Chapter 11 Problems: 11, 15, 18, 20-23, 30, 32-35, 39, 41, 43, 45, 47, 49-51, 53, 55-57, 59-61, 63, 65, 67, 70, 71, 74, 75, 78, 81, 85, 86, 93

Chapter 11 Problems: 11, 15, 18, 20-23, 30, 32-35, 39, 41, 43, 45, 47, 49-51, 53, 55-57, 59-61, 63, 65, 67, 70, 71, 74, 75, 78, 81, 85, 86, 93

Chapter 11 Properties of Solutions Types of mixtures: homogenous components are uniformly mixed heterogeneous components are not uniformly mixed Homogenous mixtures are called solutions solvent: component in the majority solute: component in the minority

Concentration Units: 1. % by mass = mass solute x 100 mass solution Problem:.892 g KCl are dissolved in 24.0 g water. Determine the % by mass. 2. Mole fraction (X) where Xa = mole a (mole a + mole b) Problem: use the data above to calculate the mole fraction of KCl Problem: What is the mole fraction of water? What is the sum of the 2 mole fractions?

3. Molarity = M = moles of solute Molar Solution Liters of solution 4. Molality = m = moles of solute molal solution kg of solvent Problem:.892 g KCl is dissolved in 24.0 g water. Determine the molality of the solution.

5. Normality N The number of equivalents per liter of solution. 1 equivalent of acid or base = mass of acid or base that will furnish 1 mole of H + or OH - H 2 SO 4 will provide 2 mole H + ions and weighs 98.0 g/mol therefore 1 equivalent of H 2 SO 4 weighs 49.0 grams. NaOH will provide 1 mole OH - ions and weighs 40.0 g/mol therefore 1 equivalent of NaOH weighs 40.0 grams. For Redox Reactions The mass or quantity of oxidizing or reducing agent that will accept or furnish 1 mole of electrons. MnO 4 - + 5e - -> Mn +2 + 4H 2 O Problem: How many grams of MnO 4 - is 1 equivalent?

Conversion between Molarity and Normality (acids and bases only) M x #H + = N M x #OH - = N What is the normality of: a) 2M HCl b) 3M H 2 SO 4 c) 3M Ca(OH) 2 Since liquids expand when heated, and contract when cooled, the Molarity of a solution is not constant. Therefore, we measure Molarity at a standard temperature of 25 o C. Density of solution must be known to convert 1 solution concentration to another.

Problem: Determine the Molarity of a.396m aqueous C 6 H 12 O 6 solution with a density of 1.16 g/ml.

Problem: The density of a 2.45 M aqueous CH 3 OH solution is.976 g/ml. Calculate the molality.

Problem: Calculate the molality of a 34.5% by mass aqueous solution of phosphoric acid.

Problem: The density of a 3.75 M H 2 SO 4 solution in a battery is 1.230 g/ml. Determine its mass % and Normality.

The Process of dissolving can be broken down into 3 steps that involve energy: 1 Step 1 & 2: Expanding the solvent-solvent and solute-solute bonds involves breaking intermolecular bonds, which is endothermic. 2 Step 3 Solute/Solvent bonds form, which is exothermic. H solution = H 1 + H 2 + H 3

Problem: When CaCl 2 is dissolved, the solution becomes warm. How do process #1,2 (expanding the solute and solvent) compare to process #3 (bonds forming between solute and solvent)? Problem: When NH 4 NO 3 dissolved, the solution cools down. Compare #1,2,3 2 Factors control whether a reaction is spontaneous or not. 1. H 2. Entropy A measure of disorder, where disorder is favorable to order. Which is more favorable (disordered), pure solvent + solute, or solution?

Factors That Affect Solubility: 1. Structure of the molecule Likes dissolve likes -- polar substances dissolve polar substances; non-polar substances dissolve non-polar substances. Miscible 2 liquids that mix in any ratio Immiscible 2 liquids that don t mix.

Perc, or perchloroethane, is used in the dry cleaning industry, and is non-polar

Immiscible 2 liquids that don t mix

Vitamin A is essentially nonpolar, thus fat soluble. Vitamin C is quite polar, thus water soluble.

Solubility maximum amount of solute that can dissolve in a given amount of solvent. Saturated Rate of dissolve = rate of re-precipitation. (Factors that affect Solubility) 2. Temperature A) Solid dissolved in a liquid generally, but not always, solubility increases with increasing temperature.

B) Gas dissolved in a liquid- Generally, but not always, solubility decreases with increased temperature.

3. Pressure Only has an effect on gases dissolved in liquids. An increase in pressure of gas over a liquid causes an increase in gas solubility.

Henry s Law: P = kc where k = const. P = Pressure C = Concentration of gas dissolved in solution Valid only for gases that dissolve, and not those that react. SCUBA diving and the Bends Problem: The solubility of N 2 in blood at 37 o C at a partial pressure of.80 atm is 5.6 x 10-4 mol/l. A SCUBA diver breathes compressed air at 132 feet deep with a partial pressure of 4.0 atm (N 2 ). The diver has 5.0 liters of blood. Calculate the moles of N 2 gas that will dissolve in the divers blood while diving at 132 feet deep. Determine the volume of gas that will exsolve when the diver surfaces. (1 Atm.)

The Colligative Properties (Collective Properties) properties of solutions that don t depend on the type of solute, just the concentration of the solute. 1. Boiling Point Elevation 2. Freezing Point Depression 3. Vapor Pressure Lowering 4. Osmotic Pressure Van t Hoff factor: i - the relationship between moles of solute dissolved and moles of particles in solution. Ideal or expected i: NaCl (s) -> Na+ + Cl - i =? C 6 H 12 O 6 (s) -> C 6 H 12 O 6 (aq) i =? Fe 2 (SO 4 ) 3 -> 2Fe +3 + 3 SO 4-2 i =?

In reality, some free ions will pair up, called ion pairs, causing actual i to be smaller than expected. i (expected) i (observed).05 M NaCl 2 1.9.05 M FeCl 3 4 3.4.05 M glucose 1 1

1. Vapor Pressure Lowering of solutions: (first colligative property) a) Non-Volatile solute (solute doesn t evaporate) The vapor pressure of a solution is lower than the pure liquid. Interference of non-volatile solute prevents a fraction of the liquid from evaporating.

Predict what these 2 beakers will look like in a week or so.

Raoults Law: (For non-volatile solute) P solution = (X solvent ) x (P o solvent ) where P solution = vapor pressure of solution X solvent = mole fraction of solvent P o solvent = vapor pressure of pure solvent Problem: Calculate the vapor pressure of a water solution at 25 o C where 158 g of sucrose, C 12 H 22 O 11 is dissolved in 643.5 ml of water. Problem: Do the same problem as above, using salt as the solute.

b) 2 liquids are mixed, and both are volatile (produce a vapor pressure) The total vapor pressure can be calculated by: P total = P A + P B = X A P Ao + X B P B o If a solution obeys Raults Law, it is called an ideal solution. In reality, very few solutions are perfectly ideal.

3 Scenarios for 2 liquids mixed to make a solution: 1. 2 pure liquids are mixed and H sol n = 0 this solution will behave ideally. Example: Benzene and Toluene 2. 2 pure liquids are mixed and H sol n is exothermic (solution warms up). This indicates that the 2 liquids form strong bonds to each other. Result: Calculated vapor pressure > Observed vapor pressure Example: Acetone and Water 3. 2 pure liquids are mixed and H sol n is endothermic (solution cools). This indicates that the 2 liquids don t bond well to each other. Result: Calculated vapor pressure < Observed vapor pressure Example: Ethanol and Hexane

Problem: 5.80 grams acetone, C 3 H 6 O is mixed with 11.9 grams of chloroform, CHCl 3 at 35 o C. The solutions vapor pressure is measured at 260. torr. Is this solution ideal? (at 35 o C, P acetone = 345 torr, P chloroform = 293 torr) What can be said about the strength of intermolecular bonds in this solution?

2. Boiling Point Elevation Boiling occurs when vapor pressure = external pressure. A solute lowers the vapor pressure of a solution, therefore the solution must be heated to a higher temperature in order for the solution to boil.

To calculate the solution boiling point: T B = k B x m where k b = constant m = molality of solution 3. Freezing Point Depression a solute lowers the freezing point of a pure solvent. T f = k f x m where k f = constant

Problem: Calculate the freezing point and boiling point of a solution composed of 651 grams of ethylene glycol, C 2 H 6 O 2 and 2505 grams of water.

Problem: What mass of ethylene glycol must be added to 10 kg water to make a solution that freezes at -10 o F?

Problem: Calculate the freezing and boiling point of a 2.0 m magnesium chloride solution.

Osmosis Flow of a liquid through a semi-permeable membrane from an area of high to low concentration. 4. Osmotic Pressure Solvent, but not solute can pass through membrane. Pressure is caused by pure solvent passing through membrane faster than solvent in solution. (The pressure that stops osmosis.)

To calculate pressure from the height of a column of solution: torr = mm height of solution x density of solution density of Mercury where density of Mercury is 13.6 g/ml Problem: Convert the pressure of a 25.3 mm column of water to mm Hg: To calculate osmotic pressure: π = MRT where π = osmotic pressure (atm) M = Molarity R = Gas Constant 0.0821 L x atm K x mol T = Kelvin Temp. Problem: 1.00 g of solute/liter causes a solution to rise 25.3 mm. Determine the molecular weight of this substance. Temperature of solution is 25 o C and the solution density is 1.00 g/ml

Osmotic Pressure is the most easily measured colligative property. The previous problem would have had a freezing point depression of.00086 o C freezing point depression. Problem: A sample of hormone weighing 0.546 grams is dissolved in 15.0 g benzene. The freezing point was determined to be 5.26 o C. Calculate the molar mass of the hormone.

Problem: Calculate the osmotic pressure of a 1.2 M NaCl solution at 25 o C.

Back to electrolyte solutions: Free ion fully hydrated ion, completely surrounded by water molecules. Ion Pair at higher concentrations, some ions won t be free ions, but rather, will form an ion pair. Ion pairs decrease conductivity and decrease the effect of the colligative property. Higher charged ions are more likely to form ion pairs

Problem: The observed osmotic pressure of a 0.10 M Fe(NH 4 ) 2 (SO 4 ) 2 solution at 25 o C is 10.8 atm. Compare the experimental vs. expected van t Hoff factor, i.

Desalination of sea water: Distill Freeze Reverse Osmosis

Colloidal Solution Suspension of relatively large uncharged particles in a solution. These particles don t settle, unlike most uncharged particles. Although the particle overall is neutral, the outside of the particle has a charge that neighboring molecules will be repelled by. This charged particle suspends itself indefinitely in a solution.

Tyndall Effect often used to detect suspended particles in a solution. Milk, a colloid, and K 2 Cr 2 O 7, a pure solution