1) 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)
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
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
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
a) 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
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
SolutionsColloidsSuspensions HomogeneousHeterogeneous Particle size: 0.01-1 nm; can be atoms, ions, molecules Particle size: 1- 1000 nm, dispersed; can be aggregates or large molecules Particle size: over 1000 nm, suspended; can be large particles or aggregates
SolutionsColloidsSuspensions 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
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
1) Solvent = the substance that does the dissolving in a solution a) Typically present in the greatest amount b) Typically a liquid c) Water is the most common or “universal” solvent
B/c water molecules are polar The hydrogen side of each water (H 2 O) 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.
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)
2) Solute = substance being dissolved in a solution a) Typically present in the least amount b) Typically a solid
SolventSoluteCommon Example Gas Diver’s tank GasLiquid Humidity GasSolidMoth ball LiquidGas Carbonation Liquid Vinegar LiquidSolid Seawater SolidGasGas stove lighter SolidLiquid Dental fillings Solid Sterling Silver (Ag + Cu) o NOT all solutions are liquids/solids! o Solutions are formed in ALL 3 states!
C. Johannesson = 1.) Solvation = the process of dissolving b.) Then- solute particles are separated and pulled into solution a.) First- solute particles are surrounded by solvent particles
2.) Dissociation = separation of an ionic solid into aqueous ions Ex: NaCl + H 2 O – 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 H 2 O particles NaCl(s) Na + (aq) + Cl – (aq)
Animation of Salt Dissolving: http://www.northland.cc.mn.us/biology/Biology1111/animations/dissolve.html
3.) Ionization = breaking apart of some polar molecules into aqueous ions Ex: HNO 3 (aq) + H 2 O(l) H 3 O + (aq) + NO 3 – (aq)
4.) Molecular Solvation =molecules stay intact Ex: C 6 H 12 O 6 (s) C 6 H 12 O 6 (aq)
1) Grinding: increases surface area exposed to solvent 2) Stirring: allows solvent continual contact with solute 3) Heating: increases kinetic energy; increases mixing
Many people know me as containing “electrolytes”… that just means I contain dissolved, dissociated ions !
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
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
4) Solutions of nonelectrolytes CANNOT conduct electric current: a) When a nonelectrolyte dissolves in water there are NO charged particles in solution. b) Ex: Solute exists as molecules
5) Weak Electrolytes a) Only a portion of dissolved molecules ionize
6) Solid ionic compounds CANNOT conduct electric current: a) Ions are present but they are NOT mobile.
Strong Electrolyte Non- Electrolyte solute exists as ions only - + salt - + sugar solute exists as molecules only - + acetic acid Weak Electrolyte solute exists as ions and molecules
1) Solubility = quantity of solute that will dissolve in specific amount of solvent at a certain temperature. (pressure must also be specified for gases). a) Ex: 204 g of sugar will dissolve in 100 g of water at 20 C b) soluble and insoluble are relative terms c) solubility should NOT be confused with the rate at which a substance dissolves
2) saturated solution = a stable solution in which the maximum amount of solute has been dissolved. a) Visual evidence: a quantity of undissolved solute remains in contact with the solution
3) solubility equilibrium = state where the solute is dissolving at the same rate that the solute is coming out of solution (crystallizing). a) Opposing processes of the dissolving and crystallizing of a solute occur at equal rates. b) solute + solvent solution
4) unsaturated solution = a solution that contains less solute than a saturated solution under existing conditions
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
SATURATED SOLUTION no more solute dissolves UNSATURATED SOLUTION more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form concentration
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
Solvent TypeSolute Type Is Solution Likely? Polar Nonpolar Polar Nonpolar b.) Solvent-Solute Combinations: YES YES NO NO “Like Dissolves Like”
2) Pressure: a) Pressure has little effect on the solubility of liquids or solids in liquid solvents. b) The solubility of a gas in a liquid solvent INCREASES when pressure increases. It is a direct relationship.
3) Temperature: a) The solubility of a gas in a liquid solvent DECREASES with an increase in temperature. b) 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.
1. What is the solubility of the following solutes in water? a) NaCl at 60ºC= b) KCl at 40ºC= c) KNO 3 at 20ºC= 38g 39g 31g
2. Are the following solutions saturated or unsaturated? Each solution contains 100 g of H 2 0. a) 31.2 g of KCl at 30ºC = b) 106g KNO 3 at 60ºC= Unsaturated Saturated (on the line)
2. Are the following solutions saturated or unsaturated? Each solution contains 100 g of H 2 0. c)40 g NaCl at 10ºC= d)150g KNO 3 at 90ºC = Unsaturated Saturated
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 KNO 3 in 100 g of water at 80ºC 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? b) 180 g of KNO 3 in 100 g of water at 20ºC 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? C) 60 g of NaCl in 100 g of water at 60ºC 39g dissolved 21g undissolved
4. A saturated solution of KNO 3 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
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
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
1.) Molarity = unit of concentration of a solution. Formula: total combined volume substance being dissolved
1) To solve for Molarity when given grams of solute, you will use gram/mole conversion and the equation for Molarity. Use the following steps: a) 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!
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.
2) Ex: What is the molarity of a solution composed of 22.4g 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 58.443 g NaCl 0.383 mole NaCl = 0.766 M NaCl 0.500 L soln Note: You need to change mL of solution to L. So move decimal 3 places to the left!
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.
a) Substitute the information and solve for moles by cross multiplying or multiplying by the reciprocal: “X” moles LiBr___ = 0.4M LiBr 0.300 L solution X = 0.12 moles of LiBr are needed
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 86.845 g LiBr = 10. g LiBr 1 mol LiBr
1.) Sometimes solutions of lower concentrations are made from existing solutions. 2.) Moles of solute remain the same. 3.) Formula: M 1 V 1 = M 2 V 2 M 1 is the original molarity concentration V 1 is the volume of the original solution (in L) M 2 is the new concentration V 2 is the amount of the new solution needed (in L)
Ex 1: What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3
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 (NH 3 ), dissolved in exactly 1 kg solvent, is a “one molal” solution.
mass of solvent only c) Note: 1 kg water = 1 L water b) Formula:
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 molalitym = moles of solute kg of solvent
3) Ex: Find the molality of a solution containing 75 g of MgCl 2 in 250 mL of water. 75 g MgCl 2 1 mol MgCl 2 95.211 g MgCl 2 = 3.2 m MgCl 2 0.250 kg water = 0.79 mol MgCl 2 0.79 mol MgCl 2
1) Use Molality equation and solve for moles 2) Then convert moles of solute to grams of solute
“X” moles NaCl___ = 1.54m NaCl 0.500 kg H 2 O X = 0.770 moles NaCl 0.770 mol NaCl 58.443 g NaCl = 45.0 g NaCl 1 mol NaCl Ex: How many grams of NaCl are required to make a 1.54m solution using 0.500 kg of water?
C. Johannesson 1.) Colligative property = A property that depends on the NUMBER of solute particles rather than the type of particle.
1.) Vapor Pressure LOWERING of a solution a) Ex: We have two beakers. One contains H 2 O, the other contains a salt solution (NaCl) in H 2 O. 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.
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 H 2 O 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 H 2 O indicates that there is an attractive interaction between the solutes and the H 2 O. 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.
The interaction of the Na + and Cl - ions for H 2 O will act to hinder the ability of the solvated H 2 O molecules to leave and go into the vapor phase (Na + and Cl - ions are non-volatile). This hindrance of the H 2 O molecules to enter the vapor phase will reduce the vapor pressure of the H 2 O 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.
The two beakers are in the same sealed container, thus, the vapor pressure above the solutions is identical. The rate of H 2 O entering the solutions by collisions from the vapor state will be identical The rate of H 2 O 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.
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.
3.) Freezing Point depression= The solution will begin to freeze at a temperature BELOW that of the pure solvent.
4.) Boiling Point elevation = the solution will begin to boil at a temperature ABOVE that of the pure solvent.
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.
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
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.
2) Water is the liquid we will be dealing with most often a) The freezing point of pure water is 0°C. b) The normal boiling point of water is 100°C. 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!
1) Freezing point depression ( T FP ) – solutions will freeze at LOWER temperature than the pure solvent a) The more solute dissolved, the greater the effect.
b) Ex: ethylene glycol (antifreeze) protects against freezing of the water in the cooling system by lowering the freezing point to about -40°C
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.
2.) Boiling point elevation ( T BP ) – 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.
1) Solution concentrations are given in molality (m) 2) Colligative properties are directly proportional to the molal concentration of a solute
3)Account for particle molality for ELECTROLYTES Ex: NaCl = 2 ions (Na + & Cl - ) Ex: MgCl 2 = 3 ions (Mg +2 and 2 Cl - ) 4) A change in the concentration (m), changes the freezing point and boiling point of a solution
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: K fp = 1.853 ˚C/m K bp = 0.515˚C/m
Other values can be found below: FREEZING Point Depression Constants Compound Freezing Point ( o C) K fp ( o C/m) water 01.853 acetic acid 16.663.90 benzene 5.535.12 p -xylene 13.264.3 naphthalene 80.296.94 cyclohexane 6.5420.0 BOILING Point Elevation Constants Compound Boiling Point ( o C) K bp ( o C/m) water 1000.515 ethyl ether 34.551.824 carbon disulfide 46.232.35 benzene 80.102.53 carbon tetrachloride 76.754.48 camphor 207.425.611
4.) Steps solving for ∆ T: a) Solve for molality (see steps from previous problems) m = moles of solute kg of solvent
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
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!
5.) At what temperature will a solution that is composed of 32.8 g of glucose (C 6 H 12 O 6 ) in 225 g of water boil? 32.8g C 6 H 12 O 6 1 mol C 6 H 12 O 6 180.155g C 6 H 12 O 6 = 0.809 m C 6 H 12 O 6 0.225 kg water = 0.182 mol C 6 H 12 O 6 0.182 mol C 6 H 12 O 6 m =m = T BP = mK bp T BP = (0.809 m) (0.515 C/m) = 0.417 C BP = 100 C + 0.417 C = 100.417 C
6.) Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 115 g water. 28.0g NaCl 1 mol NaCl 58.443g NaCl = 4.17 m NaCl 0.115 kg water = 0.479 mol NaCl 0.479 mol NaCl m =m = T FP = mK fp T FP = (8.34 m) (1.853 C/m) = 15.5 C FP = 0 C – 15.5 C = -15.5 C Particle molality = 4.17 m X 2 = 8.34m