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
Pre
ssur
e -
atm
Temperature - oC
1 atm
VP pure solvent
VP solution
Tb
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
-2
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+
Cu+Cu+
Cu+Cl-
Cl-
Cl-
Cl-
Cl-
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-
Ent
halp
y, H
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
22
100(g)soln mass
(g) solute of mass % mass PP1) 20.0% 100
g 150
g 30.0 % mass
ppm 10*2 10g 150
g 30.0 ppm 56 ppb 10*2 10
g 150
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
Pre
ssur
e -
atm
Temperature - oC
1 atm
VP pure solvent
VP solution
Tb
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
solv
soln: 3/4 soln is solvent particles, Xsolv = 0.75, then Psoln is 0.75Po
solv
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
XH2O
Psoln
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
42
222
mix gas
CH N
CO F O
2N O2 F
2 CO2 CH
4
Pt acts as semi-porous filter
Melt solids ----- mix (metallic bonding) ---- freeze
SOLUTIONS SOLID-SOLID
Alloys: brass Zn + Cu bronze Sn + Cu
