solution can be a solid, liquid, and/or a gas. 4 Copyright © 2009 by Pearson Education, Inc.

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 5

Water is the most common solvent. is a polar molecule. forms hydrogen bonds between the hydrogen atom in one molecule and the oxygen atom in a different water molecule. 6 Copyright © 2009 by Pearson Education, Inc.

Formation of a Solution Na + and Cl - ions on the surface of a NaCl crystal are attracted to polar water molecules. are hydrated in solution with many H 2 O molecules surrounding each ion. 7 Copyright © 2009 by Pearson Education, Inc.

Equations for Solution Formation When NaCl(s) dissolves in water, the reaction can be written as H 2 O NaCl(s) Na + (aq) + Cl - (aq) solid separation of ions 8

Learning Check Solid LiCl is added to water. It dissolves because A. the Li + ions are attracted to the 1) oxygen atom (  - ) of water. 2) hydrogen atom (  + ) of water. B. the Cl - ions are attracted to the 1) oxygen atom (  - ) of water. 2) hydrogen atom (  + ) of water. 9

Like Dissolves Like Two substances form a solution when there is an attraction between the particles of the solute and solvent. when a polar solvent such as water dissolves polar solutes such as sugar, and ionic solutes such as NaCl. when a nonpolar solvent such as hexane (C 6 H 14 ) dissolves nonpolar solutes such as oil or grease. 10

Water and a Polar Solute 11 Copyright © 2009 by Pearson Education, Inc.

Like Dissolves Like Solvents Solutes Water (polar) Ni(NO 3 ) 2 CH 2 Cl 2 (nonpolar) (polar) I 2 (nonpolar) 12 Copyright © 2009 by Pearson Education, Inc.

Learning Check Which of the following solutes will dissolve in water? Why? 1) Na 2 SO 4 2) gasoline (nonpolar) 3) I 2 4) HCl 13

7.2 Electrolytes and Nonelectrolytes 14 Copyright © 2009 by Pearson Education, Inc.

15 In water, strong electrolytes produce ions and conduct an electric current. weak electrolytes produce a few ions. nonelectrolytes do not produce ions. Solutes and Ionic Charge Copyright © 2009 by Pearson Education, Inc.

16 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. H 2 O 100% ions NaCl(s) Na + (aq) + Cl − (aq) H 2 O CaBr 2 (s) Ca 2+ (aq) + 2Br − (aq) Strong Electrolytes

17 Complete each for strong electrolytes in water. H 2 O A. CaCl 2 (s) 1) CaCl 2 (s) 2) Ca 2+ (aq) + Cl 2 − (aq) 3) Ca 2+ (aq) + 2Cl − (aq) H 2 O B. K 3 PO 4 (s) 1) 3K + (aq) + PO 4 3− (aq) 2) K 3 PO 4 (s) 3) K 3 + (aq) + P 3− (aq) + O 4 − (aq) Learning Check

18 A weak electrolyte dissociates only slightly in water. in water forms a solution of a few ions and mostly undissociated molecules. HF(g) + H 2 O(l) H 3 O + (aq) + F - (aq) NH 3 (g) + H 2 O(l) NH 4 + (aq) + OH - (aq) Weak Electrolytes

19 Nonelectrolytes dissolve as molecules in water. do not produce ions in water. do not conduct an electric current. Copyright © 2009 by Pearson Education, Inc.

20 Equivalents An equivalent (Eq) is the amount of an electrolyte or an ion that provides 1 mole of electrical charge (+ or -). 1 mole of Na + = 1 equivalent 1 mole of Cl − = 1 equivalent 1 mole of Ca 2+ = 2 equivalents 1 mole of Fe 3+ = 3 equivalents

21 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 − 155 mEq/L K + 4 mEq/L Ca 2+ 4 mEq/L The milliequivalents per liter of cations must equal the milliequivalents per liter of anions.

Electrolytes in Body Fluids 22 Copyright © 2009 by Pearson Education, Inc.

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

24 Chapter 7 Solutions 7.3 Solubility Copyright © 2009 by Pearson Education, Inc.

25 Solubility is the maximum amount of solute that dissolves in a specific amount of solvent. expressed as grams of solute in 100 grams of solvent, usually water. g of solute 100 g water Solubility

26 Unsaturated Solutions Unsaturated solutions contain less than the maximum amount of solute. can dissolve more solute. Copyright © 2009 by Pearson Education, Inc.

27 Saturated Solutions Saturated solutions contain the maximum amount of solute that can dissolve. have undissolved solute at the bottom of the container. Copyright © 2009 by Pearson Education, Inc.

28 At 40  C, the solubility of KBr is 80 g/100 g of H 2 O. Identify the following solutions as either 1) saturated or 2) unsaturated. Explain. A. 60 g KBr added to 100 g of water at 40  C. B. 200 g KBr added to 200 g of water at 40  C. C. 25 g KBr added to 50 g of water at 40  C. Learning Check

29 Effect of Temperature on Solubility Solubility depends on temperature. of most solids increases as temperature increases. of gases decreases as temperature increases. Copyright © 2009 by Pearson Education, Inc.

30 A. Why could a bottle of carbonated drink possibly burst (explode) when it is left out in the hot sun? B. Why do fish die in water that is too warm? Learning Check

31 Solubility and Pressure Henry’s law states 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. Copyright © 2009 by Pearson Education, Inc.

32 Chapter 7Solutions 7.4 Percent Concentration Copyright © 2009 by Pearson Education, Inc.

33 The concentration of a solution is the amount of solute dissolved in a specific amount of solution. amount of solute amount of solution Percent Concentration

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

35 Mass of Solution 8.00 g KCl g of KCl solution Add water to give g of solution Copyright © 2009 by Pearson Education, Inc.

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

37 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. 1) 15.0% (m/m) Na 2 CO 3 2) 6.38% (m/m) Na 2 CO 3 3) 6.00% (m/m) Na 2 CO 3 Learning Check

38 The volume percent (% v/v) is percent volume (mL) of solute (liquid) to volume (mL) of solution. volume % (v/v) = mL of solute x 100 mL of solution solute (mL) in 100 mL of solution. volume % (v/v) = mL of solute 100 mL of solution Volume Percent

39 The mass/volume percent (% m/v) is percent mass (g) of solute to volume (mL) of solution. mass/volume % (m/v) = g of solute x 100 mL of solution solute (g) in 100 mL of solution. mass/volume % (m/v) = g of solute 100 mL of solution Mass/Volume Percent

40 Percent Conversion Factors Two conversion factors can be written for each type of % value. Copyright © 2009 by Pearson Education, Inc.

41 Write two conversion factors for each solutions. A. 8.50% (m/m) NaOH B. 5.75% (v/v) ethanol C. 4.8% (m/v) HCl Learning Check

42 Using Percent Concentration (m/m) as Conversion Factors How many grams of NaCl are needed to prepare 225 g of a 10.0% (m/m) NaCl solution? STEP 1: Given: 225 g solution; 10.0% (m/m) NaCl Need: g of NaCl STEP 2: g solution g NaCl STEP 3: Write the 10.0% (m/m) as conversion factors g NaCl and 100 g solution 100 g solution 10.0 g NaCl STEP 4: Set up using the factor that cancels g solution.

43 How many milliliters of a 5.75% (v/v) ethanol solution can be prepared from 2.25 mL of ethanol? 1) 2.56 mL 2) 12.9 mL 3) 39.1 mL Learning Check

44 How many mL of a 4.20% (m/v) will contain 3.15 g of KCl? STEP 1: Given: 3.15 g of KCl(solute); 4.20% (m/v) KCl Need: mL of KCl solution STEP 2: Plan: g of KCl mL of KCl solution STEP 3: Write conversion factors g KCl and 100 mL solution 100 mL solution 4.20 g KCl STEP 4: Set up the problem Using Percent Concentration (m/v) as Conversion Factors

45 Learning Check How many grams of NaOH are needed to prepare 125 mL of a 8.80% (m/v) NaOH solution?

46 Chapter 7 Solutions 7.5 Molarity and Dilution Copyright © 2009 by Pearson Education, Inc.

47 Molarity (M) is a concentration term for solutions. gives the moles of solute in 1 L of solution. moles of solute liter of solution

48 Preparing a 1.0 Molar Solution A 1.00 M NaCl solution is prepared by weighing out 58.5 g of NaCl (1.00 mole) and adding water to make 1.00 liter of solution. Copyright © 2009 by Pearson Education, Inc.

49 What is the molarity of L of NaOH solution if it contains 6.00 g of NaOH? STEP 1: Given 6.00 g of NaOH in L of solution Need molarity (mole/L) STEP 2: Plan g NaOH mole NaOH molarity Calculation of Molarity

50 What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO 3 ? 1) M 2) 1.44 M 3) 1.71 M Learning Check

51 What is the molarity of 225 mL of a KNO 3 solution containing 34.8 g of KNO 3 ? 1)0.344 M 2)1.53 M 3)15.5 M Learning Check

52 Molarity Conversion Factors The units of molarity are used as conversion factors in calculations with solutions. MolarityEquality 3.5 M HCl1 L = 3.5 moles of HCl Written as Conversion Factors 3.5 moles HCl and 1 L 1 L3.5 moles HCl

53 Calculations Using Molarity How many grams of KCl are needed to prepare 125 mL of a M KCl solution? STEP 1: Given 125 mL (0.125 L) of M KCl Need g of KCl STEP 2: Plan L KCl moles KCl g KCl

54 How many grams of AlCl 3 are needed to prepare 125 mL of a M solution? 1) 20.0 g of AlCl 3 2) 16.7 g of AlCl 3 3) 2.50 g of AlCl 3 Learning Check

55 How many milliliters of 2.00 M HNO 3 contain 24.0 g of HNO 3 ? 1) 12.0 mL 2) 83.3 mL 3) 190. mL Learning Check

56 D ilution In a dilution water is added. volume increases. concentration decreases. Copyright © 2009 by Pearson Education, Inc.

57 Comparing Initial and 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

58 Dilution Calculations with Percent 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? Prepare a table: C 1 = 14.0% (m/v)V 1 = 25.0 mL C 2 = 2.00% (m/v)V 2 = ? Solve dilution equation for unknown and enter values: C 1 V 1 = C 2 V 2 V 2 =

59 Learning Check What is the percent (% m/v) of a solution prepared by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

60 Dilution Calculations with Molarity What is the molarity (M) of a solution prepared by diluting L of M HNO 3 to L? Prepare a table: M 1 = MV 1 = L M 2 = ?V 2 = L Solve dilution equation for unknown and enter values: M 1 V 1 = M 2 V 2 M 2 =

61 Learning Check What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a M solution? 1) 27.0 mL 2) 60.0 mL 3) 90.0 mL

62 Chapter 7 Solutions 7.6 Solutions in Chemical Reactions Copyright © 2009 by Pearson Education, Inc.

63 Molarity in Chemical Reactions In a chemical reaction, the volume and molarity of a solution are used to determine the moles of a reactant or product. molarity ( mole ) x volume (L) = moles 1 L if molarity (mole/L) and moles are given, the volume (L) can be determined. moles x 1 L = volume (L) moles

64 Using Molarity of Reactants How many mL of 3.00 M HCl are needed to completely react with 4.85 g of CaCO 3 ? 2 HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) STEP 1: Given 3.00 M HCl; 4.85 g of CaCO 3 Need volume in mL STEP 2: Plan g CaCO 3 mole CaCO 3 mole HCl mL HCl

65 Learning Check How many mL of a M Na 2 S solution are needed to completely react 18.5 mL of M NiCl 2 solution? NiCl 2 (aq) + Na 2 S(aq) NiS(s) + 2NaCl(aq) 1) 4.16 mL 2) 6.24 mL 3) 27.0 mL

66 Learning Check If 22.8 mL of M MgCl 2 is needed to completely react 15.0 mL of AgNO 3 solution, what is the molarity of the AgNO 3 solution? MgCl 2 (aq) + 2AgNO 3 (aq) 2AgCl(s) + Mg(NO 3 ) 2 (aq) 1) M 2) M 3) M

67 Learning Check How many liters of H 2 gas at STP are produced when Zn reacts with 125 mL of 6.00 M HCl? Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) 1) 4.20 L of H 2 2) 8.40 L of H 2 3) 16.8 L of H 2

68 Chapter 7 Solutions 7.7 Properties of Solutions Copyright © 2009 by Pearson Education, Inc.

69 Solutions contain small particles (ions or molecules). are transparent. do not separate. cannot be filtered.

70 Colloids have medium-size particles. cannot be filtered. can be separated by semipermeable membranes.

71 Examples of Colloids Examples of colloids include fog whipped cream milk cheese blood plasma pearls Copyright © 2009 by Pearson Education, Inc.

72 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.

73 Solutions, Colloids, and Suspensions Copyright © 2009 by Pearson Education, Inc.

74 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 1) solution. 2) colloid. 3) suspension.

75 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. Copyright © 2009 by Pearson Education, Inc.

76 4% starch Osmosis Suppose a semipermeable membrane separates a 4% starch solution from a 10% starch solution. Starch is a colloid and cannot pass through the membrane, but water can. What happens? semipermeable membrane 10% starch H2OH2O

77 Water Flow Equalizes The 10% starch solution is diluted by the flow of water out of the 4% and its volume increases. The 4% solution loses water and its volume decreases. Eventually, the water flow between the two becomes equal. 7% starch H 2 O

78 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.

79 Learning Check A semipermeable membrane separates a 10% sucrose solution A from a 5% sucrose solution B. If sucrose is a colloid, fill in the blanks in the statements below. 1. Solution ____ has the greater osmotic pressure. 2. Water initially flows from ___ into ___. 3. The level of solution ____will be lower.

80 Osmotic Pressure of the 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.

81 Isotonic Solutions An isotonic solution exerts the same osmotic pressure as red blood cells. is known as a “physiological solution.” of 5.0% glucose or 0.90% NaCl is used medically because each has a solute concentration equal to the osmotic pressure equal to red blood cells. H2O H2O Copyright © 2009 by Pearson Education, Inc.

82 Hypotonic Solutions A hypotonic solution has a lower osmotic pressure than red blood cells. has a lower concentration than physiological solutions. causes water to flow into red blood cells. causes hemolysis: RBCs swell and may burst. H2OH2O Copyright © 2009 by Pearson Education, Inc.

83 Hypertonic Solutions A hypertonic solution has a higher osmotic pressure than RBCs. has a higher concentration than physiological solutions. causes water to flow out of RBCs. cause crenation: RBCs shrink in size. H2OH2O Copyright © 2009 by Pearson Education, Inc.

84 Dialysis 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).

85 Learning Check Indicate if each of the following solutions is 1) isotonic, 2) hypotonic, or 3) hypertonic. A.____ 2% NaCl solution B.____ 1% glucose solution C.____ 0.5% NaCl solution D.____ 5% glucose solution

86 Learning Check When placed in each of the following, indicate if a red blood cell will 1) not change, 2) hemolyze, or 3) crenate. A.____ 5% glucose solution B.____ 1% glucose solution C.____ 0.5% NaCl solution D.____ 2% NaCl solution

87 Learning Check Each of the following mixtures is placed in a dialyzing bag and immersed in pure water. Which substance, if any, will be found in the water outside the bag? A. 10% KCl solution B. 5% starch solution C. 5% NaCl and 5% starch solutions