# Chapter 1: Fundamental Concepts - personal.tcu.edu Solutions (Chapter 12) The Solution Process Why do things dissolve? -- driving force toward more random state (entropy) -- attractive forces between solute and solvent (enthalpy) like dissolves like (ionic and polar substances tend to be water soluble) Intermolecular Forces!!! Intermolecular Forces in Solutions Solubility the amount of a substance that will dissolve in a given amount of solvent How much will dissolve?

Units are often: g solute per 100 g of solution saturated solution: maximum amount of solute is dissolved Supersaturated: more dissolved than saturated miscible: soluble in all proportions Recall dissociation equations e.g. NaCl(s) NaCl(aq) = Na+(aq) + Cl(aq) molecular form solute(undissolved) ionic form

solute(dissolved) Solubility the amount of a substance that will dissolve in a given amount of solvent How much will dissolve? Units are often: g solute per 100 g of solution saturated solution: maximum amount of solute is dissolved Supersaturated: more dissolved than saturated miscible: soluble in all proportions Recall dissociation equations e.g. NaCl(s) NaCl(aq) = Na+(aq) + Cl(aq)

molecular form solute(undissolved) ionic form solute(dissolved) Heats of Solution Hsoln = Hsolute + Hsolvent + Hmix >0 >0

<0 Hhydration Hsoln is a combination of two opposing effects: Lattice Energy (Hsolute) -- endothermic energy required to separate solid particles Hydration (solvation) Energy (Hhydration) -- exothermic energy released as gaseous solute particles are surrounded by solvent molecules e.g. K+(g) + Cl(g) K+(aq) + Cl (aq) Hhydration = -819 kJ/mol

Hsoln Hsolute + Hhydration (Hsoln can be positive or negative) Factors Affecting Solubility Effect of Temperature on Solubility Most solids are more soluble at higher temp, most gases are less soluble at higher temp Effect of Pressure on Solubility No significant effect for solid or liquid solutes, but major effect with gaseous solutes dissolved in liquid solvents Henrys Law: Sg P g P2

gases are more soluble at higher pressure (e.g. carbonated beverage) or S g = k H Pg or S1/S2 = P1/ where S = solubility, P = pressure, kH = Henrys law constant (depends on gas) Concentrations of Solutions I (See Table 12.5, p 529) Molarity (M)

M = moles of solute/liters of solution Mole Fraction (and mole percent) XA = moles A/[moles A + moles B + ] mole % = XA x 100% Mixtures of gases: XA nA PA (at constant temp) where X = mol fraction, n = mol, P = pressure so, XA = PA/Ptotal (i.e. the mol fraction will equal the pressure fraction!) Mass %, parts per million, etc ppm, ppb can be by mass or by volume, e.g. mass solute x multiplication factor mass solution

Multiplication factor = 100 Multiplication factor = 106 Multiplication factor = 109 mass % ppm ppb Concentrations of Solutions II Weight Fraction (and weight percent) WFA = mass A/mass of solution Wt % = WFA x 100% e.g. a 5.00% (by weight) solution of NaCl contains: 5 g NaCl in 100 g of solution (5.00 g NaCl and 95.00 g H2O) molality (m) -- dont confuse it with Molarity (M)!!!

m = moles solute/kg of solvent Independent of temperature e.g. molality of above 5.00% NaCl solution? m = [(5.00 g NaCl x (1 mole NaCl/58.4 g NaCl)]/(0.09500 kg H2O) = 0.90 mol NaCl/kg = 0.901 m NaCl Conversions Between Concentration Methods Example: Commercial hydrobromic acid, HBr, is 40.0% by weight. The density of this solution is 1.38 g/mL. Calculate the molality, molarity, and mole fraction of this HBr solution. 40.0% HBr means that 100 g of solution contains: 40.0 g HBr and 60.0 g H2O Moles HBr = 40.0 g x (1 mole/80.9 g) = 0.494 mol Moles H2O = 60.0 g x (1 mole/18.0 g) = 3.333 mol XHBr = 0.494/(0.494 + 3.333) = 0.129 (or 12.9 mole %)

m = moles HBr/kg H2O = 0.494 mole/0.0600 kg = 8.23 m To find molarity, need volume of solution (from density): Volume of 100 g of solution = 100 g x (1 mL/1.38 g) = 72.5 mL M = mole HBr/L soln = 0.494 mol/0.0725 L = 6.82 M Sample Problem Commercial sulfuric acid is 96.0% H2SO4 (formula mass = 98.07 g/mole) by weight and has a density of 1.85 g/mL. Calculate the molarity (M) and the molality (m) of the H 2SO4 solution. Colligative Properties (depend on number of solute particles) Vapor Pressure (related to mole fraction of solvent) Vapor pressure of solution is always less than the pure solvent

For solutions of non-volatile solutes, Raoults Law applies: Psolution = Xsolvent Psolvent where Psolution = vapor pressure of soln, Xsolvent = mol fraction of solvent, Psolvent = vapor pressure of pure solvent OMIT -- mixtures of two or more volatile components (p 539-541) Freezing and Boiling Points Freezing Point Depression and Boiling Point Elevation (related to molality of the solution) Change in freezing and boiling points: Tf = Kfm Tb = Kbm where Kf and Kb are properties of the solvent:

Kf = molal freezing point depression constant Kb = molal boiling point elevation constant e.g. for water: Kf = 1.86 C/m Kb = 0.51 C/m Example Problem A solution of 6.400 g of an unknown compound in 100.0 g of benzene (C6H6) boils at 81.7 C. Determine the molecular mass of the unknown. Data for benzene: Kf = 5.07 C/m Tf = 5.07 C Kb = 2.53 C/m

Tb = 80.2 C Tb = Kbm Tb = 81.7 - 80.2 = 1.5 C m = Tb/Kb = (1.5 C)/(2.53 C/m) = 0.593 m 0.593 mol cmpd = kg benzene 0.593 mol cmpd kg benzene = 0.0593 moles xcmpd 0.1000

kg benzene Molecular mass = g/mole = 6.400 g cmpd 0.0593 mol cmpd 102 g/mol = 1.1 x Sample Problem Automobile antifreeze is a concentrated aqueous solution of ethylene glycol, C2H6O2 (formula mass = 62.0 g/mol). Calculate the weight percent of an antifreeze solution that would have a freezing point of -25 C (equivalent to -13 F). The Kf constant for water is 1.86 C/m and the freezing point of water is 0.00 C.

Osmotic Pressure (related to molarity) osmosis -passage of solvent through a semipermeable membrane into a solution osmotic pressure () -- back pressure required to stop osmosis M (at constant temp) vant Hoff equation: V = nRT since n/V = M, then = MRT Used for determining MM of unknowns, especially large

molecules, e.g. polymers, proteins, etc. Important in medical solutions; cell walls are semipermeable membranes! hyperosmotic ( > body), hypoosmotic (< body), isosmotic, or isotonic ( = body) Osmotic Pressure Measurement Real Solutions Strong electrolytes do not always dissociate 100%. vant Hoff factors correct for ion pairing and other effects. i= mol of particles in solution mol of formula units dissolved

Corrected equations Tf = imKf Tb = imKb = iMRT e.g., NaCl(s) --> Na+(aq) + Cl(aq) # moles of ions = 2 x (moles of NaCl) so, colligative properties are about twice as large

Tyndall effect: scattering of light by a colloidal dispersion Micelle Formation how soaps work -- micelle formation long hydrocarbon "tail" hydrophobic O O - Na + -O C 2

CO 2 - -O C 2 CO 2 - -O C 2 -O C 2 -O C 2 anionic "head" hydrophilic

CO 2 - CO 2 CO 2 CO 2 CO 2 micelle How Soaps Work Sample Problem A sample of a protein is dissolved in water to give a solution that contains 5.00 mg of protein per 1.00 mL. At 20.0 C, this solution is found to have an osmotic pressure of 0.760 torr. Calculate the molecular mass of the protein.