1 Chapter 12 Solutions 12.1 Solutions. 2 Solute and Solvent Solutions Are homogeneous mixtures of two or more substances. Consist of a solvent and one.

1 Chapter 12 Solutions 12.1 Solutions

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

3 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. Nature of Solutes in Solutions

4 Examples of Solutions The solute and solvent can be a solid, liquid, and/or a gas. Table 12.3

5 Identify the solute in each of the following solutions: A. 2 g sugar (1) and 100 mL water (2) B. 60.0 mL of ethyl alcohol(1) and 30.0 mL of methyl alcohol (2) C. 55.0 mL water (1) and 1.50 g NaCl (2) D. Air: 200 mL O 2 (1) and 800 mL N 2 (2) Learning Check

6 Identify the solute in each of the following solutions: A. 2 g sugar (1) B. 30.0 mL of methyl alcohol (2) C. 1.5 g NaCl (2) D. 200 mL O 2 (1) Solution

7 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.

8 Formation of a Solution Na + and Cl - ions On the surface of a NaCl crystal are attracted to polar water molecules. In solution are hydrated as several H 2 O molecules surround each.

9 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 Equations for Solution Formation

10 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. Learning Check

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

12 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. Like Dissolves Like

13 Water and a Polar Solute

14 Like Dissolves Like Solvents Solutes Water (polar) Ni(NO 3 ) 2 CH 2 Cl 2 (nonpolar) (polar) I 2 (nonpolar)

15 Will the following solutes dissolve in water? Why? 1) Na 2 SO 4 2) gasoline (nonpolar) 3) I 2 4) HCl Learning Check

16 Will the following solutes dissolve in water? Why? 1) Na 2 SO 4 Yes, ionic 2) gasoline No, nonpolar 3) I 2 No, nonpolar 4) HClYes, polar Most polar and ionic solutes dissolve in water because water is a polar solvent. Solution

17 12.2 Electrolytes and Nonelectrolytes Chapter 12 Solutions

18 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

19 Strong electrolytes Dissociate in water producing positive and negative ions. Produce 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

20 Complete each of the following equations: 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

21 Complete each of the following equations: H 2 O A. CaCl 2 (s) 3) Ca 2+ (aq) + 2Cl − (aq) H 2 O B. K 3 PO 4 (s) 1) 3K + (aq) + PO 4 3− (aq) Solution

22 A weak electrolyte Dissociates only slightly in water. In water forms a solution of only 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

23 Nonelectrolytes Dissolve as molecules in water. Do not produce ions in water. Do not conduct an electric current.

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

26 Effect of Temperature on Solubility Solubility Depends on temperature. Of most solids increases as temperature increases. Of gases decreases as temperature increases. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

27 Unsaturated Solutions Unsaturated solutions Contain less than the maximum amount of solute. Can dissolve more solute. Dissolved solute Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

28 Saturated Solutions Saturated solutions Contain the maximum amount of solute that can dissolve. Have undissolved solute at the bottom of the container. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

29 At 40  C, the solubility of KBr is 80 g/100 g 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

30 A. 2 Amount of 60 g KBr in 100 g water is less than the solubility of 80 g KBr in 100 g water. B. 1 In 100 g of water, 100 g KBr exceeds the solubility of 80 g KBr water at 40  C. C. 2 This is the same as 50 g KBr in 100 g water, which is less than the solubility of 80 g KBr in 100 g water at 40  C. Solution

31 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

32 A. The pressure in a bottle increases as the gas leaves solution as it becomes less soluble at high temperatures. As pressure increases, the bottle could burst. B. Because O 2 gas is less soluble in warm water, fish cannot obtain the amount of O 2 required for their survival. Solution

33 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 © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

34 Soluble and Insoluble Salts Ionic compounds that Dissolve in water are soluble salts. Do not dissolve in water are insoluble salts. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

35 Solubility Rules Soluble salts Typically contain at least one ion from Groups 1A(1) or NO 3 −, or C 2 H 3 O 2 − (acetate). Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings Table 12.3

37 Use the solubility rules to determine if each salt is (1) soluble or (2) insoluble. A. ______ Na 2 SO 4 B. ______ MgCO 3 C. ______ PbCl 2 D. ______ MgCl 2 Learning Check

38 A. Na 2 SO 4 (1) Soluble; contains Na + B. MgCO 3 (2) Insoluble; contains carbonates C. PbCl 2 (2) Insoluble; insoluble chloride D. MgCl 2 (1) Soluble; only chlorides of Pb 2+, Ag +, and Hg 2 2+ are insoluble Solution

40 Equations for Forming Solids A molecular equation shows the formulas of the compounds. Pb(NO 3 )(aq) + 2NaCl(aq) PbCl 2 (s) + 2NaNO 3 (aq) An ionic equation shows the ions of the compounds. Pb 2+ (aq) + 2NO 3 − (aq) + 2Na + (aq) + 2Cl − (aq) PbCl 2 (s) + 2Na + (aq) + 2NO 3 − (aq) A net ionic equation shows only the ions that form a solid. Pb 2+ (aq) + 2Cl − (aq) PbCl 2 (s)

41 Add GPS Guide to Writing Net Ionic Equations for the Formation of an Insoluble Salt Basic 2e p 388

42 Equations for the Insoluble Salt STEP 1 Observe the ions in the reactants. Pb 2+ (aq) + 2NO 3 − (aq) 2Na + (aq) + 2Cl − (aq) STEP 2 Determine if any new ion combinations are insoluble salts. Yes. PbCl 2 (s) STEP 3 Ionic equation with insoluble salt product. Pb 2+ (aq) + 2NO 3 − (aq) + 2Na + (aq) + 2Cl − (aq) PbCl 2 (s) + 2Na + (aq) + 2NO 3 − (aq) STEP 4 Net ionic equation. Pb 2+ (aq) + 2Cl − (aq) PbCl 2 (s)

43 Learning Check Write the formula of any insoluble salt, if any. Write the net ionic equation for any of the following which form insoluble salts: A. BaCl 2 (aq) + Na 2 SO 4 (aq) B. AgNO 3 (aq) + KCl(aq) C. KNO 3 (aq) + NaCl(aq)

44 Solution A. BaSO 4 (s) Ba 2+ (aq) + SO 4 2− (aq) BaSO 4 (s) B. AgCl(s) Ag + (aq) + Cl − (aq) AgCl(s) C. None; all combinations of ions are soluble salts.

46 The concentration of a solution Is the amount of solute dissolved in a specific amount of solution. amount of solute amount of solution As percent concentration describes the amount of solute that is dissolved in 100 parts of solution. amount of solute 100 parts solution Percent Concentration

47 The mass percent (%m/m) Concentration is the percent by mass of solute in a solution. mass percent (%m/m) = g of solute x 100 g of solute + g of solvent Is the g of solute in 100 g of solution. mass percent = g of solute x 100 100 g of solution Mass Percent

49 Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution). g of KCl = 8.00 g g of solvent (water) = 42.00 g g of KCl solution = 50.00 g 8.00 g KCl (solute) x 100 = 16.0% (m/m) 50.00 g KCl solution Calculating Mass Percent

50 A solution is prepared by mixing 15.0 g 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

51 3) 6.00% (m/m) Na 2 CO 3 mass solute = 15.0 g Na 2 CO 3 mass solution= 15.0 g + 235 g = 250. g mass %(m/m) = 15.0 g Na 2 CO 3 x 100 250. g solution = 6.00% Na 2 CO 3 solution Solution

52 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

54 Write two conversion factors for each solutions: A. 8.50%(m/m) NaOH B. 5.75%(v/v) ethanol Learning Check

55 A. 8.50%(m/m) NaOH 8.50 g NaOH and 100 g solution 100 g solution 8.50 g NaOH B. 5.75%(v/v) ethyl alcohol 5.75 mL alcohol and 100 mL solution 100 mL solution 5.75 mL alcohol Solution

56 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. 10.0 g NaCl and 100 g solution 100 g solution 10.0 g NaCl STEP 4 Set up problem to calculate g NaCl. 225 g solution x 10.0 g NaCl = 22.5 g NaCl 100 g solution Using Percent Factors

57 How many grams of NaOH are needed to prepare 75.0 g of 14.0%(m/m) NaOH solution? 1)10.5 g NaOH 2) 75.0 g NaOH 3)536 g NaOH Learning Check

58 1)10.5 g NaOH 75.0 g x 14.0 g NaOH = 10.5 g NaOH 100 g 14.0 % (m/m) factor Solution

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

60 3) 39.1 mL 2.25 mL ethanol x 100 mL solution 5.75 mL ethanol 5.75 %(v/v) inverted = 39.1 mL solution Solution

62 Molarity (M) Molarity (M) is A concentration term for solutions. The moles of solute in 1 L solution. moles of solute liter of solution

63 Preparing a 1.0 Molar Solution A 1.00 M NaCl solution is prepared By weighing out 58.5 g NaCl (1.00 mol) and Adding water to make 1.00 liter of solution. Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

64 Add GPS Guide to Calculating Molarity Basic 2e p 394

65 What is the molarity of 0.500 L NaOH solution if it contains 6.00 g NaOH? STEP 1 Given 6.00 g NaOH in 0.500 L solution Need molarity (mol/L) STEP 2 Plan g NaOH mol NaOH molarity STEP 3 Conversion factors 1 mol NaOH = 40.00 g 1 mol NaOH and 40.00 g NaOH 40.00 g NaOH 1 mol NaOH Calculation of Molarity

66 STEP 4 Calculate molarity. 6.00 g NaOH x 1 mol NaOH = 0.150 mol 40.00 g NaOH 0.150 mol = 0.300 mol = 0.300 M NaOH 0.500 L 1 L Calculation of Molarity (cont.)

67 What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO 3 ? 1) 0.557 M 2) 1.44 M 3) 1.71 M Learning Check

68 3) 1.71 M 46.8 g NaHCO 3 x 1 mol NaHCO 3 = 0.557 mol NaHCO 3 84.01 g NaHCO 3 0.557 mol NaHCO 3 = 1.71 M NaHCO 3 0.325 L Solution

69 What is the molarity of a solution if 225 mL contains 34.8 g KNO 3 ? 1)0.0775 M 2)1.53 M 3)15.5 M Learning Check

70 2) 1.53 M 34.8 g KNO 3 x 1 mol KNO 3 = 0.344 mol KNO 3 101.11g KNO 3 M = mol = 0.344 mol KNO 3 = 1.53 M L 0.225 L In one setup 34.8 g KNO 3 x 1 mol KNO 3 x 1 = 1.53 M 101.11g KNO 3 0.225 L Solution

71 Molarity Conversion Factors The units of molarity are used as conversion factors in calculations with solutions. Table 2.6 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

72 Molarity in Calculations How many grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution? STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl Need Grams of KCl STEP 2 Plan L KCl mol KCl g KCl

73 Molarity in Calculations (cont.) STEP 3 Conversion factors 1 mol KCl = 74.55 g 1 mol KCl and 74.55 g KCl 74.55 g KCl 1 mol KCl 1 L KCl = 0.720 mol KCl 1 L and 0.720 mol KCl 0.720 mol KCl 1 L STEP 4 Calculate grams. 0.125 L x 0.720 mol KCl x 74.55 g KCl = 6.71 g KCl 1 L 1 mol KCl

74 How many grams of AlCl 3 are needed to prepare 125 mL of a 0.150 M solution? 1) 20.0 g AlCl 3 2) 16.7g AlCl 3 3) 2.50 g AlCl 3 Learning Check

75 Solution 3) 2.50 g AlCl 3 0.125 L x 0.150 mol x 133.3 g = 2.50 g AlCl 3 1 L 1 mol

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

77 3) 190. mL 24.0 g HNO 3 x 1 mol HNO 3 x 1000 mL 63.02 g HNO 3 2.00 mol HNO 3 Molarity factor inverted = 190. mL HNO 3 Solution

79 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 equation M 1 V 1 = M 2 V 2 initial diluted

80 Add GPS Guide to Calculating Dilution Quantities Basic 2e p 398

81 Dilution Calculations What is the molarity if 0.180 L of 0.600 M KOH is diluted to a final volume of 0.540 L? STEP 1 Prepare a table: M 1 = 0.600 MV 1 = 0.180 L M 2 = ?V 2 = 0.540 L STEP 2 Solve dilution equation for unknown. M 1 V 1 = M 2 V 2 M 1 V 1 / V 2 = M 2 STEP 3 Set up and enter values: M 2 = M 1 V 1 = (0.600 M)(0.180 L) = 0.200 M V 2 0.540 L

82 Learning Check What is the final volume if 15.0 mL of a 1.80 M KOH is diluted to give a 0.300 M solution? 1) 27.0 mL 2) 60.0 mL 3) 90.0 mL

83 Solution STEP 1 Prepare a table: M 1 = 1.80 MV 1 = 15.0 mL M 2 = 0.300MV 2 = ? STEP 2 Solve dilution equation for unknown. M 1 V 1 = M 2 V 2 V 2 = M 1 V 1 / M 2 STEP 3 Set up and enter values: V 2 = M 1 V 1 = (1.80 M)(15.0 mL) = 90.0 mL M 2 0.300 M

85 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. volume (L) x molarity ( mol ) = moles 1 L If molarity (mol/L) and moles are given, the volume (L) can be determined mol x 1 L = volume (L) mol

86 Add GPS Guide to Calculations Involving Solutions in Chemical Reactions Basic 2e p 401

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

88 Using Molarity of Reactants (cont.) 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) STEP 3 Equalitites 1 mol CaCO 3 = 100.09 g; 1 mol CaCO 3 = 2 mol HCl 1000 mL HCl = 3.00 mol HCl STEP 4 Set Up 4.85 g CaCO 3 x 1 mol CaCO 3 x 2 mol HCl x 1000 mL HCl 100.09 g CaCO 3 1 mol CaCO 3 3.00 mol HCl = 32.3 mL HCl required

89 Learning Check How many mL of a 0.150 M Na 2 S solution are needed to react 18.5 mL of 0.225 M NiCl 2 solution? NiCl 2 (aq) + Na 2 S(aq) NiS(s) + 2NaClaq) 1) 4.16 mL 2) 6.24 mL 3) 27.8 mL

90 Solution 3) 27.8 mL 0.0185 L x 0.225 mol NiCl 2 x 1 mol Na 2 S x 1000 mL 1 L 1 mol NiCl 2 0.150 mol = 27.8 mL Na 2 S solution

91 Learning Check What is the molarity if 15.0 mL of AgNO 3 solution reacts with 22.8 mL of 0.100 M MgCl 2 ? MgCl 2 (aq) + 2AgNO 3 (aq) 2AgCl(s) + Mg(NO 3 ) 2 (aq) 1) 0.0760 M 2) 0.152 M 3) 0.304 M

92 Solution 3) 0.304 M AgNO 3 0.0228 L x 0.100 mol MgCl 2 x 2 mol AgNO 3 x 1 = 1 L 1 mol MgCl 2 0.0150 L = 0.304 mol/L = 0.304 M AgNO 3

93 Learning Check How many liters of H 2 gas at STP are produced when 12.5 g Zn react with 20.0 mL of 1.50 M HCl? Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) 1) 4.28 L H 2 2) 0.336 L H 2 3) 0.168 L H 2

94 Solution 2) 0.336 L H 2 Limiting Reactant 12.5 g Zn x 1.00 mol Zn x 1 mol H 2 x 22.4 L = 4.28 L H 2 65.41 g Zn 1 mol Zn 1 mol 0.0200 L x 1.50 mol HCl x 1 mol H 2 x 22.4 L 1 L 2 mol HCl1 mol = 0.336 L H 2 gas

1 Chapter 12 Solutions 12.1 Solutions. 2 Solute and Solvent Solutions Are homogeneous mixtures of two or more substances. Consist of a solvent and one.

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1 Chapter 12 Solutions 12.1 Solutions. 2 Solute and Solvent Solutions Are homogeneous mixtures of two or more substances. Consist of a solvent and one.

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Presentation on theme: "1 Chapter 12 Solutions 12.1 Solutions. 2 Solute and Solvent Solutions Are homogeneous mixtures of two or more substances. Consist of a solvent and one."— Presentation transcript:

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