6/22/11 Last time: Orientation Contact information What is a solution. Enthalpy change in dissolution process Chem 103 Effect of temperature on solubility of solids and gases. Today: Lecture 1b* 1. Henry’s Law 2. Units of concentration 3. Colligative properties: Solutes and Solutions a) Freezing point depression *Week 1 Wednesday(b) b) Boiling point elevation c) Vapor pressure lowering d) Osmotic pressure Henry’s Law Concentration units Review of concentration units: Solubility of a gas in a liquid is proportional to the partial pressure of the gas on the liquid surface: 1 molar (M) = 1 mole solute/ liter soln sg = kHPg where sg= solubility of gas; Pg = partial P of gas 1 molal (m) = 1 mole solute/ kg solvent -3 -1 Example: O2 has kH = 1.26 x 10 M atm (at 25°C) 1 % (m/v) = (1 g solute/g soln )x 100% If the O2 mole fraction (XO2) in air is 0.21, a) what is P if the atmospheric pressure is 1.0 atm? (same as =g solute/100g soln) O2 6 PO2 = XO2Patm=0.21(1.0 atm) = 0.21 atm. 1 ppm = (1 g solute/g soln )x 10 6 b) what is the molar solubility of O2 in an aquarium at Patm =1.0 atm? (same as =g solute/10 g soln) -3 -1 -4 sO2 = kO2PO2 = 1.26 x 10 M atm (0.21 atm)=2.65 x 10 M (for dilute aqueous solutions: 1 ppm = 1mg solute/L soln) Mole fraction, X1 = #mol 1 / total moles = n1 /(Σni) Sample Problem Example Problem #2 What is the molality (m) of a 1.20 M NaCl solution whose density (d) is What is the ppm Ca2+ in a 2.50 L solution containing 1.20g CaCl ? 1.0200 g/mL? (FW of NaCl = 58.5 g/mol) 2 (MW’s: Ca=40.1 g/mol; CaCl = 111.1 g/mol) 2 Solution: Solution: recall that ppm ≈ 1 mg/L Recall that M = mol solute/L soln, m = mol solute/kg solvent Assume 1.0 L: therefore, #mol NaCl = 1.20, g Ca2+ = 1.20g CaCl (1mol)(1 mol Ca2+)(40.1g Ca2+) = 0.433 g Also: mass soln = 1.0200 g/mL(1000mL/L)(1kg/1000g) = 1.0200 kg 2 111.1g mol CaCl mol Ca2+ kg solvent = kg solution – kg solute = kg solution – kg NaCl 2 But kg NaCl = 1.20 mol (58.5 g/mol)(1 kg/1000 g) = 0.0702 kg ppm Ca2+ = 0.433g Ca2+ (1000 mg) = 173 mg Ca2+ = 173 ppm Ca2+ So, kg solvent = 1.0200 kg - 0.0702 kg = 0.9498 kg; 2.50L g L Hence, m = 1.20 mol/.9498 kg = 1.26 m 1 6/22/11 Colligative properties Boiling point elevation Colligative properties = physical properties of solutions ∆Tb = ikbmsolute which depend upon the concentration of solute Where ∆Tb = change in boiling point particles. k = molal boiling pt elev’n const. 4 main examples we are concerned with in this class: b m = molality of solute particles a) Boiling point elevation b) Freezing point depression i = Van’t Hoff factor (accounts for #ions) c) Vapor Pressure lowering So, to get boiling pt, Tb: Tb = Tb° + ∆Tb d) Osmotic pressure For water, kb = 0.512 K/m ; Tb° = 100.°C Checking your understanding: a) Is kb a constant of the solvent or the solute? kb is a constant of the solvent. b) What’s i =? for NaCl solution? i = 2 for NaCl solution because there are 2 moles of ions per mole of NaCl, namely 1 mole each of Na+ and Cl-. Freezing point depression Sample problems Solvent A normally freezes at 10.0°C. A 3.0 m solution of B (nonionic) in A freezes at 4.0°C. ∆Tf = ikfmsolute What is kf=? Where ∆Tf = change in freezing pt; -1 Answer: kf = ΔTf/m = (10.0-4.0)°C/3.0m = 2.0 °Cm i = van’t Hoff factor (remember?). kf = molal freezing pt depression constant What is the freezing point, Tf, of a 2.0 m solution of and, m = molality of solute particles a solute D (nonionic) in A? To get Tf : Tf = T°f - ∆Tf -1 Answer: ΔT = kfm = 2.0 °Cm (2.0m) = 4.0°C For water, kf = 1.853 K/m So T = 10°C - 4°C = 6°C f Sample probs continued… Vapor pressure lowering What is MW of solute D if a sol’n of 10.0g of D in 20 g of -1 solvent A has Tf = 6.0°C? (kf= 2.0 °Cm )? P1 = X1P1° (Raoult’s Law) Answer: first determine m: from ΔTf = kfm; m= ΔTf/kf -1 m = (10.0-6.0)°C /2.0°Cm = 2.0 m Where P = vapor pressure of a solvent (1) in a Since m =2.0 m = 2.0 mol D/kg A = 10.0g D/MW / 0.020kg A 1 Solve for MW: MW = (10.0g D)/ (2.0 mol D/kg A )(0.020kg A) liquid solution. MW = 250 g D/mol D X1 = mole fraction of that solvent in the solution P1° = vapor P of solvent when it is pure solvent. (N.b. for a solution, X1 < 1 , and so, P1 < P1° ) 2 6/22/11 Sample problem… Osmotic pressure P1 = X1P1° (Raoult’s Law) π = iMRT (across semiperm.memb.) Consider 2 liquids, A and B. The vapor pressures of Where π = osmotic pressure (π = pi = Greek p) pure A and pure B are 25.0 torr and 45.0 torr i = “ van’t Hoff factor”, a correction factor to respectively. What is the total vapor pressure account for actual # particles in solution inside an airtight chamber containing only 2 M = molarity of solute liquids, 1.5 moles of A and 2.5 moles of B? R = 0.0821 atmL/molK Answer: Ptotal = PA + PB = XAPA + XBPB T = temperature in K Where: XA = 1.5/(1.5+2.5)=0.375 and XB=2.5/(1.5+2.5)=0.625 Ptot = (0.375)(25.0 torr) + (0.625)(45.0 torr) =9.38 torr + 28.1 torr = 37.5 torr Osmotic pressure problem Red blood cells and π The equation is: π = i c R T where c = moles/L (molarity) Red blood cells (RBC’s) are “semipermeable bags”, which (can be derived from pV=nRT => p =(n/V)RT ) Example. A solution prepared by adding 50. g of solute to must maintain the same concentration within and without, or make 1.0 L solution at 300 K has π = .821 atm. What is the else! (i.e. solution surrounding it must be isotonic) MW of the solute (assuming it is a nonelectrolyte)? Water leaves cell: crenation g /MW ! = cRTi = solute solute RTi If solution is more concentrated (hypertonic) than the internal V solution concentration, what happens? g or, MW = solute RTi "Vsolution If solution is less concentrated (hypotonic) than the internal 50.g atm # L concentration, what happens? MW = (0.0821 )(300K)(1) (.821atm)(1.0L) mol # K Water enters cell:hemolysis 3 MW =1.50x10 g/mol Reverse osmosis: desalination If the applied pressure is Applied P=100 high enough (and the 1$/8,4 atm > π=25 atm membrane robust enough!), it is possible to reverse the 8,4,%0,)3$+1 Example: desalination plants - common in middle east, Florida,… Pure H2O Salt water 3 .