Chapter 14 The Chemistry of Solutes and Solutions. Solute-Solvent Interactions. Solute-Solvent Interactions. Solute-Solvent Interactions

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1 John W. Moore Conrad L. Stanitski Peter C. Jurs Solubility & Intermolecular Forces Solution = homogeneous mixture of substances. It consists of: solvent - component in the greatest amount. Chapter 14 The Chemistry of Solutes and Solutions solute - all other components (may be >1). Solvent-solute interactions determine if a substance will dissolve in a particular solvent. Stephen C. Foster Mississippi State University Solubility & Intermolecular Forces Solutions: Exist in all 3 physical states. Can be mixtures of solids, liquids and gases. Type of Solution Examples Gas in gas Air Gas in liquid Carbonated drinks. Gas in solid Hydrogen in Pd metal. Liquid in liquid Motor oil, vinegar. Solid in liquid Ocean water, sugar-water. Solid in solid Bronze, pewter, 14K gold. Solute-Solvent Interactions Like dissolves like If solute and solvent intermolecular forces are: Similar the pair will be soluble. polar dissolves polar; non-polar dissolves non-polar. Dissimilar insoluble. Solute-Solvent Interactions Substances dissolve when: solvent-solute attraction > solvent-solvent attraction, and > solute-solute attraction. Solute-Solvent Interactions Miscible liquids dissolve in all proportions. e.g. ethanol and water (both H-bonded polar liquids). Immiscible liquids form distinct separate phases. e.g. gasoline (non-polar) and water (polar). CCl 4 NiCl 2 (aq) NiCl 2 (aq) C 7 H 16 and C 7 H 16 CCl 4 Carefully layered Stir settle 1

2 Solute-Solvent Interactions Name Formula Solubility (g/100g H 2 C) methanol CH 3 OH miscible ethanol C 2 H 5 OH miscible 1-propanol C 3 H 7 OH miscible 1-butanol C 4 H 9 OH pentanol C 5 H 11 OH hexanol C 6 H 13 OH 0.6 London forces increasing Water solubility decreases as alcohols grow larger: Solute-solute attraction grows. Solute-solvent attraction stays constant. Solute-Solvent Interactions Alternate view Alcohols: polar head ( OH) on a non-polar tail. The head is hydrophilic ( water loving ) The tail is hydrophobic ( water hating ) As the tail gets bigger, it is harder and harder to dissolve. Enthalpy, Entropy and Dissolving Solutes ΔH soln = ΔH step a + ΔH step b + ΔH step c Enthalpy, Entropy and Dissolving Solutes On mixing, solute and solvent molecules spread out over a larger V. This increases entropy (S). Processes with ΔS > 0 tend to be product-favored. For NH 4 NO 3 + H 2 O: ΔH soln > 0 (unfavorable). But ΔS soln > 0 (drives solution formation). Solubility and Equilibrium Solutions are either: Unsaturated [Solute] < solubility. All added solid is dissolved. Can dissolve more solute. Saturated No more solute will dissolve. Undissolved solid is often present. Solubility and Equilibrium At saturation, [solute] remains constant, but the solute is in dynamic equilibrium: solute + solvent solution Solute constantly moves in and out of solution. Supersaturated 2

3 Solubility and Equilibrium Supersaturated solutions have more than the equilibrium amount dissolved. They can be produced by: Dissolving solute at high T. solubility usually increases with T. Slowly lowering T. it can take a long time to occur by chance. Too much remains temporarily dissolved. Supersaturation seed crystal of sodium acetate Supersaturated sodium acetate solution Excess sodium acetate begins to crystallize around the seed The solution warms heat is released as crystals form More and more crystallizes until a saturated solution remains. Dissolving Ionic Solids in Liquids When an ionic compound dissolves in water, the ions overcome the forces holding them in the lattice. become hydrated (surrounded by water). Lattice energy = E required to overcome the forces holding the ions together in a crystal. Enthalpy of hydration (ΔH hydration ) = E released as an ion becomes hydrated. Dissolving Ionic Solids in Liquids ΔH soln = -Lattice E + ΔH hyd (cations) + ΔH hyd (anions) ΔH soln may be: positive (endothermic) cold packs (NH 4 NO 3 ) negative (exothermic) hot packs (CaCl 2 ) Ionic solids are insoluble in nonpolar solvents Ionic compound lattice energies are large. need energy to break them apart. Nonpolar solvents cannot hydrate ions. do not release E to overcome the large lattice E. Entropy and Dissolving Ionic Solids in Water ΔS lattice > 0 when a crystal lattice breaks down. ΔS hyd < 0 when H 2 O molecules order around the ions. If ΔS = ΔS lattice + ΔS hyd > 0; dissolving is favored. Temperature and Solubility Solubility of Gases Le Chatelier s principle: Gas + solvent sat. solution usually ΔH soln < 0 Gas solubility almost always decreases as T increases. +1 & -1 ion compounds are often soluble (ΔS > 0). H 2 O can be highly organized around small +2 and +3 ions (ΔS hyd << 0). So ΔS < 0 (favors insolubility). e.g. CaO = slightly soluble (0.13 g/100 ml at 10 C) Al 2 O 3 = insoluble. 3

4 Temperature and Solubility Solubility of Solids Usually increases as T increases. Pressure and Dissolving Gases More gas dissolves when P of the gas above a liquid increases. Gas + solvent saturated solution Le Chatelier: More gas on the reactant side shift toward products (more gas in solution). Henry s Law Gas solubility is directly proportional to P of the gas. S g = k H P g Solubility Henry s law constant Henry s Law Soft drinks fizz when opened. Calculate [CO 2 ] when a drink is bottled (P = 4.5 atm), and after it s opened. For CO 2 in water k H = 3.4 x 10-2 M/atm. The partial P of CO 2 in the atmosphere = 3.1 x 10-4 atm. S g = k H P g When P g = 4.5 atm S g = 3.4 x 10-2 M (4.5 atm) = 0.15 M atm When opened S g = 3.4 x 10-2 M (3.1 x 10-4 atm) = 1.1 x10-5 M atm Solution Concentration: Units Mass fraction = Mass Solute Total Mass of Solution Weight percent = Mass fraction x 100% Example Saline solutions, NaCl(aq), are often used in medicine. What is the weight percent of NaCl in a solution of 4.6 g of NaCl in 500. g of water? mass fraction = 4.6 g 500. g g = weight percent = x 100% = 0.91 % Parts per Million, Billion and Trillion mass solute % = Parts per hundred = x 10 mass solution 2 Very dilute = low mass fraction. Units for high dilution (trace solute): mass solute Parts per million (ppm) = x 10 mass solution 6 mass solute Parts per billion (ppb) = mass solution x 10 9 Parts per Million, Billion and Trillion 1 ppm of solute in water = 1 mg / 1000 g of solution Since 1L of water has a mass 1000 g. 1 ppm 1 mg/l Similarly 1 ppb 1 μg/l 1 ppt 1 ng/l Parts per trillion (ppt) mass solute = mass solution x

5 Molarity and Molality Molarity = M = moles of solute liters of solution Molality is another concentration scale: Molality = m = moles of solute kilograms of solvent A mass-based unit. Uses solvent mass (not solution). It is T independent (unlike molarity). Converting Units Commercial 30.0% hydrogen peroxide has density = 1.11 g/ml at 25 C. What is its molarity? 30.0 % H 2 O 2 = 30.0 g H 2 O 2 in g of solution Moles of H 2 O 2 = Volume of solution = [H 2 O 2 ] = 30.0 g = mol g mol g 1.11 g/ml mol L = 9.79 M = ml Converting Units Sea water is 10,600 ppm Na +. Calculate the mass fraction and molarity of sodium ions in sea water. The density of sea water is 1.03 g/ml. Mass fraction 10,600 ppm = 10,600 g Na + in 10 6 g of solution Mass fraction = 10,600 g Na g of solution = Mass percent = 1.06 % Converting Units 10,600 ppm Na +. Calculate the molarity of Na +. Density = 1.03 g/ml. Molarity Moles of Na + in 10 6 g of solution = (1.06 x 10 4 g)/(22.99 g mol -1 ) = mol Volume of 10 6 g of solution = (10 6 g/1.03 g ml -1 ) = x 10 5 ml = L [Na + ] = (461.1 mol Na + )/(970.9 L) = M Vapor Pressure Lowering Solvent vapor P drops if non-volatile solute is added. Raoult s law: P 1 = X 1 P 1 vapor pressure of solvent over the solution mole fraction of the solvent X 1 = moles of 1 = n 1 total moles (n 1 +n 2 + ) Lower purity solvent = lower vapor P. vapor pressure of pure solvent Vapor Pressure Lowering A solution of urea in water has a vapor P of torr. The vapor P of pure water is torr. Calculate the mole fraction of urea in the solution. Urea is non-volatile. Water obeys Raoult s law: P water = X water P water torr = X water (355.1 torr) X water = 291.2/355.1 = X urea = = (Remember: X 1 + X 2 + = 1) 5

6 Boiling Point Elevation Non-volatile solutes increase the b.p. of a solvent. Why? Solvent vapor P is lowered. Higher T is needed to get the vapor P = external P. Freezing Point Lowering A non-volatile solute lowers the f.p. of a solvent: Quantitatively ΔT f = K f m solute Quantitatively ΔT b = K b m solute K f is a constant for the solvent. Note: the K b value only depends on the solvent, the molality, m depends on the solute. Ethylene glycol (antifreeze) is added to car radiators to lower the freezing point of water (stops freezing). Freezing Point Lowering Calculate the f.p. of an aqueous 30.0% ethylene glycol mixture. For water K f = 1.86 C kg mol -1. Boiling Point Elevation At what T will m aqueous solutions of urea, NaCl and sucrose boil? For water K b = 0.51 C kg mol g of 30% mix: 30.0 g C 2 H 2 (OH) g H 2 O n glycol = 30.0 g / g mol -1 = mol m glycol = ( mol / kg) = molal ΔT f = 1.86 C kg mol -1 (6.904 mol/kg) = 12.8 C Freezing point = 0.00 C 12.8 C = C Freezing points are lowered b.p. of aq. urea = b.p. of aq. sucrose ΔT b = K b m solute = 0.51 C kg mol -1 (0.100 mol/kg) = C b.p. = C C = C m aqueous NaCl has a higher b.p Colligative Properties of Electrolytes Vapor pressure lowering, b.p. elevation, and f.p. depression are colligative properties: Depend upon the number of particles in solution. The type of particle is unimportant. 1 M sugar and 1 M urea aqueous solutions have the same effect. 1 M NaCl is different. NaCl yields 2 particles in solution (Na + and Cl - ). Sugar and urea do not dissociate in solution. Colligative Properties of Electrolytes In aqueous solution: 1 mol sucrose 1 mol particles 1 mol NaCl 2 mol particles (Na + and Cl - ) 1 mol CaCl 2 3 mol (Ca 2+ and 2 Cl - ) Modify the formulas by replacing m solute with i solute m solute where i = the number of particles per formula unit. = van t Hoff factor 6

7 Boiling Point Elevation At what T will molal NaCl(aq) boil? K b = 0.51 C kg mol -1 The b.p. of molal NaCl: ΔT b = K b i solute m solute = 0.51 C kg mol -1 (2)(0.100 mol/kg) = 0.10 C b.p. = C C = C Colligative Properties of Electrolytes A simple correction, but only works at low [ion]. Ions attract each other in solution. Only act as separate units at very low [ion]. i is smaller than expected in many solutions: [MgSO 4 ] i expected i observed M M M Colligative Properties of Electrolytes Arrange these solutions in increasing b.p order: 0.10 m BaCl 2 (aq) 0.12 m K 2 SO 4 (aq) 0.12 m KBr(aq) Similar molalities, so similar deviations from i expected m i expected : BaCl 2 = 0.10(3) = 0.30 K 2 SO 4 = 0.12(3) = 0.36 KBr = 0.12(2) = 0.24 b.p order: KBr < BaCl 2 < K 2 SO 4 Osmotic Pressure of Solutions Semipermeable membrane Allows passage of small particles. Stops large particles. Ion + H 2 O coordination sphere too large to pass through net solvent flow e.g. animal bladders and cell membranes. Large molecule cannot pass. Osmosis Movement of solvent through a semipermeable membrane from dilute to more concentrated solution. Osmotic Pressure Osmotic pressure = P that must be applied to stop osmosis. molarity of solution Pure water semipermeable bag of 5% sugar water P = c R T i gas constant absolute T Water enters the bag, increasing the P particles / formula unit Height of the column of solution is a measure of P. Osmotic Pressure P = c R T i Easy to remember. Very similar to the ideal gas law: P = n R T = V c R T 7

8 Osmotic Pressure A cell can be exposed to 3 kinds of solution: Reverse Osmosis Used to purify water Applied Pressure Isotonic [solute] out = [solute] in No net flow. Hypertonic [solute] out > [solute] in Net flow out. Hypotonic [solute] out < [solute] in Net flow in. Water Brine Osmotic pressure P Normal Osmosis Water molecules cross the membrane diluting the brine. Reverse Osmosis Flow is reversed if P > P is applied. Flow stops when P = P is applied. Pure water can be separated from brine. 8

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