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SOLUTION CHEMISTRY.

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Presentation on theme: "SOLUTION CHEMISTRY."— Presentation transcript:

1 SOLUTION CHEMISTRY

2

3 I. Mixtures: mixture = a blend of two or more kinds of matter, each of which retains its own identity and properties can be physically separated (filtration, evaporation, decanting, magnetism, etc)

4 I. MIXTURES: a.) homogeneous mixture = a mixture that is uniform in composition throughout Ex: Food coloring and water b.) heterogeneous mixture = a mixture that is NOT uniform in composition throughout Ex: Oil and water

5 A. Types of Mixtures: 1) solution = a homogeneous mixture

6 Heterogeneous mixture Ex: Sand and water
A. Types of Mixtures: 2) suspension = a mixture in which the particles are so large that they settle out unless the mixture is constantly stirred or agitated Heterogeneous mixture Ex: Sand and water

7 Heterogeneous Mixture
A. Types of Mixtures: 3) colloid = a mixture consisting of particles that are intermediate in size between those in solutions and those in suspensions Heterogeneous Mixture Ex: Milk, mayonnaise, smog, butter, whipped cream

8 3) Colloids What property of a colloid helps to prevent colloid particles from settling out of a mixture? Brownian motion = the random continuous motions of colloidal particles

9 3) Colloids b) Tyndall effect = visible pattern caused by the reflection of light from suspended particles in a colloid (or from suspended particles in a suspension if the particles have not settled out) Ex: visibility of a headlight beam on a foggy night

10 B. Comparison of solution, colloids, and suspensions
Solutions Colloids Suspensions Homogeneous Heterogeneous Particle size: nm; can be atoms, ions, molecules Particle size: nm, dispersed; can be aggregates or large molecules Particle size: over 1000 nm, suspended; can be large particles or aggregates

11 B. Properties of solution, colloids, and suspensions
Solutions Colloids Suspensions Do not separate on standing Particles settle out Cannot be separated by filtration Can be separated by filtration Do not scatter light Scatter light (Tyndall effect) May scatter light, but are not transparent

12 C. Determining if a mixture is a true solution, a colloid, or a suspension:
1.) If particles settle or can be filtered out = suspension 2.) If particles do not settle or filter out shine a beam of light (Tyndall effect) through the mixture If the Tyndall effect is observed = colloid If the Tyndall effect is not observed = solution

13 II. Nature of Solutions

14 II. THE NATURE OF SOLUTIONS:
1) Solvent = the substance that does the dissolving in a solution Typically present in the greatest amount Typically a liquid Water is the most common or “universal” solvent

15 B/c water molecules are polar
The hydrogen side of each water (H2O) molecule carries a slight positive electric charge, while the oxygen side carries a slight negative electric charge. water can dissociate ionic compounds into their positive and negative ions. The positive part of an ionic compound is attracted to the oxygen side of water while the negative portion of the compound is attracted to the hydrogen side of water.

16 Water won't dissolve or won't dissolve well
Water won't dissolve or won't dissolve well. If the attraction is high between the opposite-charged ions in a compound, then the solubility will be low. Ex: hydroxides exhibit low solubility in water. Ex: nonpolar molecules don't dissolve very well in water (fats and waxes)

17 II. THE NATURE OF SOLUTIONS:
2) Solute = substance being dissolved in a solution Typically present in the least amount Typically a solid

18 A. 9 Possible Solutions Combinations:
Solvent Solute Common Example Gas Diver’s tank Liquid Humidity Solid Moth ball Carbonation Vinegar Seawater Gas stove lighter Dental fillings Sterling Silver (Ag + Cu) NOT all solutions are liquids/solids! Solutions are formed in ALL 3 states!

19 B. Solvation 1.) Solvation = the process of dissolving
a.) First- solute particles are surrounded by solvent particles b.) Then- solute particles are separated and pulled into solution C. Johannesson

20 B. Solvation NaCl(s)  Na+(aq) + Cl–(aq)
2.) Dissociation = separation of an ionic solid into aqueous ions Ex: NaCl + H2O – the Na ion and Cl ion become hydrated and gradually move away from the crystal into solution. Each ion in the solution acts as though it were present alone: So there is only a solution containing Na+ and Cl- ions uniformly mixed with H2O particles NaCl(s)  Na+(aq) + Cl–(aq)

21 Animation of Salt Dissolving:

22 B. Solvation 3.) Ionization = breaking apart of some polar molecules into aqueous ions Ex: HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq)

23 4.) Molecular Solvation =molecules stay intact
B. Solvation 4.) Molecular Solvation =molecules stay intact Ex: C6H12O6(s)  C6H12O6(aq)

24 C. Factors Affecting the Rate of Dissolving (Increases Solution Rate):
Grinding: increases surface area exposed to solvent Stirring: allows solvent continual contact with solute Heating: increases kinetic energy; increases mixing

25 C. Factors Affecting the Rate of Dissolving (Increase Solution Rate):

26 III. Electrolytes & Nonelectrolytes

27 Many people know me as containing “electrolytes”… that just means I contain dissolved, dissociated ions!

28 A. Electrolytes and Nonelectrolytes
1) electrolyte = a substance that dissolves in water to give a solution that conducts electric current 2) nonelectrolyte = a substance that dissolves in water to give a solution that does NOT conduct an electric current

29 A. Electrolytes and Nonelectrolytes
3) Solutions of electrolytes can conduct electric current: a) The positive ions and the negative ions disassociate (separate) in solution. The mobile ions can move a charge from one point in the solution to another point

30 A. Electrolytes and Nonelectrolytes
4) Solutions of nonelectrolytes cannot conduct electric current: When a nonelectrolyte dissolves in water there are NO charged particles in solution. Ex: Solute exists as molecules

31 A. Electrolytes and Nonelectrolytes
5) Weak Electrolytes a) Only a portion of dissolved molecules ionize

32 A. Electrolytes and Nonelectrolytes
6) Solid ionic compounds cannot conduct electric current: a) Ions are present but they are NOT mobile.

33 A. Electrolytes and Nonelectrolytes
- + sugar - + acetic acid - + salt Non- Electrolyte Weak Electrolyte Strong Electrolyte solute exists as molecules only solute exists as ions and molecules solute exists as ions only

34 IV. Solubility

35 A. SOLUBILITY: 1) Solubility = quantity of solute that will dissolve in specific amount of solvent at a certain temperature. (pressure must also be specified for gases).  Ex: 204 g of sugar will dissolve in 100 g of water at 20C  soluble and insoluble are relative terms  solubility should NOT be confused with the rate at which a substance dissolves

36 A. SOLUBILITY: 2) saturated solution = a stable solution in which the maximum amount of solute has been dissolved. Visual evidence: a quantity of undissolved solute remains in contact with the solution

37 A. SOLUBILITY: 3) solubility equilibrium = state where the solute is dissolving at the same rate that the solute is coming out of solution (crystallizing). Opposing processes of the dissolving and crystallizing of a solute occur at equal rates. solute solvent solution

38 A. SOLUBILITY: 4) unsaturated solution = a solution that contains less solute than a saturated solution under existing conditions

39 A. SOLUBILITY: 5) supersaturated solution = a solution that temporarily contains more than the saturation amount of solute than the solvent can hold Unstable – if disturbed, the excess solute will crystallize out of solution

40 A. Solubility UNSATURATED SOLUTION more solute dissolves
no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration

41 B. 3 FACTORS EFFECTING SOLUBILITY:
1) Nature of solute and solvent a) “Like dissolves like” = rule of thumb for predicting whether or not one substance dissolves in another “Alikeness” depends on: Intermolecular forces Type of bonding Polarity or nonpolarity of molecules: ionic solutes tend to dissolve in polar solvents but not in nonpolar solvents

42 B. 3 FACTORS EFFECTING SOLUBILITY:
b.) Solvent-Solute Combinations: Solvent Type Solute Type Is Solution Likely? Polar Nonpolar “Like Dissolves Like” YES NO NO YES

43 B. 3 FACTORS EFFECTING SOLUBILITY:
2) Pressure: Pressure has little effect on the solubility of liquids or solids in liquid solvents.  The solubility of a gas in a liquid solvent increases when pressure increases. It is a direct relationship.

44 B. 3 FACTORS EFFECTING SOLUBILITY:
3) Temperature: The solubility of a gas in a liquid solvent decreases with an increase in temperature.  The solubility of a solid in a liquid solvent MOST OFTEN increases with an increase in temperature. However, solubility changes vary widely with temperature changes sometimes decreasing with temperature increases.

45 V. Solubility Graph

46 A. SOLUBILITY CURVE: Saturated = any point on the line or ABOVE the line 2) Unsaturated = any point BELOW the line

47

48 1. What is the solubility of the following solutes in water?
a) NaCl at 60ºC= b) KCl at 40ºC= c) KNO3 at 20ºC= 38g 39g 31g

49 Saturated (on the line)
2. Are the following solutions saturated or unsaturated? Each solution contains 100 g of H20. a) 31.2 g of KCl at 30ºC = b) 106g KNO3 at 60ºC= Unsaturated Saturated (on the line)

50 c)40 g NaCl at 10ºC= Saturated Unsaturated d)150g KNO3 at 90ºC =
2. Are the following solutions saturated or unsaturated? Each solution contains 100 g of H20. c)40 g NaCl at 10ºC= d)150g KNO3 at 90ºC = Saturated Unsaturated

51 169g dissolved 11g undissolved
3. For each of the following solutions, explain how much of the solute will dissolve and how much will remain undissolved at the bottom of the test tube? a) 180 g of KNO3 in g of water at 80ºC 169g dissolved 11g undissolved

52 31g dissolved 149g undissolved
3. For each of the following solutions, explain how much of the solute will dissolve and how much will remain undissolved at the bottom of the test tube? b) 180 g of KNO3 in g of water at 20ºC 31g dissolved 149g undissolved

53 C) 60 g of NaCl in 100 g of water at 60ºC
3. For each of the following solutions, explain how much of the solute will dissolve and how much will remain undissolved at the bottom of the test tube?  C) 60 g of NaCl in g of water at 60ºC 39g dissolved  21g undissolved

54 4. A saturated solution of KNO3 is formed from one hundred grams of water. If the saturated solution is cooled from 90°C to 30°C, how many grams of precipitate are formed? (200g-45g) = 155g

55 5. A saturated solution of KCl is formed from one hundred grams of water. If the saturated solution is cooled from 90°C to 40°C, how many grams of precipitate are formed? (53g-39g) = 14g

56 VI. Concentration: Molarity & Molality

57 A. Concentration 1.) Concentration = The amount of solute in a solution. a.) Describing Concentration: % by mass - medicated creams % by volume - rubbing alcohol ppm, ppb - water contaminants molarity - used by chemists molality - used by chemists

58 substance being dissolved
A. Molarity 1.) Molarity = unit of concentration of a solution. Formula: substance being dissolved total combined volume

59 A. Molarity 2M HCl What does this mean?

60 B. Molarity Calculations
To solve for Molarity when given grams of solute, you will use gram/mole conversion and the equation for Molarity. Use the following steps: Change grams of solute to moles. To do this you need to use the following conversion: Grams of solute 1 mole of solute =moles of solute Molar mass of solute* *make sure formulas are correct; Use ion list!    

61 B. Molarity Calculations
b) Change the volume of solution to liters (L) by moving the decimal 3 places to the LEFT. Remember that 1000 mL = 1 L. Water will most often be the solvent in a solution. c) Substitute the information into the molarity equation and solve.

62 22.4 g of NaCl 1 mole NaCl =0.383 moles NaCl 58.443 g NaCl
2) Ex: What is the molarity of a solution composed of g of sodium chloride dissolved in enough water to make 500 mL of solution? 22.4 g of NaCl 1 mole NaCl =0.383 moles NaCl g NaCl 0.383 mole NaCl = M NaCl 0.500 L soln Note: You need to change mL of solution to L. So move decimal 3 places to the left!

63 B. Molarity Calculations
Use the Molarity equation to solve for grams of solute. You will need to solve the Molarity equation for moles of solute. Then convert moles of solute to grams of solute.

64 Ex: How many grams of lithium bromide are present in 300mL of a 0
Ex: How many grams of lithium bromide are present in 300mL of a 0.4M lithium bromide solution? a) Substitute the information and solve for moles by cross multiplying or multiplying by the reciprocal: “X” moles LiBr___ = 0.4M LiBr L solution X = 0.12 moles of LiBr are needed

65 Ex: How many grams of lithium bromide are present in 300mL of a 0
Ex: How many grams of lithium bromide are present in 300mL of a 0.4M lithium bromide solution? b) Convert moles to grams using the following conversion: Moles of solute molar mass of solute = grams of solute 1 mole of solute The substitution would look like: 0.12 moles LiBr g LiBr = 10. g LiBr 1 mol LiBr

66 C. Dilution Calculations
1.) Sometimes solutions of lower concentrations are made from existing solutions. 2.) Moles of solute remain the same. 3.) Formula: M1V1 = M2V2 M1 is the original molarity concentration V1 is the volume of the original solution (in L) M2 is the new concentration V2 is the amount of the new solution needed (in L)

67 C. Dilution Calculations
Ex 1: What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution? GIVEN: M1 = 15.8M V1 = ? M2 = 6.0M V2 = 250 mL WORK: M1 V1 = M2 V2 (15.8M) V1 = (6.0M)(250mL) V1 = 95 mL of 15.8M HNO3

68 D. Molality 1.) Molality (m) = a unit of concentration of a solution expressed in moles of solute per kilogram of solvent. a) Ex: A solution that contains 1 mol of solute, ammonia (NH3), dissolved in exactly 1 kg solvent, is a “one molal” solution.

69 D. Molality b) Formula: mass of solvent only
c) Note: 1 kg water = 1 L water

70 E. Calculations: Solving for Molality
1) If needed convert: grams of solute moles of solute grams of solvent kg of solvent (move decimal 3 places to the left!) 2) Solve for molality m = moles of solute kg of solvent

71 3) Ex: Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water.
75 g MgCl2 1 mol MgCl2 g MgCl2 = 0.79 mol MgCl2 0.79 mol MgCl2 = 3.2m MgCl2 0.250 kg water

72 F. Calculations: Using the Molality equation to solve for grams of solute:
1) Use Molality equation and solve for moles 2) Then convert moles of solute to grams of solute

73 Ex: How many grams of NaCl are required to make a 1
Ex: How many grams of NaCl are required to make a 1.54m solution using kg of water? “X” moles NaCl___ = 1.54m NaCl kg H2O X = moles NaCl mol NaCl g NaCl = 45.0 g NaCl 1 mol NaCl

74 VII. Colligative Properties

75 A. Colligative Property
1.) Colligative property = A property that depends on the number of solute particles rather than the type of particle. C. Johannesson

76 B. Examples of colligative properties:
1.) Vapor Pressure LOWERING of a solution a) Ex: We have two beakers. One contains H2O, the other contains a salt solution (NaCl) in H2O. Both beakers are placed into a sealed chamber: DAY 1 DAY 3: If we leave it for a couple of days and then come back and take a look, this is what it might look like.

77 B. Examples of colligative properties:
b) WHAT HAPPENED?: Na+ and Cl- ions are non-volatile ions. In other words, they will not leave and go into the vapor phase. Those H2O molecules with enough kinetic energy will leave the surface of the solutions and enter into the vapor phase. The fact that Na+ and Cl- ions are dissolved in H2O indicates that there is an attractive interaction between the solutes and the H2O. DAY 1 DAY 3: If we leave it for a couple of days and then come back and take a look, this is what it might look like.

78 B. Examples of colligative properties:
The interaction of the Na+ and Cl- ions for H2O will act to hinder the ability of the solvated H2O molecules to leave and go into the vapor phase (Na+ and Cl- ions are non-volatile). This hindrance of the H2O molecules to enter the vapor phase will reduce the vapor pressure of the H2O in the salt-containing solution. DAY 1 DAY 3: If we leave it for a couple of days and then come back and take a look, this is what it might look like.

79 B. Examples of colligative properties:
The two beakers are in the same sealed container, thus, the vapor pressure above the solutions is identical. The rate of H2O entering the solutions by collisions from the vapor state will be identical The rate of H2O leaving the liquid phase and entering the vapor phase is slower for the NaCl- containing solution DAY 1 DAY 3: If we leave it for a couple of days and then come back and take a look, this is what it might look like.

80 B. Examples of colligative properties:
2.) Osmotic Pressure = If two solutions, with different solute concentrations, are separated by a semi-permeable membrane, there can be a net flow of solvent across the membrane.

81 B. Examples of colligative properties:
3.) Freezing Point depression= The solution will begin to freeze at a temperature BELOW that of the pure solvent.

82 B. Examples of colligative properties:
4.) Boiling Point elevation = the solution will begin to boil at a temperature ABOVE that of the pure solvent.

83 C. How are solutions different than pure liquids?
1.) One of the ways in which they are different, is that when you add a solute to a liquid both the freezing point and boiling point of the solution CHANGE.

84 C. How are solutions different than pure liquids?
a) Freezing point – In a solution, the solute particles interfere with the attractive forces among the solvent particles. This prevents the solvent from entering the solid state at its normal freezing point

85 C. How are solutions different than pure liquids?
b) Boiling point- When the temperature of a solution containing a nonvolatile solute is raised to the boiling point of the pure solvent, the resulting vapor pressure is still less than the atmospheric pressure and the solution will not boil. Thus, the solution must be heated to a higher temperature to supply the additional KE needed to raise the vapor pressure to atmospheric pressure.

86 C. How are solutions different than pure liquids?
2) Water is the liquid we will be dealing with most often The freezing point of pure water is 0°C. The normal boiling point of water is 100°C. But if you make a solution using water as the solvent, the freezing point of that solution will not be 0°C nor will the boiling point be 100°C!

87 D. Freezing Point Depression & Boiling Point Elevation:
1) Freezing point depression (TFP) – solutions will freeze at lower temperature than the pure solvent a) The more solute dissolved, the greater the effect.

88 D. Freezing Point Depression & Boiling Point Elevation:
b) Ex: ethylene glycol (antifreeze) protects against freezing of the water in the cooling system by lowering the freezing point to about -40°C

89 D. Freezing Point Depression & Boiling Point Elevation:
c) Ex: making of homemade ice cream- The ice cream mix is put into a metal container which is surrounded by crushed ice. Then salt is put on the ice to lower its melting point. This gives a temperature gradient across the metal container into the saltwater-ice solution which is lower than 0°C. The heat transfer out of the ice cream mix allows it to freeze.

90 D. Freezing Point Depression & Boiling Point Elevation:
2.) Boiling point elevation (TBP) – solutions will boil at higher temperatures than the pure solvent a) The boiling point of pure water is 100°C, but that boiling point can be elevated by the addition of a solute such as a salt. b) The more solute dissolved, the greater the effect.

91 E. Calculating Freezing Point Depression & Boiling Point Elevation:
1) Solution concentrations are given in molality (m)  2) Colligative properties are directly proportional to the molal concentration of a solute

92 E. Calculating Freezing Point Depression & Boiling Point Elevation:
 3)Account for particle molality for ELECTROLYTES Ex: NaCl = 2 ions (Na+ & Cl-) Ex: MgCl2 = 3 ions (Mg+2 and 2 Cl-) 4) A change in the concentration (m), changes the freezing point and boiling point of a solution

93 E. Calculating Freezing Point Depression & Boiling Point Elevation:
a) You will need to figure out the change in temperature for the normal freezing point (FP) or boiling point (BP) and adjust them with the following steps:  ∆T = mK ∆T = change in temperature m = moles solute/kg solvent (MOLALITY) K = constant You will mostly use the constants for water: Kfp = ˚C/m Kbp = 0.515˚C/m

94 E. Calculating Freezing Point Depression & Boiling Point Elevation:
Other values can be found below: Freezing Point Depression Constants Compound Freezing Point (oC) Kfp (oC/m) water 1.853 acetic acid 16.66 3.90 benzene 5.53 5.12 p-xylene 13.26 4.3 naphthalene 80.29 6.94 cyclohexane 6.54 20.0 Boiling Point Elevation Constants Compound Boiling Point (oC) Kbp(oC/m) water 100 0.515 ethyl ether 34.55 1.824 carbon disulfide 46.23 2.35 benzene 80.10 2.53 carbon tetrachloride 76.75 4.48 camphor 207.42 5.611

95 E. Calculating Freezing Point Depression & Boiling Point Elevation:
4.) Steps solving for ∆T: a) Solve for molality (see steps from previous problems) m = moles of solute kg of solvent

96 E. Calculating Freezing Point Depression & Boiling Point Elevation:
b) Find your K constant from the chart listed in your notes. Be sure to find the correct K constant for what you are solving for-- either FP or BP!! c) Solve for ∆T ∆T = mK

97 E. Calculating Freezing Point Depression & Boiling Point Elevation:
d) If solving for: Freezing point: Take the normal freezing point of the solvent and subtract the ∆T value Round final answer to 3 significant figures! Boiling point: Take the normal boiling point of the solvent and add the ∆T value Round final answer to 6 significant figures!

98 E. Calculating Boiling Point Elevation:
5.) At what temperature will a solution that is composed of g of glucose (C6H12O6) in 225 g of water boil? 32.8g C6H12O6 1 mol C6H12O6 g C6H12O6 = mol C6H12O6 0.182 mol C6H12O6 = m C6H12O6 m = 0.225 kg water TBP = mKbp TBP = (0.809 m) (0.515 C/m) = 0.417C BP = 100C C = C

99 E. Calculating Freezing Point Depression:
6.) Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 115 g water. 1 mol NaCl 58.443g NaCl 28.0g NaCl = mol NaCl 0.479 mol NaCl = 4.17m NaCl m = 0.115 kg water Particle molality = 4.17m X 2 = 8.34m TFP = mKfp TFP = (8.34 m) (1.853 C/m) = 15.5C FP = 0C – 15.5 C = -15.5C


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