1 / 28

CH. 13 PROPERTIES OF MIXTURES

CH. 13 PROPERTIES OF MIXTURES. Relate actions of inter- forces in solns. Equations. Henry’s Law S gas = K H * P gas. Various concentration expressions. van’t Hoff factor freezing pt depression D T = K f * m boiling pt elevation

sitara
Télécharger la présentation

CH. 13 PROPERTIES OF MIXTURES

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CH. 13 PROPERTIES OF MIXTURES Relate actions of inter- forces in solns Equations Henry’s Law Sgas = KH * Pgas Various concentration expressions van’t Hoff factor freezing pt depression DT = Kf*m boiling pt elevation DT = Kb*m Raoult’s Law Psoln = (X*Po)solvent

  2. RECALL Mixture: composition varied, retains properties of individual parts Solution: homogeneous mixture, 1 phase, uniform properties Colloid: heterogeneous mixture, 2+ phases (not easily seen) SolutionColloid particles: individual atoms, lrg molecules or sm molecules ions, sm molecules not separate out “LIKE DISSOLVES LIKE” subst w/ similar inter- forces will dissolve in each other interaction bet solute & solvent Solute: subst being dissolved Solvent: subst doing the dissolvine H2O: universal solvent Miscible: subst dissolve in each other Electroylte: subst dissoc into ion, conducts electrical current all ionic cmpds, strong acids Nonelectroylte: no dissoc or very little %, not conduct current weak acids, covalent molecules

  3. COLLIGATIVE PROPERTIES OF SOLUTIONS Look at: Colligative Properties properties of dilute solns of nonvolatile solute in volatile solvent 1. Lowering vp 2. Elevate bp 3. Lower fp 4. Osmotic P

  4. CONCENTRATION EXPRESSIONS Molarity (M) mols solute/L soln (mols/L) affected by DT molality (m) mols solute/Kg solvent (mols/Kg) mass not D by T Parts by mass^ (w/w%) (mass solute/mass soln)*100 Part by vol^ (v/v%) (vol solute/vol soln)*100 (w/v%) (mass solute/vol soln)*100 Mole Fraction (X) [mols solute/(mols solute + mols solvent)] mol % = X * 100 ^ ppm(v) * by 106 (not 100) ppb(v) * by 109 (not 100) Conversions convert mol to mass: use molar mass convert mass to vol: use density

  5. 1 atm VP pure solvent Pressure - atm VP solution bp pure H2O bp solution Temperature - oC PHASE DIAGRAM PURE H2O DTb

  6. Solution prepared by dissolving 18.00 g glucose in 150.0 g H2O. Solution has a bp of 100.34oC Calculate the molecular wt of glucose. find DT DT = Tb,solution - Tb,solvent = 100.34 - 100.00 = 0.34oC Plan glucose = 1 ion DT = Kb*m find: DT, msolute, nglucose Kb: 0.51 0C-Kg/mol find molality solute m = DT/Kb = (0.34)/(0.51) = 0.67 mol/Kg find quantity solute, mols KgH2O = 150 g = 0.1500 Kg nsolute = (0.67 mol/Kg)*(0.1500 Kg) = 0.10 mol Analysis 0.10 mol of glucose weighs 18.00 g, as 1 mol glucose (C6H12O6) weighs 180 g. So, 10 * 18.00 g = 180 g

  7. 23.6 % HF by mass, states: 23.6 g HF/100 g soln 0.050 ppmstates: 0.050 g solute in million (106) g soln, or 0.050 mg/Kg also ≈ 0.050 mg/L 50 ppbstates: 50 g solute in billion (109) g soln 50 g/Kg => 50 g/L

  8. 1 atm VP pure solvent Pressure - atm VP solution bp pure H2O bp solution Temperature - oC fp solution fp pure H2O PHASE DIAGRAM PURE H2O DTf DTb

  9. Van’t Hoff Factor Plan ^convert mass% to m ^DT = Kf*m*i ^assume 100 g sample, so, 1.00 g NaCl & 99.0 g H2O ^Kf H2O = 1.86oC/m Find m & van’t Hoff factor, i 1.00 mass % NaCl, freezing pt = -0.593oC mNaCl= mols solute/Kg solvent mols NaCl = 1.00 g*(1mol/58.5 g) = 0.017 mols sovlent H2O = 99.0 g * (1 Kg/1000 g) = 0.099 Kg mNaCl= 0.017 mols/0.099 Kg = 0.1717 m iis close to 2, as NaCl dissoc. into 2 ions Na+1 & Cl-1 DT = 0.00 - (-0.593) = 0.593 i = DT/(Kf*m) = (0.593)/(1.86*0.1717) = 1.86

  10. Now, for you find m & van’t Hoff factor, i 0.750 mass % H2SO4, freezing pt = -0.423oC DT = Kf*m*i mH2SO4 = [0.750 g H2SO4/0.09925 Kg H2O]/(98.1 g) = 0.0770 m DTf = Tf – Ti = [0.00 – (-0.423)] = 0.423oC i = DTf/Kf*m = (0.423oC)/[1.86oC/m * 0.0770 m] = 2.95 i = 3, so 3 ions in solution, sulfuric acid is a strong acid so will dissociate completely into ions; 2H+1 & SO4-2 What can be said if i = 1.67 for CuCl, which would form 2ions? Why isn’t i closer to 2? Ion pairing; usually less in dilute solutions Cl- Cl- Cu+ Cu+ Cl- Cu+ Cl- Cl- Cl- Cu+ Cu+ Cu+

  11. SOLUTION TYPES Forces in Solution Ion-Dipole Hydration shell: ion surrounded by H2O molecules; attraction of H2O H-Bonding imprt in aq solns; prime reason for solubility in H2O; factor of solubility for many biological & organic subst Dipole-Dipole factor for solubility of polar-polar molecules Dispersion factor in NP-NP subst

  12. LIQUID SOLUTIONS & POLARITY Li+1+ Cl-1 + H2O Ions of solid attracted to dipole of H2O; attraction as strong as ion-attraction; H2O “substitutes” bet ions CH4 + H2O NP + Polar NP attraction too weak for H2O sub H-bonding bet H2O molecules too strong CH3COOH + H2O P+P (H-bonding) H-bonding attraction forces similiar H2O sub C6H14 + C8H18 NP + NP Dispersion force attraction; Sub bet the 2 molecules

  13. Trend in organic series: solubility in H2O & hexane molecule has +/- dipoles and NP end, as NP end incr in length, NP influences incr T 13.2 pg 492 H2O - polar C6H14 - NP CH3OH Y less CH3CH2OH Y Y CH3(CH2)2OH Y Y CH3(CH2)3OH less Y CH3(CH2)4OH less Y CH3(CH2)5OH less Y Which is more soluble. Why CH3(CH2)3OH or HO(CH2)5OH in H2O CHCl3 or CCl4 in H2O polar - polar match more H-bonding capability How to dissolve CCl4? use NP solvent; hexane

  14. NP Gases: low bp, weak inter- attraction, no solubility in H2O as weak forces, incr solubility - bp incr GAS - SOLID SOLUTIONS All gases soluble w/ other gases, Air: 18 gases in varied % amts, optimize yield by vary T & P

  15. Gas dissolves by filling void spaces bet particles in solid O2 F2 CO2 CH4 Pt acts as semi-porous filter N2 sm molecule to pass thru void space bet Pt atoms while other atoms too lrg to pass remain on Pt surface

  16. Alloys: brass Zn + Cu bronze Sn + Cu Melt solids ----- mix (metallic bonding) ---- freeze

  17. Henry’s Law: Sgas = KH * Pgas gas solubility KH; Henry’s Law const; value for given gas @ given T Ex: 78% of air is nitrogen. What is the solubility in H2O @ 250C & 1 atm. KH = 7*10-4 mol/L-atm SN2 = (7*10-4 mol/L-atm) * (1 atm) * (0.78) = 5.5*10-4 mol/L Ex: a soda bottle at 25oC contains CO2 gas at 5.0 atm over the liquid. Assume partial PCO2 is 4.0*10-4 atm. Calculate [CO2] before & after bottle opened. Pressure Effect relation gas P & concen dissolved gas Sgas = KH* Pgas unopened: P = 5.0 atm SCO2 = (5.0 atm)*(0.032 mol/L-atm) = 0.16 mol/L opened: P = 4.0*10-4 atm SCO2=(4.0*10-4 atm)*(0.032 mol/L-atm) = 1.3*10-5 mol/L Notice, large D in concen, why pop goes “flat”

  18. MOLARITY -- MOLALITY Find M of a soln w/ 8.70 g lithium nitrate in 505 mL. mols = (8.98 g)/(68.9 g/mol) = 0.130 mols M = 0.130 mol/.505 L = 0.258 M Find m of soln w/ 164 g HCl in 753 mL H2O. mols = (164 g)/(36.5 g/mol) = 4.49 mol m = (4.49 mol)/(0.753 Kg) = 5.96 m

  19. (w/w)% (v/v)% (w/v)% Patient is given 30.0 g glucose in 150 mL solution, what is the w/v%? Patient is given 30.0% glucose solution, what is the mass glucose in 250.0 g H2O? 30.0% is 30.0 g solute in 100.0 g solution. So, 30 g glucose for 70.0 g H2O

  20. SOLUTION PROCESS +DH to separate particles -DH to mix & attract particles 3 process occur 1: solute particles sep apart solute + heat ----> separate DH > 0 2: some solvent particles sep solvent + heat ----> separate DH > 0 gains room for solute particles 3: solute/solvent particles mix solute + solvent -----> soln + heat DH < 0 DHsoln: total DH when soln forms from solute/solvent DHsoln = DHsolute + DHsolvent + DHmix small + -large = -DHsoln more positive DHsoln, solubility decr to 0 Strength of forces, solution formation, between: solute-solute solute-solvent solvent-solvent Na+I- solution forms  H2O-H2O

  21. ENTHALPY DIAGRAM DissolveNaI(s) in H2O; exo- Na+1 (g); I-1(g) DHsolute DHlattice DHhydra Enthalpy, H Hinitial NaI(s) Na+1 (aq); I-1(aq) Hfinal Spont; process occurs if E of sys decr DHsoln = “-”

  22. TREND col: charg =, size , CD , DHhydra row: charge , size , CD , DHhydra IONIC SOLIDS Solute surrounded by solvent, termed as Solvation process in water, Hyrdation (gaining of H2O) DHsoln = DHsolute + DHsolvent + DHmix DHhydra DHhydra based on charge density higher CD --> more neg DHhydra incr charge, stronger attraction to H2O DHlattice: E needed to sep ionic solid into gaseous ions very +++ DH > 0 Ionic Solids + H2O ===> we have, + DHlattice + DHhysra cation + DHhydra anion

  23. measure of disorder in a sys various ways sys distr E relates to freedom of particle movement Solid ------------ Liquid ----------- Gas Ssolid < Sliq Sliq < Sgas DS > 0 DS > 0 Ssoln > Ssolute or Ssolvent more interactions occur in soln, so more ways to distr E & more freedom of movement Natural Tendency: is to form soln Manfacturing facilities need larg amts E to produce pure subst, as reverse natural way (spontaneous) sys tends toward: low DH & high DS 2nd Basic Principle Spont process at const T occur as entropy (randomness) incr

  24. SOLUBILITY IN EQUILIBRIUM Solubility: max amt solute that dissolves in given amt solvent @ specific Temp Saturated: rate solute dissolves = rate solute undissolves Unsaturated: limit to which soln has dissolved all solute possible Supersaturated: soln able to dissolve excess solute above a T & stays dissolved

  25. Temp vs Solubility solid more soluble @ higher T heat absorbed to form soln solute + solvent + heat <---> sat soln DHsoln > 0 incr T, incr rate dir -----> gasDHsolute = 0 since gas particles already sep DHhydra < 0 heat released for gases in H2o solute + H2O <----> sat soln + heatDHsoln < 0 incr R, decr gas solubility rate can lead to thermal pollution pressurelittle effect on solids or liq but MAJOR effect on gases. @ equilibrium; # gas particles leave soln = # gas particles reenter incr P (decr V) then incr # gas particles collide w/ surface, so # particles entry > # particles leave

  26. VAPOR PRESSURE OF SOLNS Raoult’s Law Psoln = (X*Po)solvent Note relation: soln: 50/50% solute/solvet molecules, Xsolv = 0.5, then Psoln is 0.5Posolv soln: 3/4 soln is solvent particles, Xsolv = 0.75, then Psoln is 0.75Posolv Idea behind this: nonvolatile solute just dilutes the solvent Remember!!! Ideal Gas obeys ideal-gas eqn and ideal soln obeys Raoult’s Law Real soln approx ideal when……. low [ ], similar molecular sizes, similar inter- attractions A solution prepared, dissolve 35.8 g NaClin 643.5 cm3 H2O at 25oC. DH2O = 0.9971 & vp = 23.76 torr

  27. plan: determine X fraction for H2O mols NaCl g H2O -------------------> mols H2O XH2O Psoln Raoult’s Law Psoln = (X*Po)H2O 58.5 g/mol (17.9 g/58.5)*2 ions = 0.612 mol 643.5*0.997 = 641.6 g 641.6 g/18.0 = 35.6 mol = (molH2O)/(molH2O+molNaCl) =35.6/(35.6+0.612) = 0.983 = (0.983)*(23.76 torr) = 23.36 torr

  28. Solution prepared by mixing 35.0 g Na2SO4(s), [142.1], w/ 175 g H2O @ 25oC. Find vp. vpH2O = 0.031 atm Find XH2O nH2O = 175 g/18.0= 9.72 mol nNa2SO4 = 35.0 g/142.1= 0.246 mol Na2SO4----> 2 Na+ + SO4-2 : 3 ions nNa2SO4 = 3(0.246) = 0.783 mol XH2O = 9.72/(9.72+0.783) = 0.929 Psoln = XH2OPoH2O = 0.929*0.031 atm = 0.0288 atm

More Related