CH. 13 PROPERTIES OF MIXTURES

transcript

CH. 13 SOLUTIONS 13.1-.5

Process at molecular levelSolubility- solute/solvent/solutionFactors affect solubilityConcentration ExpressionsColligative Properties

EquationsEquationsHenry’s Law Sgas = KH * Pgas

Various concentration expressions

van’t Hoff factorfreezing pt depressionT = Kf*mi boiling pt elevationT = Kb*mi

Raoult’s Law Psoln = (X*Po)solvent

RECALLMixture: composition varied, retains properties of individual partsSolution: homogeneous mixture, 1 phase, uniform propertiesColloid: heterogeneous mixture, 2+ phases (not easily seen)

Solution Colloid particles: individual atoms, lrg molecules or sm molecules ions, sm molecules not separate out

Solute: subst being dissolvedSolvent: subst doing the dissolveH2O: universal solvent

“LIKE DISSOLVES LIKE”subst w/ similar inter- forces will dissolve in each otherinteraction bet solute & solvent

Miscible: subst dissolve in each other

Electrolyte: subst dissoc into ion, conducts electrical current all ionic cmpds, strong acidsNonelectroylte: no dissoc or very little %, not conduct current weak acids, covalent molecules

COLLIGATIVE PROPERTIES OF SOLUTIONS

Depends on number of solute particles in solution.Properties of dilute solns of nonvolatile solute in volatile solvent

1. Lowering vp2. Elevate bp 3. Lower fp4. Osmotic P

Calculating: 1) VP of sovent 2) FP depression & BP elevation of solvent 3) molar mass, M, of solute from FP data 4) value of mole number, i, of solute from FP data

Boiling Point Elevation, Freezing Point Depression,Vapor Pressure Reduction (Raoult’s)

Depends on [solute]Examine [solute] affects property of pure solvent

Soln vs Pure Solvent BP FP VP incr decr decr

COLLIGATIVE PROPERTIES

FREEZING POINT DEPRESSION&

BOILING POINT ELEVATIONSOF SOLUTIONS

Tb/f = kf*m*I i: moles solute particles m: molality mols solute/ Kg solvent

FP Tf = kf*m*iTf FP depression (FP pure solvent ) – (FP solvent in soln)

kf FP constant; solvent specific

BP Tb = kb*m*iTb BP elevation (BP solvent in soln) – (BP pure solvent)

kb BP constant; solvent specific

PHASE DIAGRAM PURE H2O

Temperature - oC

VP pure solvent

VP solution

bp pure H2O bp solution

Tb= kbmi kb = 1.22 oC/m BP: 78.4 oC

Ex. Find BP & FP of the solvent ethanol in sln when dissolve 18.0 g C6H12O6 in 200.0 g C2H5OH

moles fructose = 18.0 g * (1 mol/180 g) = 0.100 mol

Kg ethanol = 200.0 g * (1 Kg/1000 g) = 0.200 Kg i = 1

m = 0.100/0.200 = 0.500 m

Tb= (1.22 oC/m)*(0.500 m)*(1) = 0.610 oC BP = Pure + Tb = 78.4 + 0.61 = 79.0 oC

Now, find FP depressionTf= kfmi kf = 1.99 oC/m FP: -114.6 oC

Same calculations needed as for BPmoles fructose = 18.0 g * (1 mol/180 g) = 0.100 mol

Kg ethanol = 200.0 g * (1 Kg/1000 g) = 0.200 Kg i = 1

m = 0.100/0.200 = 0.500 m

Tf= (1.99 oC/m)*(0.500 m)*(1) = 0.995 oC FP = Pure - Tf = -114.6 - 0.995 = -115.6 oC

Plan glucose = 1 ion DTb = Kb*mfind: DTb, msolute, nglucose

Kb: 0.51 0C-Kg/mol find molality solutem = DTb/Kb

= (0.34)/(0.51) = 0.67 mol/Kg

Solution prepared by dissolving 18.00 g glucose in 150.0 g H2O. Solutionhas a bp of 100.34oC Calculate the molecular wt of glucose.

find DTb

DTb = Tb,solution - Tb,solvent

= 100.34 - 100.00 = 0.34oC

find quantity solute, molsKgH2O = 150 g = 0.1500 Kgnsolute = (0.67 mol/Kg)*(0.1500 Kg) = 0.10 mol

Analysis0.10 mol of glucose weighs 18.00 g, as 1 mol glucose (C6H12O6)weighs 180 g.

Molar Mass from BP Data

Van’t Hoff Factor - “i”

Find m & van’t Hoff factor, i1.00 mass % NaCl, freezing pt = -0.593oC

Plan ^convert mass% to m ^Tf = Kf*m*i ^assume 100 g sample, so, 1.00 g NaCl & 99.0 g H2O ^Kf H2O = 1.86oC/m

mNaCl= mols solute/Kg solvent

mols NaCl = 1.00 g*(1mol/58.5 g) = 0.017 molssovlent H2O = 99.0 g * (1 Kg/1000 g) = 0.099 Kg

mNaCl= 0.017 mols/0.099 Kg = 0.1717 m

Tf = 0.00 - (-0.593) = 0.593iTf/(Kf*m) = (0.593)/(1.86*0.1717) = 1.86

i is close to 2,as NaCl dissoc. into 2 ions Na+1 & Cl-1

Now, for you find m & van’t Hoff factor, i0.750 mass % H2SO4, freezing pt = -0.423oC

mH2SO4 = [0.750 g H2SO4/0.09925 Kg H2O]/(98.1 g) = 0.0770 m

Tf = Tf – Ti = [0.00 – (-0.423)] = 0.423oC i Tf/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

Tf = Kf*m*i

What can be said if i = 1.67 for CuCl, which would form 2 ions?Why isn’t i closer to 2? Ion pairing; usually less in dilute solutions

Cu+Cu+

Cu+Cl-

IONIC SOLIDS Solute surrounded by solvent, termed as Solvation process in water, Hyrdation (gaining of H2O)Hsoln = Hsolute + Hsolvent + Hmix

Hlattice: E needed to sep ionic solid into gaseous ions very +++ H > 0Ionic Solids + H2O ===> we have, + Hlattice + Hhydra cation + Hhydra anion

Hhydra based on charge density higher CD --> more neg Hhydra

incr charge, stronger attraction to H2O

Hhydra

3 process occur3 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

+DH to separate particles-DH to mix & attract particles

Strength of forces, solution formation, between:solute-solute solute-solvent solvent-solvent Na+I- solution forms H2O-H2O

SOLUTION PROCESS

Hsoln: total H when soln forms from solute/solventHsoln = Hsolute + Hsolvent + Hmix

more positive Hsoln, solubility decr to 0

small + -large = -Hsoln

3 components - 2 break & 1 form

H1: solute/solute H2: solvent-solvent H3: solute-solventHsoln = H1 + H2 + H3

(>0) + (>0) + (<0)

H < 0, exo, spontH >>>>> 0, too endo, no soln forms

ENTHALPY DIAGRAMDissolve NaI(s) in H2O; exo-

NaI(s)

Hsolute

Hlattice

Na+1 (g); I-1(g)

Hhydra

Na+1 (aq); I-1(aq)Hfinal

Hinitial

Hsoln = “-”

From this?1st Basic Principle

Spont; process occurs if E of sys decr

measure of disorder in a sysvarious ways sys distr Erelates to freedom of particle movement

Ssoln > Ssolute or Ssolvent

more interactions occur in soln, so moreways to distr E & more freedom of movement

Solid ------------ Liquid ----------- Gas Ssolid < Sliq Sliq < Sgas

S > 0 S > 0

S ENTROPY

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 H & high S

2nd Basic Principle Spont process at const T occur as entropy (randomness) incr

1) gas & gas mixture2) NaCl bonds

1 - gases spont mix & expand; more space2 - strong bond holds ions together, not spont dissolve in gasoline

Solubility: max amt solute that dissolves in given amt solvent @ specific Temp

Saturated: rate solute dissolves = rate solute undissolves; soln in equilibrium

Unsaturated: limit to which soln has dissolved all solute possible; < saturated amt

Supersaturated: soln able to dissolve excess solute above a T & stays dissolved

SOLUBILITY IN EQUILIBRIUM

Solute-Solvent H2O + CH3OH ----> P + P ----> mixmiscible: subst easily mix together

immisicble: sust not dissolve together C2H6 + H2O ----> NP + P ----> X

FACTORSdissolve depends on nature solute/solvent; T; P(gases)

Greater solute-solvent attraction, greater solubility

Recall “like dissolves like” solvent + solute ------> solution P + P -------> soln dipole + dipole interaction NP + NP ------> soln dispersion forces

Trend in organic series: solubility in H2O & hexane

molecule has +/- dipoles and NP end, as NP end incr in length, NP influences incrT 13.2 pg 539

H2O - polar C6H14 - NP CH3OH Y less CH3CH2OH Y YCH3(CH2)2OH Y YCH3(CH2)3OH less 0.11 YCH3(CH2)4OH less 0.030 YCH3(CH2)5OH less0.0058 Y

Which is more soluble. Why CH3(CH2)3OH or HO(CH2)5OH in H2O CHCl3 or CCl4 in H2O

more H-bonding capability polar - polar match

How to dissolve CCl4? use NP solvent; hexane

PRESSURE EFFECT

d2 = 0.5(d1) then P2 = ? P1

Gas molecules enter soln incr as P is incr

pressure little 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

KH; Henry’s Law const; values for given gas @ given T

Henry’s Law gas solubility Sgas = KH * Pgas

M = const * atm

Ex 1: 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 2: 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.KH,CO2 = 0.032 mol/L-atm (from table)

Pressure Effect relation gas P & concen dissolved gas Sgas = KH* Pgas

unopened: P = 5.0 atm CCO2 = (5.0 atm)*(0.032 mol/L-atm) = 0.16 mol/Lopened: P = 4.0*10-4 atm CCO2=(4.0*10-4 atm)*(0.032 mol/L-atm) = 1.3*10-5 mol/L

Notice, large in concen, why pop goes “flat”

Practice ProblemIf CO2 partial pressure is 3.0*10-4 atm, what is the [CO2] at 250C?

Temp vs Solubility

solidsolid more soluble @ higher T more soluble @ higher T heat absorbed to form soln solute + solvent + heat <---> sat soln DHsoln > 0 incr T, incr rate dir -----> gas DHsolute = 0 since gas particles already sep DHhydra < 0 heat released for gases in H2o solute + H2O <----> sat soln + heat DHsoln < 0 incr T, decr gas solubility rate can lead to thermal pollution

TEMPERATURE EFFECT ON SOLUBILITY OF VARIOUS SUBSTANCES

total soln = solute + solvent

Mass % = [mass subst in soln/total mass soln]*100 (%) ppm: * 106 ppb: * 109

Mole Fraction (X) = mols solute/mols soln (use in Raoult’s Law) (mol % = X * 100)

Molarity (M) = mols solute/L soln (mols/L) vol affected by T

molality (m) = mols solute/Kg solvent (mols/Kg) mass not affected by T

CONCENTRATION EXPRESSIONS

Conversionsconvert mol to mass: use molar massconvert mass to vol: use density

23.6 % HF by mass, states: 23.6 g HF/100 g soln

0.050 ppm states: 0.050 g solute in million (106) g soln, or 0.050 mg/Kg also ≈ 0.050 mg/L

50 ppb states: g solute in billion (109) g soln 50 g/Kg => 50 g/L

PP1) Patient is given 30.0 g glucose in 150 mL solution, what is the mass %? ppm? ppb?

PP2) Patient is given 30.0% glucose solution, what is the mass glucose in 250.0 g H2O?

MASS % PRACTICE PROBLEM

MOLARITY -- MOLALITYPP3) Find M of a soln w/ 8.98 g lithium nitrate in 505 mL

PP4) Find m of soln w/ 164 g HCl in 753 mL H2O

MASS %

30.0% is 30.0 g solute in 100.0 g solution.So, 30 g glucose for 70.0 g H2O

answer unit egiven valu PP2) facorunit

glucose g 107 OH g 70.0

glucose g 30.0 OH g 250.0

100(g)soln mass

(g) solute of mass % mass PP1) 20.0% 100

g 30.0 % mass

ppm 10*2 10g 150

g 30.0 ppm 56 ppb 10*2 10

g 30.0 ppb 89

MOLARITY -- MOLALITY

mols = (8.98 g)/(68.9 g/mol) = 0.130 mols M = 0.130 mol/.505 L = 0.258 M

mols = (164 g)/(36.5 g/mol) = 4.49 mol m = (4.49 mol)/(0.753 Kg) = 5.96 m

M (L) vol mols g PP3) wt.form.

m (Kg)solvent mols g PP4) wt.form.

PHASE DIAGRAM PURE H2O

Temperature - oC

VP pure solvent

VP solution

bp pure H2O bp solution

fp solution fp pure H2OTf

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 0.5Po

soln: 3/4 soln is solvent particles, Xsolv = 0.75, then Psoln is 0.75Po

Idea behind this: nonvolatile solute just dilutes the solvent

Remember!!! Ideal Gas obeys ideal-gas eqn and ideal soln obeys Raoult’s LawReal soln approx ideal when……. low [ ], similar molecular sizes, similar inter- attractions

VP of soln containing nonvolatile solutes given by: RAOULT’S LAW

Example:1 mol glucose, result => lower VP same as 0.5 mol NaCl

Reasoning behind this ???> both nonvolatile> form 1 mol of particles

1 mol C6H12O6

NaCl dissolves into: 0.5 mol Na+1 + 0.5 mol Cl-1

Raoult’s Law Psoln = (X*Po)H2O

plan: determine X fraction for H2Omols NaCl

g H2O -------------------> mols 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

A solution prepared, dissolve 17.9 g NaCl in 643.5 cm3 H2O at 25oC.DH2O = 0.9971 & vp = 23.76 torr

Solution prepared by mixing 35.0 g Na2SO4(s), [142.1mw], 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----> 2Na+ + 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

PRACTICE PROBLEMA nonelectrolyte solution is prepared by dissolving 0.250 g in 40.0 g of CCl4.The normal soln BP is increased 0.357oC. Find the molecular wt. of the solute.

Know:i = 1, nonelectrolyteTb = 0.357oC

Find:Kb CCl4: 5.02oC/mKg solvent = 0.040 Kg CCl4

Calculation:Tb = Kb*m*i m = Tb/(Kb*i )

m = (0.357oC)/(5.02oC/m) = 0.0711 m

So, soln contains: 0.0711 mol solute/0.040 Kg solvent = 6.25 g/Kgmeans:0.0711 mol = 6.25 g

then 1 mol: (6.25 g)/(0.0711 mol) = 87.9 g or M = 87.9 amu

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

SOLUTION TYPES

Forces in Solution

Li+1+ Cl-1 + H2OIons 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

LIQUID SOLUTIONS & POLARITY

SOLUTIONS LIQLIQ

LIQSOLID

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

NP Gases: low bp, weak inter- attraction, no solubility in H2O as weak forces, incr solubility - bp incr

All gases soluble w/ other gases,Air: 18 gases in varied % amts,optimize yield by vary T & P

SOLUTIONS LIQ-GAS

GAS - SOLID SOLUTIONS

SOLUTIONS GAS-GAS

Gas dissolves by filling void spaces bet particles in solid

N2 sm molecule to pass thru void space bet Pt atoms while other atoms too lrg to passremain on Pt surface

SOLUTIONS SOLID-GAS

mix gas

CO F O

2N O2 F

2 CO2 CH

Pt acts as semi-porous filter

Melt solids ----- mix (metallic bonding) ---- freeze

SOLUTIONS SOLID-SOLID

Alloys: brass Zn + Cu bronze Sn + Cu

CH. 13 PROPERTIES OF MIXTURES

Documents