Solutions. Chapter 14 Solutions. Ion-Ion Forces (Ionic Bonding) Attraction Between Ions and Permanent Dipoles. Covalent Bonding Forces

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Solutions Chapter 14 1 Brief Review of Major Topics in Chapter 13, Intermolecular forces Ion-Ion Forces (Ionic Bonding) 2 Na + Cl - in salt These are the strongest forces. Lead to solids with high melting temperatures. NaCl,, mp = 800 o C MgO,, mp = 2800 o C Covalent Bonding Forces 3 Attraction Between Ions and Permanent Dipoles 4 C=C, 610 kj/mol C H, 413 kj/mol C C, C, 346 kj/mol CN, 887 kj/mol water -δ O dipole H H +δ Water is highly polar and can interact with positive ions to give hydrated ions in water. Attraction Between Ions and Permanent Dipoles 5 Attraction Between Ions and Permanent Dipoles 6 water -δ O dipole H H +δ Water is highly polar and can interact with positive ions to give hydrated ions in water. Many metal ions are hydrated. This is the reason metal salts dissolve in water.

Attraction Between Ions and Permanent Dipoles Attraction between ions and dipole depends on ion charge and ion-dipole distance. Measured by H H for M n+ + H 2 O --> > [M(H 2 O) x ] n+ 7 Dipole-Dipole Forces Such forces bind molecules having permanent dipoles to one another. 8 δ- O H δ- H O H H Mg2+ δ+ δ+ Na + Cs + δ- H O H δ+ -1922 kj/mol -405 kj/mol -263 kj/mol Dipole-Dipole Forces Influence of dipole-dipole forces is seen in the boiling points of simple molecules. Compd Mol. Wt. Boil Point N 2 28-196 C CO 28-192 C Br 2 160 59 o C ICl 162 97 o C 9 Hydrogen Bonding A special form of dipole-dipole attraction, which enhances dipole-dipole attractions. 10 H-bonding is strongest when X and Y are N, O, or F Hydrogen Bonding in H 2 O H-bonding is especially strong in water because the O H O H bond is very polar there are 2 lone pairs on the O atom Accounts for many of water s s unique properties. 11 FORCES INVOLVING INDUCED DIPOLES How can non-polar molecules such as O 2 and I 2 dissolve in water? The water dipole INDUCES a dipole in the O 2 electric cloud. Dipole-induced induced dipole 12

FORCES INVOLVING INDUCED DIPOLES 13 FORCES INVOLVING INDUCED DIPOLES 14 Solubility increases with mass the gas Process of inducing a dipole is polarization Degree to which electron cloud of an atom or molecule can be distorted in its polarizability. IM FORCES INDUCED DIPOLES Consider I 2 dissolving in ethanol, CH 3 CH 2 OH. I-I -δ O R H +δ The alcohol temporarily creates or INDUCES a dipole in I 2. I-I -δ +δ -δ O R H +δ 15 FORCES INVOLVING INDUCED DIPOLES Formation of a dipole in two nonpolar I 2 molecules. Induced dipole- induced dipole 16 17 18 Intermolecular Forces Summary Liquids In a liquid molecules are in constant motion there are appreciable intermolecular forces molecules close together Liquids are almost incompressible Liquids do not fill the container

Liquids The two key properties we need to describe are EVAPORATION and its opposite CONDENSATION LIQUID evaporation--- ---> Add energy VAPOR break IM bonds make IM bonds Remove energy <---condensation 19 Liquids Evaporation To evaporate, molecules must have sufficient energy to break IM forces. Breaking IM forces requires energy. The process of evaporation is endothermic. 20 21 22 Phase Diagrams Phase Diagrams Solutions: Definitions A solution is a One constituent is usually HOMOGENEOUS regarded as the SOLVENT mixture of 2 or more and the others as SOLUTES. substances in a single phase. Definitions Solute(s): That which is dissolved, generally in lower concentration Solvent: The medium doing the dissolving of the solute 23 Dissolving An Ionic Solid 24 Solutions can be classified as unsaturated or saturated. A saturated solution contains the maximum quantity of solute that dissovles at that temperature. SUPERSATURATED SOLUTIONS contain more than is possible and are unstable. Writing Equations: NaCl(s) Na + (aq) + Cl - (aq) You do CaCl 2 and Al 2 (SO 4 ) 3

Energy of the Solution Process 25 Energy of the Solution Process 26 27 28 Lattice Energy E to form lattice (note negative sign) Hydration Energy E evolved when gaseous ions are dissolved Energy - KF If the enthalpy of formation of the solution is more negative that that of the solvent and solute, the enthalpy of solution is negative. The solution process is exothermic! 29 30 Supersaturated Sodium Acetate Supersaturated Sodium Acetate One application of a supersaturated solution is the sodium acetate heat pack. Sodium acetate has an ENDOthermic heat of solution. Sodium acetate has an ENDOthermic heat of solution. NaCH 3 CO 2 (s) + heat ----> Na + (aq) ) + CH 3 CO - 2 (aq) Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC. Na + (aq) ) + CH 3 CO - 2 (aq) ---> NaCH 3 CO 2 (s) + heat

Concentration Units An IDEAL SOLUTION is one where the properties depend only on the concentration of solute. Need conc. units to tell us the number of solute particles per solvent particle. The unit molarity does not do this! 31 Concentration Units X A = mol fraction A = MOLE FRACTION, X For a mixture of A, B, and C mol A mol A + mol B + mol C MOLALITY, m mol solute m of solute = kilograms solvent WEIGHT % = grams solute per 100 g solution 32 Calculating Concentrations 33 Calculating Concentrations 34 Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate mol fraction, molality,, and weight % of glycol. Consider a solution that is 20.00 g of NaOH dissolved in 250 ml of total aqueous solution (the final density of solution is 1.10 g/ml) Calculate M, m, mole fraction NaOH and w% Calculating Concentrations 35 Calculating Concentrations 36 Consider a solution that is 20.00 g of NaOH dissolved in 250 ml of total aqueous solution (the final density of solution is 1.10 g/ml) Consider a solution that is 20.00 g of NaOH dissolved in 250 ml of total aqueous solution (the final density of solution is 1.10 g/ml) Calculate M, m, mole fraction NaOH and w% Calculate M, m, mole fraction NaOH and W%

What Affects Solubility?? 37 Dissolving Gases & Henry s Law 38 Pressure Inc. P over gas, increase solubility Gas + Solvent Solution Temperature Reactions are exothermic ( H = negative) or endothermic ( H = positive). Endo: NaNO 3 (s) + Heat Na + (aq) + NO 3- (aq) Increase temperature increase solubility (most salts) Gas solubility (mol/l) = k H P gas k H for H 2 = 1.07 x 10-6 M/mmHg Exo: NH 4 Cl (s) NH 4+ (aq) + Cl - (aq) + Heat Increase temperature decrease solubility When P gas drops, solubility drops. What is solubility of H 2 at 1 atm?? And 0.33 atm? Colligative Properties On adding a solute to a solvent, the props. of the solvent are modified. Vapor pressure decreases Will Melting point Boiling point decreases increases Osmosis is possible (osmotic pressure) These changes are called COLLIGATIVE PROPERTIES. They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles. 39 To understand colligative properties, study the LIQUID-VAPOR EQUILIBRIUM for a solution. Understanding Colligative Properties 40 Understanding Colligative Properties VP of H 2 O over a solution depends on the number of H 2 O molecules per solute molecule. P solvent proportional to X solvent 41 Raoult s Law An ideal solution is one that obeys Raoult s law. P A = X A P o A 42 P solvent = X solvent P o solvent VP of solvent over solution = (Mol frac solvent) (VP pure solvent) RAOULT S S LAW Because mole fraction of solvent, X A, is always less than 1, then P A is always less than P o A. The vapor pressure of solvent over a solution is always LOWERED!

Raoult s Law Assume the solution containing 62.1 g of glycol in 250. g of water is ideal. What is the vapor pressure of water over the solution at 30 o C? (The VP of pure H 2 O is 31.8 mm Hg; see App. E.) 43 Changes in Freezing and Boiling Points of Solvent VP Pure solvent VP solvent after adding solute 1 atm P 44 BP solution BP pure solvent T Vapor Pressure Lowering 45 Elevation of Boiling Point Elevation in BP = T BP = K BP m (where K BP is characteristic of solvent) 46 Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the BP of the solution? K BP = +0.512 o C/molal for water (see Table 14.3). Figure 14.14 Elevation of Boiling Point Elevation in BP = T BP = K BP (where K BP is characteristic of solvent) P VP Pure solvent 1 atm BP pure solvent VP solvent after adding solute BP solution T BP m 47 Change in Freezing Point Pure waterethylene glycol/water solution The freezing point of a solution is LOWER than that of the pure solvent. FP depression = T FP = K FP m Calculate the FP of a 4.00 molal glycol/water solution. FP = -1.86 o C/molal (Table 14.4) K FP 48

Freezing Point Depression 49 Freezing Point Depression 50 Calculate the FP of a 4.00 molal glycol/water solution. K FP = -1.86 o C/molal (Table 14.4) How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 o C?. K FP = (-1.86( o C/molal) Freezing Point Depression 51 Freezing Point Depression 52 How much NaCl must be dissolved in 4.00 kg of water to lower FP to -10.00 o C?. Solution How much NaCl must be dissolved in 4.00 kg of water to lower FP to - 10.00 o C?. Solution Boiling Point Elevation and Freezing Point Depression T T = K m i A general equation i = van t Hoff factor = number of particles produced per formula unit. Compound Theoretical Value of i glycol 1 NaCl 2 CaCl 2 3 53 Osmosis Solvent Solution Semipermeable membrane The semipermeable membrane allows only the movement of solvent molecules. Solvent molecules move from pure solvent to solution in an attempt to make both have the same concentration of solute. OSMOTIC PRESSURE, = crt (c is conc. in mol/l) 54

Osmosis at the Particulate Level 55 Process of Osmosis 56 Figure 14.17 Osmosic Pressure, Equilibrium is reached when pressure the OSMOTIC PRESSURE, produced by extra solution Osmotic counterbalances pressure of pressure solvent molecules moving thru the membrane. = crt (c is conc. in mol/l) 57 Osmosis Calculating a Molar Mass Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. Π measured to be 10.0 mm Hg at 25 C. Calc. molar mass of hemoglobin. 58 Osmosis Calculating a Molar Mass 59 Osmosis 60 Dissolve 35.0 g of hemoglobin in enough water to make 1.00 L of solution. Π measured to be 10.0 mm Hg at 25 C. Calc. molar mass of hemoglobin. Osmosis of solvent from one solution to another can continue until the solutions are ISOTONIC they have the same concentration.

Osmosis and Living Cells 61 Reverse Osmosis Water Desalination 62 Water desalination plant in Tampa