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Gen Chem--Chapter 14 Lecture Notes

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Gen Chem--Chapter 14 Lecture Notes 6/1/12 Solutes and Solution Saturation and Equilibrium n The first rule of solubility is “likes dissolve likes” n Polar or ionic substances are soluble in polar solvents n Non-polar substances are soluble in non- polar solvents Solutes and Solution Solutes and Solution n There must be a reason why a substance n The forces that drive the dissolution of a is soluble in a solvent: solute usually involve both enthalpy and n either the solution process lowers the overall entropy terms enthalpy of the system ( H < 0) Δ rxn n ΔHsoln < 0 for most species n Or the solution process increases the overall n The creation of a solution takes a more entropy of the system (ΔSrxn > 0) ordered system (solid phase or pure liquid phase) and makes more disordered system n Entropy is a measure of the amount of disorder in a system—total entropy must (solute molecules are more randomly increase for any spontaneous change distributed throughout the solution) Saturation and Equilibrium Saturation and Equilibrium n If we have enough solute available, a solute + solvent ↔ solution KC solution can become saturated—the point n The magnitude of KC indicates how soluble when no more solute may be accepted a solute is in that particular solvent into the solvent n If KC is large, the solute is very soluble n Saturation indicates an equilibrium n If K is small, the solute is only slightly between the pure solute and solvent and C soluble the solution solute + solvent ↔ solution KC 1 6/1/12 Saturation and Equilibrium Saturation and Equilibrium Examples: + - NaCl(s) + H2O(l) ↔ Na (aq) + Cl (aq) KC = 37.3 A saturated solution of NaCl has a [Na+] = 6.11 M and [Cl-] = 6.11 M + - AgCl(s) + H2O(l) ↔ Ag (aq) + Cl (aq) -10 KC = 1.8 x 10 A saturated solution of AgCl has a [Ag+] = 1.34 x 10-5 M and [Cl-] 1.34 x 10-5 M Solubility of Gases Solubility of Gases n To understand the effects of temperature n When a gas is dissolved in a solvent, the process is usually exothermic: on solubility, we can use LeChatlier’s Principle: gas + solvent ↔ solution ΔHrxn < 0 gas + solvent ↔ solution + heat n In solution, the gas molecules are now in close proximity to the solvent molecules n Since heat appears on the product side of which increases the effects of the the chemical equilibrium, increasing the intermolecular forces and lowers the temperature will move the equilibrium overall enthalpy of the system toward the reactants and the gas will be less soluble Solubility of Gases Solubility of Gases n Pressure has no effect on the solubility of n Henry’s Law: solids and liquids, but it has a significant Sg = kH Pg effect on the solubility of gases Sg = solubility of gas in solution n The higher the pressure of the gas over kH = Henry’s Law constant for that the solution, the higher the solubility of gas/solvent combination the gas in solution Pg = partial pressure of gas over the n Henry’s Law: solution Sg = kH Pg 2 6/1/12 Solubility of Gases Solubility of Gases Example: What is the molar concentration of N2 in water when the partial pressure of N2 is 600. Torr? -7 kH = 8.4 x 10 mol/L⋅mm Hg -7 SN2 = (8.4 x 10 mol/L⋅mm Hg)(600. mm Hg) = 5.04 x 10-4 M What is the concentration at 20. Torr? -7 SN2 = (8.4 x 10 mol/L⋅mm Hg)(20. mm Hg) = 1.7 x 10-5 M Concentration Units Concentration Units Molarity: defined as moles of solute per Molality: defined as moles of solute per liter of solution mass of solvent moles solute mol moles solute mol M = = m = = liters of solution L kilograms of solvent kg The volume used is the total volume of the In this case, the mass is only the mass of solution, not just the volume of the solvent the solvent, not the total mass of the Usually the volume of the solute is negligible solution compared to the solvent volume Concentration Units Colligative Properties Mole fraction: mole fraction is defined as n Colligative properties are a set properties the number of moles solute divided by that depend only on the amount of solute the total number of moles of all species in in a solution, and not on the chemical the solution identity of the solute n Colligative properties include: moles solute nA X A = = n Vapor pressure lowering total moles in solution ntot n Freezing point depression Mole fraction is a unitless number n Boiling point elevation n Osmotic pressure 3 6/1/12 Colligative Properties Colligative Properties Vapor pressure lowering Vapor pressure lowering n As a consequence, the vapor pressure of n When solute is added to a pure solvent, solvent molecules are “tied up” in the solution is lowered keeping the solute molecules in solution n Raoult’s Law states: o n Because solvent molecules are more P1 = X1 P1 strongly attracted to the solute than to P1 = vapor pressure of the solution themselves, it requires more energy to X = mole fraction of solvent remove them from the solution compared 1 P o = vapor pressure of the pure solvent to the pure solvent 1 Colligative Properties Colligative Properties Vapor pressure lowering Vapor pressure lowering Example: What is the vapor pressure of a Example: What is the vapor pressure of a o saturated NaCl solution at 25 C? saturated NaCl solution at 25 oC? o P = 23.76 Torr Step 1—Determine mole fraction of solution ρH2O = 0.99707 g/mL n 35.7 g NaCl per 100 mL H2O H2O XH O = Step 1—Determine mole fraction of solution 2 n n n Na+ + Cl− + H2O 35.7 g NaCl/58.443 g/mol =0.611 mol NaCl 5.53 mol (100 mL)(.9971 g/mL)/(18.0152 g.mol) = = .819 .611 mol + .611 mol + 5.53 mol =5.53 mol H2O Colligative Properties Colligative Properties Vapor pressure lowering Boiling Point Elevation Example: What is the vapor pressure of a saturated NaCl solution at 25 oC? n Because the vapor pressure of solution is lower than the vapor pressure of the pure Step 2—Determine vapor pressure of solution solvent, the solution’s boiling point will o PH O = X P = (.819)(23.76 Torr) be elevated 2 H2O = 19.5 Torr n Remember that the boiling point is that The vapor pressure over a saturated NaCl temperature where the vapor pressure of solution is 18% lower than that of pure water the solution is equal to the pressure over the solution 4 6/1/12 Colligative Properties Colligative Properties Boiling Point Elevation Boiling Point Elevation ΔT = T (soln) – T (solvent) Example: What is boiling point of a NaCl solution b b b that is saturated at 25 oC? = Kb m Kb(H2O) = 0.512 K kg/mol Kb = boiling point elevation constant m = (1.22 mol solute)/(.09771 kg H2O) m = molality of solute = 12.5 m ΔTb = (12.5 mol/kg)(0.512 K kg/mol) = 6.40 K o Tb = 106.4 C Colligative Properties Colligative Properties Freezing Point Depression Freezing Point Depression Example: Determine the freezing point of a n The freezing point of a solution will be lower than that of the pure solvent solution that is 40% by volume ethylene glycol in water because the solute molecules interrupt K (H O) = 1.86 K kg/mol the crystal structure of the solid solvent f 2 ρ(C2H6O2) =1.109 g/mL ΔTf = Kf m Kf = freezing pt depression constant m = molality of solute Colligative Properties Colligative Properties Freezing Point Depression Freezing Point Depression Example: Determine the freezing point of a Example: Determine the freezing point of a solution that is 40% by volume ethylene glycol solution that is 40% by volume ethylene glycol in water in water Step 1—Assume we have 1.00 L of the solution; Step 2—Determine freezing point depression determine molality of ethylene glycol ΔTf = (11.9 m)(1.86 K kg/mol) (400 mL C2H6O2)(1.109 g/mL)/(62.069 g/mol) = 22.1 K = 7.15 mol C H O o 2 6 2 Tf = -22.1 C m = (7.15 mol)/(.600 kg H2O) =11.9 m 5 6/1/12 Colligative Properties Colligative Properties Osmotic Pressure Osmotic Pressure n A solution is initially n The solvent wants to have separated from a reservoir equal concentrations on solution of pure solvent by a semi- each side of the system pure solution permeable membrane solvent pure n The solvent will flow from solvent n A semi-permeable the pure solvent side to the membrane allows the solution side of the system solvent to move through in to minimize the flow of solvent either direction, but does concentration difference not allow the solute species between each side of the to pass through semi-permeable membrane system semi-permeable membrane Colligative Properties Colligative Properties Osmotic Pressure Osmotic Pressure n The solution side is now n The osmotic pressure of the higher than the solvent side solution is the pressure solution solution which creates a pressure exerted by the solution difference pure when the system reaches pure solvent equilibrium—balance solvent n The increased pressure on the solution side tends to between force exerted by push solvent back to the pure solvent to equalize pure solvent side of the concentration and force system exerted by increased height semi-permeable membrane of solution semi-permeable membrane Colligative Properties Colligative Properties Osmotic Pressure Osmotic Pressure n The osmotic pressure is Example: Determine the osmotic pressure of a given by: 1.00 M solution of NaCl solution Π = c R T i pure + - where solvent c = 1.00 mol/L i = 2 (Na and Cl ) c = molar concentration R = gas constant Π = 2(1.00 mol/L)(.08206 L atm/mol K)(298 K) T = temperature (in K) = 48.9 atm i = solute particles per unit formula semi-permeable membrane 6 .
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