1. CH. 13 SOLUTIONS 13.1-.5 Equations Process at molecular level
Solubility- solute/solvent/solution
Factors affect solubility
Concentration Expressions
Colligative Properties
Henry’s Law
Sgas = KH * Pgas
Various concentration expressions
van’t Hoff factor
freezing pt depression
DT = Kf*mi
boiling pt elevation
DT = Kb*mi
Raoult’s Law
Psoln = (X*Po)solvent 1

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)
Solution Colloid
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 dissolve
H2O: universal solvent
Miscible: subst dissolve in each other
Electrolyte: 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
Depends on number of solute particles in solution.
Properties of dilute solns of nonvolatile solute in volatile solvent
1. Lowering vp
2. Elevate bp
3. Lower fp
4. 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

4. COLLIGATIVE PROPERTIES
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

5. Tb/f = kf*m*I FP Tf = kf*m*i BP Tb = kb*m*i
FREEZING POINT DEPRESSION
&
BOILING POINT ELEVATIONS
OF SOLUTIONS
Tb/f = kf*m*I
i: moles solute particles
m: molality
mols solute/ Kg solvent
FP Tf = kf*m*i
Tf FP depression
(FP pure solvent ) – (FP solvent in soln)
kf FP constant;
solvent specific
BP Tb = kb*m*i
Tb BP elevation
(BP solvent in soln) – (BP pure solvent)
kb BP constant;
solvent specific

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

7. Ex. Find BP & FP of the solvent ethanol
in sln when dissolve 18.0 g C6H12O6 in
200.0 g C2H5OH
Tb= kbmi kb = 1.22 oC/m BP: 78.4 oC
moles fructose = 18.0 g * (1 mol/180 g) = mol
Kg ethanol = g * (1 Kg/1000 g) = Kg
i = 1
m = 0.100/0.200 = m
Tb= (1.22 oC/m)*(0.500 m)*(1) = oC
BP = Pure + Tb = = 79.0 oC

8. Tf= kfmi kf = 1.99 oC/m FP: -114.6 oC
Now, find FP depression
Tf= kfmi kf = 1.99 oC/m FP: oC
Same calculations needed as for BP
moles fructose = 18.0 g * (1 mol/180 g) = mol
Kg ethanol = g * (1 Kg/1000 g) = Kg
i = 1
m = 0.100/0.200 = m
Tf= (1.99 oC/m)*(0.500 m)*(1) = oC
FP = Pure - Tf = = oC

9. Molar Mass from BP Data
Solution prepared by dissolving g glucose in g H2O. Solution
has a bp of oC Calculate the molecular wt of glucose.
Plan
glucose = 1 ion
DTb = Kb*m
find: DTb, msolute, nglucose
Kb: C-Kg/mol
find DTb
DTb = Tb,solution - Tb,solvent
= =
0.34oC
find molality solute
m = DTb/Kb
= (0.34)/(0.51) = mol/Kg
find quantity solute, mols
KgH2O = 150 g = Kg
nsolute = (0.67 mol/Kg)*( Kg) = mol
Analysis
0.10 mol of glucose weighs g, as 1 mol glucose (C6H12O6)
weighs 180 g.

10. Van’t Hoff Factor - “i” Plan ^convert mass% to m ^DTf = 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 = oC
mNaCl= mols solute/Kg solvent
mols NaCl = 1.00 g*(1mol/58.5 g) = mols
sovlent H2O = 99.0 g * (1 Kg/1000 g) = Kg
mNaCl= mols/0.099 Kg = m
DTf = (-0.593) = 0.593
i = DTf/(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

11. mH2SO4 = [0.750 g H2SO4/0.09925 Kg H2O]/(98.1 g) = 0.0770 m
Now, for you find m & van’t Hoff factor, i
0.750 mass % H2SO4, freezing pt = oC
DTf = Kf*m*i
mH2SO4 = [0.750 g H2SO4/ Kg H2O]/(98.1 g) = m
DTf = Tf – Ti = [0.00 – (-0.423)] = 0.423oC
i = DTf/Kf*m = (0.423oC)/[1.86oC/m * 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 2 ions?
Why isn’t i closer to 2?
Ion pairing; usually less in dilute solutions
Cl-
Cl-
Cu+
Cl-
Cu+
Cu+
Cl-
Cl-
Cl-
Cu+
Cu+
Cu+

12. 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
TREND
col: charge =, size , CD , DHhydra
row: charge , size , CD , DHhydra
DHlattice: E needed to sep ionic solid into gaseous ions
very +++ DH > 0
Ionic Solids + H2O ===>
we have, + DHlattice + DHhydra cation + DHhydra anion

13. 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
Strength of forces, solution formation, between:
solute-solute solute-solvent solvent-solvent
Na+I  solution forms  H2O-H2O

14. DHsoln: total DH when soln forms from solute/solvent
DHsoln = DHsolute + DHsolvent + DHmix
small large = -DHsoln
more positive DHsoln, solubility decr to 0
3 components - 2 break & 1 form
DH1: solute/solute DH2: solvent-solvent DH3: solute-solvent
DHsoln = DH1 + DH2 + DH3
(>0) + (>0) + (<0)
H < 0, exo, spont
H >>>>> 0, too endo,
no soln forms

15. ENTHALPY DIAGRAM Dissolve NaI(s) in H2O; exo- Na+1 (g); I-1(g)
DHsolute
DHlattice
DHhydra
Enthalpy, H
NaI(s)
Hinitial
From this?
1st Basic Principle
Na+1 (aq); I-1(aq)
Hfinal
DHsoln = “-”
Spont; process occurs
if E of sys decr

16. 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 > 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

17. 1) gas & gas mixture
2) NaCl bonds
1 - gases spont mix & expand; more space
2 - strong bond holds ions together, not spont dissolve in gasoline

18. SOLUBILITY IN EQUILIBRIUM
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

19. FACTORS dissolve depends on nature solute/solvent; T; P(gases)
Solute-Solvent H2O + CH3OH ----> P + P ----> mix
miscible: subst easily mix together
immisicble: sust not dissolve together
C2H6 + H2O ----> NP + P ----> X
Greater solute-solvent attraction, greater solubility
Recall “like dissolves like”
solvent + solute > solution
P P > soln dipole + dipole interaction
NP + NP > soln dispersion forces

20. 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 539
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

21. PRESSURE EFFECT d2 = 0.5(d1) then P2 = ? P1 d1 d2 P1 P2
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

22. Henry’s Law gas solubility
Sgas = KH * Pgas
M = const * atm
KH; Henry’s Law const; values for given given T
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

23. Ex 2: a soda bottle at 25oC contains CO2 gas at 5
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 = 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/L
opened: 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 D in concen, why pop goes “flat”
Practice Problem
If CO2 partial pressure is 3.0*10-4 atm, what is the [CO2] at 250C?

24. Temp vs Solubility
solid more 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

25. TEMPERATURE EFFECT ON SOLUBILITY
OF VARIOUS SUBSTANCES

26. CONCENTRATION EXPRESSIONS
total soln = solute + solvent
Mass % = [mass subst in soln/total mass soln]* (%)
ppm: * 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 DT
molality (m) = mols solute/Kg solvent (mols/Kg)
mass not affected by DT
Conversions
convert mol to mass: use molar mass
convert mass to vol: use density

27. 23.6 % HF by mass, states: 23.6 g HF/100 g soln
0.050 ppm states: g solute in million (106) g soln, or
0.050 mg/Kg also ≈ mg/L
50 ppb states: g solute in billion (109) g soln
50 g/Kg => 50 g/L

28. MASS % MOLARITY -- MOLALITY PRACTICE PROBLEM
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 g H2O?
MOLARITY -- MOLALITY
PP3) 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

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

30. MOLARITY -- MOLALITY mols = (8.98 g)/(68.9 g/mol) = 0.130 mols
M = mol/.505 L = M
mols = (164 g)/(36.5 g/mol) = 4.49 mol
m = (4.49 mol)/(0.753 Kg) = 5.96 m

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

32. 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

33. 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 mol Cl-1

34. A solution prepared, dissolve 17.9 g NaCl in 643.5 cm3 H2O at 25oC.
DH2O = & vp = torr
plan: determine X fraction for H2O
mols NaCl
g H2O > mols H2O
XH2O
Psoln
58.5 g/mol
(17.9 g/58.5)*2 ions = mol
643.5*0.997 = g
641.6 g/18.0 = 35.6 mol
= (molH2O)/(molH2O+molNaCl)
=35.6/( ) = 0.983
= (0.983)*(23.76 torr)
= torr
Raoult’s Law
Psoln = (X*Po)H2O

35. Solution prepared by mixing 35. 0 g Na2SO4(s), [142
Solution prepared by mixing 35.0 g Na2SO4(s), [142.1mw], w/ 175 g 25oC.
Find VP VPH2O = atm
Find XH2O
nH2O = 175 g/18.0 = 9.72 mol
nNa2SO4 = 35.0 g/142.1 = mol
Na2SO4----> 2Na+ + SO4-2 : 3 ions
nNa2SO4 = 3(0.246) = mol
XH2O = 9.72/( ) = 0.929
Psoln = XH2OPoH2O
= (0.929)*(0.031 atm) = atm

36. PRACTICE PROBLEM
A nonelectrolyte solution is prepared by dissolving 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, nonelectrolyte
DTb = 0.357oC
Find:
Kb CCl4: 5.02oC/m
Kg solvent = Kg CCl4
Calculation:
DTb = Kb*m*i 
m = DTb/(Kb*i )
m = (0.357oC)/(5.02oC/m) = m
So, soln contains:
mol solute/0.040 Kg solvent
= 6.25 g/Kg
means:
mol = 6.25 g
then 1 mol:
(6.25 g)/( mol) = 87.9 g
or M = 87.9 amu

37. 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

38. 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 + C8H NP + NP
Dispersion force attraction;
Sub bet the 2 molecules

39. GAS - SOLID SOLUTIONS 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

40. 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

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