Chapter 13 Properties of Solutions

Transcription

1 Chapter 13 Properties of Solutions 13.1 The Solution Process - Solutions are homogeneous mixtures of two or more pure substances. - In a solution, the solute is dispersed uniformly throughout the solvent. The Effect of Intermolecular Forces on Solution Formation In order for a solution to form: - Solute molecules must be separated ΔH+ - Solvent molecules must be separated ΔH+ - Solute and solvent molecules interact ΔH Sum = ΔH sol The overall process may be exo or endothermic. If endothermic, the increase in disorder (ΔS) drives the process. Solution Formation and Chemical Reactions Dissolving is a physical process: NaCl (s) H 2 O But it can also be due to a chemical rxn: Na + (aq) + Cl - (aq) We can recover the NaCl (evaporate the water). Ni (s) + 2 HCl (aq) NiCl 2 (aq) + H 2 (g) We can't recover the Ni Saturated Solutions and Solubility The solubility of a solute in a solvent is the maximum amount that can dissolve under a certain set of conditions. Ex. The solubility of KCl in water at 20 C is: 34g KCl/100g H 2O. If you combine 50g of KCl + 100g H 2O at 20 C, only 34g KCl will dissolve, and 16g KCl (s) will remain. Types of Solutions Saturated solution contains the maximum amount of solute. Unsaturated solution contains less than the needed amount for saturation. Supersaturated solution contains more than the maximum amount of solute, but is unstable (the excess will crystallize). Supersaturated solutions are unstable. (Metastable system.) 13.3 Factors Affecting Solubility 1. Intermolecular Forces Chemists use the axiom like dissolves like. (The more similar the intermolecular attractions, the more likely one substance is to be soluble in another.) Polar substances tend to dissolve in polar solvents. Nonpolar substances tend to dissolve in nonpolar solvents. Miscible liquids that are soluble in any proportions (immiscible is the antonym). a. Water and ethanol b. oil and gasoline c. oil and water Some molecules have a polar part and a nonpolar part. Longer polar part makes them less soluble in H 2O. More electronegative atoms and/or more H-bonding groups make them more soluble in H 2O. (Look at the ratio of C s to H- bonding when comparing.) Network covalent solids are not soluble in anything. Metals are only soluble in other metals (alloys). 1

2 Problem 1 Do these combinations mix? CCl 4 and Hexane (C 6H 14) CCl 4 and water Benzene (C 6H 6) and MgSO 4 Hexane (C 6H 14) and Heptane (C 7H 16) Ethyl alcohol (C 2H 5OH) and heptanol (C 7H 15OH) Octane (C 8H 18) and methyl alcohol (CH 3OH) Problem 2 Arrange in order of increasing solubility in H 2O. a. CH 3CH 2CH 2CH 2CH 3 b. HOCH 2CH 2CH 2CH 2CH 2OH c. CH 3CH 2CH 2CH 2CH 2OH d. CH 3CH 2CH 2CH 2CH 2Cl Problem 3 Which is more soluble in H 2O? a. CH 3CH 2CH 2CH 2CH 2OH vs. CH 3CH 2CH 2OH b. CH 2(OH)CH(OH)CH(OH)CH(OH)CH 2(OH) vs. CH 3CH 2CH 2OH Vitamins There are fat soluble and water soluble vitamins. Factors Affecting Solubility 2. Pressure. - The solubility of liquids and solids does not change appreciably with pressure. - The solubility of a gas in a liquid is directly proportional to its pressure. Henry s Law S g = kp g Where: S g is the solubility of the gas, k is the Henry s Law constant for that gas in that solvent, and P g is the partial pressure of the gas above the liquid. CO 2 dissolved comes out when the bottle is opened (and the pressure drops). See p. 525 deep sea diving. (Cavitation) 2

3 Factors Affecting Solubility 3. Temperature It s effect is different for every substance, and is usually unpredictable. In general: - Most solid substances become more soluble as temperature rises. - The solubility of gases decreases as temperature rises. Thermal Pollution: - Water is used as a coolant (it s very efficient why?). - Water from roads and parking lots Expressing Solution Concentration Mass Percentage mass % A = mass of A in solution 5.0 g of NaCl 100 ; 5.0 % of NaCl = total mass of solution 100 g solution Parts per Million (ppm) ppm == mass of A in solution total mass of solution ppm Pb 2+ = 2 g Pb g solution or 2 mg Pb 2+ kg solution Since a dilute solution is mostly water (d = 1g/mL). 2 mg Pb 2+ L solution Parts per Billion (ppb) ppb == mass of A in solution total mass of solution ppm Pb 2+ = ppm and ppb are used for very dilute solutions (toxins). 2 g Pb g solution 2 μg Pb 2+ L solution or 2 μg Pb 2+ kg solution Mole Fraction (X): X A = moles of A total moles of solution Moles in solution = moles of solute + moles of solvent. Mole fractions of all components add up to 1. Molarity (M) M = mol of solute L of solution ; 3.0 M HCl = 3.0 mol HCl 1 L of solution Since volume is temperature-dependent, molarity can change with temperature; furthermore, volumes often change upon mixing (they are not additive). 3

4 Molality (m) m = mol of solute kg of solvent ; 1.5 M HCl = 1.5 mol HCl 1 kg of solvent Since both moles and mass do not change with temperature, molality (unlike molarity) is not temperature-dependent. Changing Molarity to Molality If we know the density of the solution, we can calculate the molality from the molarity and vice versa. Mass of solvent Molality (mol/kg solvent) + Mass of solute Molar mass Moles of solute Mass of solution Density Volume of solution Molarity (mol/l solution) Problem 4 You dissolve g of sugar (C 12H 22O 11) in g of water. Calculate the mole fraction, molality and mass percent of the solution. (Why can t we calculate M?) Problem 5 The maximum allowable concentration of As in drinking water in the US is ppm. a. If a 2.5 L water sample is found to contain mg As, is this within the allowable limit? b. What is the allowable mass of As in 1 large glass of water (500 ml)? 4

5 Problem 6 A sucrose solution is 25.0% sucrose (C 12H 22O 11) by mass. Calculate the molality of the solution. Problem 7 An aqueous solution of urea ((NH 2) 2CO) is 3.42 m. Calculate M for this solution. (d sol = g/ml) Problem 8 An aqueous solution of urea is 2.00 M and has a density of g/ml. What is its molality? M and ppm 1 ppm = 1 g X g solution = g X 1 g solution = 1 μg X 1 g solution For water: d = 1 g 1 ml ; 1 ppm X = 1 μg X 1 ml solution ; M[=] mol L 1 μg X 1,000 μg X 1 mg X 1 ppm = = = 1 L 1 ml solution ( 1000 ml ) 1 L solution 1 L solution ppm M Since: 1 ppm = 1 mg X 1 L solution 5

6 To change ppm s to M, convert the mg s to moles. To change M to ppm s, convert the moles to mass in grams, then to milligrams. Odor threshold value (OTV) is defined as the most minimal concentration of a substance that can be detected by a human nose; it can be expressed as a concentration of water or concentration in air. ppm M The major aromatic constituent of bell pepper (2-isobutyl-3-methoxypyrazine) can be detected at a concentration of 0.01 nm (OTV). What is this concentration in ppm? CH 3 MM = 9C + 14 H + 2 N + 1 O = g mol HC N C CH 2 CH CH mol ( g 1000 mg ) ( mol 1 g ) 0.01 nm = = ppm L HC N C OCH 3 How low is this? An Olympic swimming pool has a volume of 2,500 m 3 or 2,500,000 L. 1,000 ml 20 drops 2,500,000 L ( ) ( 1 L 1 ml ) = drops ppm = = drops( ) = 0.1 drops Like detecting a fraction of a drop in an Olympic swimming pool!! 13.5 Colligative Properties Changes in colligative properties depend only on the number of solute particles present, not on the identity of the solute particles. Among colligative properties are: Vapor pressure lowering Boiling point elevation Freezing point depression Osmotic pressure 1. Vapor pressure lowering Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase. Raoult s Law P Solution = X SolventP Solvent P A = X AP A where P A is the vp of solution X A is the mole fraction of solvent, and P A is the normal vapor pressure of solvent at that temperature. 6

7 If both, the solute and the solvent, are volatile: P T = P A + P B = X AP A + X BP B It works when the solution is ideal and A and B are very similar (total uniformity of interaction). The more volatile liquid will be richer in the vapor. Fractional distillation used to separate liquids with different bp s. (See box on p. 532.) Problem g of sugar (C 12H 22O 11) are dissolved in 225 ml of water at 60 C. What is the vapor pressure of the solution? (vp water at 60 C = torr.) 2. Boiling point elevation BP of solution > BP pure solvent At any T, the solution s vp is lower, so the solution must be heated to a higher T to boil. T b = K b m (Assume a nonvolatile solute) On the equation T b = K b m T b is added to the normal boiling point of the solvent. K b - elevation constant (depends on the solvent). m - is the molality of the solute, and it should be the concentration of solute particles in solution. Ionic substances dissociate to form more particles: 1 mol NaCl 1 mol Na + + 1mol Cl = 2 mol particles. 0.2 mol Na 2SO mol Na mol SO 4 2 = 0.6 mol particles. Ionic compounds do not ionize completely (ions form ion pairs; the effect is higher at higher concentrations, and it also increases with the charge). *Estimate values of colligative properties assuming ionic compounds ionize completely. (See Box on page 540 on the van t Hoff Factor.)* Problem 10 Estimate the BP of m Al(NO 3) 3(aq). 3. Freezing point depression FP of solution > FP pure solvent T f = K f m K f is the molal freezing point depression constant of the solvent. T f is subtracted from the normal boiling point of the solvent. 7

8 Vapor pressure (mmhg) Applications: Antifreeze in car radiators prevents freezing in winter and boiling over in summer. Salting roads in winter melts ice. A mixture of salt and ice is used to provide the low temperatures needed to make old-fashioned hand-cranked ice cream. Problem 11 Calculate the BP and FP of a 25.0% (by mass) solution of ethylene glycol (C 2H 6O 2), an antifreeze, in H 2O. Problem 12 Which aqueous solution has the lowest freezing point? mol CaCl 2/kg/solution mol NaCl/kg solution mol glucose/kg solution mol methyl alcohol/kg solution Problem 13 The diagram below shows plots of vapor pressure versus temperature for water and 1.0 mole of CaCl 2 dissolved in 1.0 kg of solution. If a vapor pressure versus temperature plot for 1.0 mole of NaCl dissolved in 1.0 kg of solution is superimposed on this diagram, the curve would occur 1. above the red curve. 2. below the green curve. 3. between the red and green curves. 4. on top of the green curve. Temperature ( o C) 4. Osmotic Pressure Osmosis is the diffusion across a semipermeable membrane. Whatever can get across the membrane will in order to equalize the concentrations on each side. In biological systems, most semipermeable membranes allow water to pass through, but solutes are not free to do so. Osmosis In osmosis, there is net movement of solvent from the area of higher solvent concentration (lower solute concentration) to the area of lower solvent concentration (higher solute concentration). Osmotic Pressure The pressure required to stop osmosis is known as: osmotic pressure,. Osmotic Pressure M is the molarity of the solution. R = L atm/mol K T = Kelvin Higher concentration, higher. Osmotic Pressures are usually large enough to be easily measured. π = ( n ) RT = MRT V 8

9 Osmosis Application (Diffusion across cell membranes): Isotonic: If the osmotic pressure is the same on both sides of a membrane (i.e., the concentrations are the same). Hypotonic: If the solute concentration outside the cell is lower than that inside the cell. Hypertonic: If the solute concentration outside the cell is greater than that inside the cell. In (a) an isotonic solution, 0.30 M, the blood cells are normal in appearance. The cells in (b) a hypotonic solution are swollen because of water gain, and may burst, a process called hemolysis. Those in (c) a hypertonic solution are shriveled because of water loss, this process is called crenation. IV solutions must be isotonic to blood. In a very salty or very sugary environment, bacterial cells cannot survive (used in preserving). Problem 14 A solution is made by mixing 20.0g glucose (C 6H 12O 6) in enough H 2O to make 300. ml. a. Is this solution isotonic, hypotonic or hypertonic to body fluids? b. What will happen if red-blood cells are placed in it? c. Calculate at body temperature, 37 C. We can determine MM using Colligative Properties. Problem 15: 144 mg of aspartame in 25 ml solution has an of 364 mmhg at 25 C. Determine MM of aspartame. Problem 16 A solution of 0.85 g of a nonvolatile organic compound in g benzene has a FP = 5.16 C. What is the MM of the solute? Benzene (C 6H 6): K f = 5.12 C/m; FP = 5.5 C Colloids Suspensions of particles larger than individual ions or molecules ( nm), but too small to be settled out by gravity are called colloids. (Not true solutions.) 9

10 Tyndall Effect Colloidal suspensions can scatter rays of light. (Cloudy or opaque.) This phenomenon is known as the Tyndall effect. Colloids in Biological Systems Some molecules like soaps have a polar, hydrophilic (water-loving) end and a non-polar, hydrophobic (water-hating) end. (AKA Amphiphiles) Soaps and Detergents These molecules can aid in the emulsification of fats and oils in aqueous solutions. Form micelles. Removal of Colloidal Particles Semipermeable membranes can be used. Ions can pass through, but the colloidal particles (like proteins) cannot. 10