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First Mid-quarter Examination

MQ-1 on Monday, Jan. 29

at 6:30 pm

Covering Chapters 10, 11, and 13

room TBA

For students with a scheduled-class conflict

with the first Mid-Quarter Exam:

(1) Send me an email today with a copy of your class schedule, and

(2) Let me know when you want to take themakeup exam:

a) during the last week of classes.b) early (5:00-6:18 pm) on Mon, Jan 29.c) late (8:00-9:18 pm) on Mon, Jan 29.

MQ-1 on Monday, Jan. 29at 6:30 pm

Covering Chapters 10, 11, and 13

Help session on Saturday, Jan 27at 4:00 – 6:00 pm

in MP1000

Week 4 Sections 13.1- 13.6—Properties of Solutions

13.1 The Solution ProcessEnergy Changes, Entropy, Rxns

13.2 Saturated Solutions and Solubility13.3 Factors Affecting Solubility

Solute-Solvent InteractionsPressure EffectsTemperature Effects

13.4 Ways of Expressing ConcentrationMass Percentage, ppm, and ppbMole Fraction, Molarity, and MolalityConversion of Concentration Units

13.5 Colligative Properties13.6 Colloids

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endothermic

endothermic

exothermic

Figure 13.3

e.g. MgSO4 e.g. NH4NO3

Figure 13.4

Realize there is an inherent tendency for the two isolated materials to form solution,regardless of the energetics!!! This represents an “entropy” factor.

Factors that FAVOR solubility:

1. Strong solute-solvent interactions

2. Weak solute-solute interactions

3. Weak solvent-solvent interactions

More often we’ll settle for the solute-solvent interactionsbeing similar to the solute-solute and solvent-solventinteractions.

A general rule: Like dissolves like.

i.e. polar and polar

non-polar and non-polar

• Dissolution: solute + solvent → solution.• Crystallization: solution → solute + solvent.• Saturation: crystallization and dissolution are in

equilibrium.• Solubility: amount of solute required to form a saturated

solution.• Supersaturation: a solution formed when more solute is

dissolved than in a saturated solution.

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Fig 13.12 Structure of glucose—note red O atoms in OH groups

which can interact nicely with water.

Consider gas solubilities in water at 20 oC with 1 atm gas pressure (~Table 13.2)

Solubility/M

He 0.40 x 10-3

N2 0.69 x 10-3

CO 1.04 x 10-3

O2 1.38 x 10-3

Ar 1.50 x 10-3

Kr 2.79 x 10-3

CO2 3.1 x 10-2

NH3 ~ 53 Fig 13.14 Henry’s Law, Cg = k Pg Note Table 13.2 again for k

Consider N2 dissolved in water at 4.0 atm. Note k = 0.69 x 10-3 mol/L-atm

Cg = k Pg

= (0.69 x 10-3 mol/L-atm)(4.0 atm) = 2.76 x 10-3 mol/L

at normal atmospheric conditions, however, Pg = 0.78 atm

Cg = (0.69 x 10-3 mol/L-atm)(0.78 atm)

= 0.538 x 10-3 mol/L

Note that (2.76 - 0.54) x 10-3 mol/L = 2.22 x 10-3 mol/l

Thus for 1.0 L of water, 0.0022 mol of nitrogen wouldbe released = 0.0022 x 22.4L = 0.049 L = 49 mL !

To read about nitrogen narcosis, seehttp://www.scuba-doc.com/narked.html and about the bends,see http://www.diversalertnetwork.org/medical/articles/index.asp

Solubility of GASES as a Function of Temperature

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Are these exothermic or endothermic processes?

And of

SALTS

Ways of expressing concentration:

a) percent, ppm, ppb usually mass/mass

b) mole fraction = XA , XB sum of Xi = 1

c) molarity = M or mol/Lsolution

depends on T and density of solnpreparation requires dilution

d) molality = m or mol/kgsolvent

independent of Teasily prepared

(a) Prepare 0.500 L of a 0.100 M aqueous solution of KHCO3 (MW = 100.12 g)

Recall M=n/L or n = (M)(L)therefore we need

5.01 g of KHCO3 dissolved and diluted to 0.500 L

(b) Use this solution as a ‘stock’ solution to prepare a final solution of 0.0400 M concentration.

What is the final volume of this solution?

Since n = M V, M1V1 = M2V2

and V2 = M1V1/M2 = (0.100 M)(0.500 L) / (0.0400 M)

= 1.25 L

Consider a solution prepared by dissolving 22.4 g MgCl2in 0.200 L of water. Assume the density of water is 1.000 g/cm3 and the density of the solution is 1.089 g/cm3.

Calculatemole fractionmolaritymolality

MQ-1 on Monday, Jan. 29at 6:30 pm

Covering Chapters 10, 11, and 13

Help session on Saturday, Jan 27at 4:00 – 6:00 pm

in MP1000

5

For students with a scheduled-class conflict

with the first Mid-Quarter Exam:

(1) Send me an email today with a copy of your class schedule, and

(2) Let me know when you want to take themakeup exam:

a) during the last week of classes.b) early (5:00-6:18 pm) on Mon, Jan 29.c) late (8:00-9:18 pm) on Mon, Jan 29.

A 9.386 M solution of H2SO4 has a density of 1.509 g/cm3.

Calculatemolality% by massmole fraction of H2SO4

Week 4 Sections 13.1- 13.6

13.3 Factors Affecting Solubility

13.4 Ways of Expressing ConcentrationMass Percentage, ppm, and ppbMole Fraction, Molarity, and MolalityConversion of Concentration Units

13.5 Colligative PropertiesLowering the Vapor PressureBoiling-Point ElevationFreezing-Point DepressionOsmosisDetermination of Molar Mass

13.6 Colloids

Colligative Properties

Solution properties that depend only on thetotal # of ‘particles’ present.

Vapor Pressure

Boiling Point

Freezing Point

Osmotic Pressure

Note that VP of a solution is lower than that of pure solvent.

A fascinating and somewhat surprising observation:

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Raoult’s Law PA = XA PAo

PA = vapor pressure over solution

XA = mole fraction of component A (solvent)

PAo = vapor pressure of pure

component A (solvent)

also PA = (1 – XB) PAo

where XB = mol fraction of B (solute)

(Recall also Dalton’s Law: PA = XA Ptotal )At first, we consider only nonvolatile solutes.

At 20 oC, the vapor pressure of benzene (cmpd A, MW=78)is 0.1252 atm,i.e. PA

o = 0.1252 atm. If 6.40 g of naphthalene(cmpd B, C10H8, 128.17 g/mol) is dissolved in 78.0 g of benzene calculate the vapor pressure of benzene over the solution.

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As shown on page 549, we also can considersolutions with two volatile components.

Consider a liquid soln containing 1.0 mol benzeneand 2.0 mol of toluene at 20 oC. This yields

Xbenzene = 0.33 and Xtoluene = 0.67

This can be coupled with the fact thatPo

benzene = 75 torrPo

toluene = 22 torr

Apply Raoult’s law to each separately, to obtainPbenzene = XbenzenePo

benzene = 25 torrPtoluene = Xtoluene Po

toluene = 15 torrand PT = Pb + Pt = 40 torr

But with Pb = 25 torr and Pt = 15 torr

We also can calculate the concentrationsof the two in the gas phase!

Xbgas = 25/40 = 0.63

and Xtgas = 15/40 = 0.37

Boiling Point Elevation and Freezing Point Depression(directly related to Raoult’s Law)

Boiling Point Elevation

∆Tb = Tbfinal– Tb

initial= Kb m => + quantity

where m is the molal concentration

Freezing Point Depression

∆Tf = Tffinal – Tf

initial = - Kf m => - quantity

where m is the molal concentration.

[note the definition and the negative sign!!!]

for H2O, Kb = 0.052 oC/m and Kf = 1.86 oC/m

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Consider a water solution which has 0.500 molof sucrose in 1.000 kg of water. Therefore ithas a concentration of 0.500 molal or 0.500 mol/kg.

recall Kb = 0.52 oC/m and Kf = 1.86 oC/m

What is the boiling point and freezing point of this solution?

(a) When 5.50 g of biphenyl (C12H10, 154.2 g/mol) is dissolved in 100 g benzene (C6H6, 78.0 g/mol), theBP increases by 0.903 oC. Calculate Kb for benzene.

(a) When 5.50 g of biphenyl (C12H10, 154.2 g/mol) is dissolved in 100 g benzene (C6H6, 78.0 g/mol), theBP increases by 0.903 oC. Calculate Kb for benzene.

Kb = 2.53 K kg/mol

(b) When 6.30 g of an unknown hydrocarbon is dissolvedin 150.0 g of benzene, the BP of the solution increasesby 0.597 oC.

What is the MW of the unknown substance?

A sample of sea water contains the following in 1.000 Lof solution. Estimate the freezing point of this solution.

Na+ = 4.58 mol Cl- = 0.533 molMg2+ = 0.052 mol SO4

2- = 0.028 molCa2+ = 0.010 mol HCO3

- = 0.002 molK+ = 0.010 Br- = 0.001 molneutral species = 0.001 mol

Sum of species = 1.095 mol

But recall we said Colligative Properties depend on thetotal concentration of ‘species’.

List the following aqueous solutions in increasing order of their expected freezing points.

0.050 m CaCl2 0.05 m Ca+2 and 0.10 m Cl-10.15 m NaCl 0.30 m total0.10 m HCl 0.20 m total0.050 m HOAc between 0.05 and 0.10 m total0.10 m C12H22O11 0.10 m total

List the following aqueous solutions in increasing order of their expected freezing points.

0.050 m CaCl2 x 3 = 0.1500.15 m NaCl x 2 = 0.300.10 m HCl x 2 = 0.200.050 m HOAc x 1 = 0.0500.10 m C12H22O11 x 1 = 0.10

These calculations assume total dissociation of thesalts and zero dissociation of the last two.

The van’t Hoft “i factor”

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This effect of the dissociation of electrolytes is usuallytaken into account through the van’t Hoff i factor, which can be stated formally as

∆Tb = i Kb m

Note that i may be defined as

∆Tf(actual) Kf meffective meffectivei = -------------- = ------------- = ------------∆Tf(ideal) Kf mideal mideal

In real systems, these i factors are NOT integers,but rather fractions whose values depend on concentration.

Focus, for example on NaCl. Notice the limiting valueas well as the values at higher concentrations.

But how can i be less than 2.00 for NaCl?

Through this partial

association.

Recall this observation:This slide from earlier offers a useful intro toOsmotic Pressure.

OsmoticPressure:a fascinatingbehavior.

Yet it is theresult of avery simpletendency toequalize theconcentrationsof solutions.

With the additionof the

semipermeablemembrane—

which permits onlysolvent particles

to move from oneside to the other.

The critical part is the membrane!!!

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A VERY practical application/consequenceof Osmotic Pressure.

Hypertonic soln

crenation(shrivels)

Hypotonic soln

hemolysis(bursts)

Examples:– Cucumber placed in NaCl solution loses water to

shrivel up and become a pickle.– Limp carrot placed in water becomes firm because

water enters via osmosis.– Salty food causes retention of water and swelling of

tissues (edema).– Water moves into plants through osmosis.– Salt added to meat or sugar to fruit prevents bacterial

infection (a bacterium placed on the salt or honey will lose water through osmosis and die).

Osmotic Pressure: π V = n R T

or π = (n/V) R T or π = M R T

(or π = ρ g h,

where ρ = density of solutiong = 9.807 m s-2

h = height of columnbut be careful of the units with this form.)

π = ρ g h, where ρ = density of solutiong = 9.807 m s-2

h = height of column

If h = 0.17 meters of a dilute aqueous soln with ρ = 1.00 g/cm3

π = [(1.00g/cm3)(10-3 kg/g)(106 cm3/m3)](9.807 m/s2)(0.17 m)

= 1.7 x 103 kg m-1 s-1 = 1.7 x 103 Pa

or = (1.7 x 103 Pa) / (1.013 x 105 Pa / atm) = 0.016 atm

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Sample Exercise 13.13

A chemist dissolves 3.50 mg of protein in waterto make 5.00 mL of solution.

water. The observed osmotic pressure is 1.54 torrat 25 oC. What is the MW of the protein?

M = pi/RT = [(1.54 torr)(1 atm/760 torr)]/(0.0821 L-atm/mol-K)(298K)

= 8.28 x 10-5 mol/L⇒ 4.14 x 10-7 moles of protein

or a MW of 8.45 x 103 g/mol

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ColloidsColloidsHydrophilic and Hydrophobic Colloids

• Focus on colloids in water.• “Water loving” colloids: hydrophilic.• “Water hating” colloids: hydrophobic.• Molecules arrange themselves so that hydrophobic

portions are oriented towards each other.• If a large hydrophobic macromolecule (giant molecule)

needs to exist in water (e.g. in a cell), hydrophobic molecules embed themselves into the macromolecule leaving the hydrophilic ends to interact with water.

Hydrophilic and Hydrophobic Colloids• Typical hydrophilic groups are polar (containing C-O, O-

H, N-H bonds) or charged.• Hydrophobic colloids need to be stabilized in water.• Adsorption: when something sticks to a surface we say

that it is adsorbed.• If ions are adsorbed onto the surface of a colloid, the

colloids appears hydrophilic and is stabilized in water.• Consider a small drop of oil in water.• Add to the water sodium stearate.

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Hydrophilic and Hydrophobic Colloids

ColloidsColloids

Hydrophilic and Hydrophobic Colloids• Sodium stearate has a long hydrophobic tail

(CH3(CH2)16-) and a small hydrophobic head (-CO2-Na+).

• The hydrophobic tail can be absorbed into the oil drop, leaving the hydrophilic head on the surface.

• The hydrophilic heads then interact with the water and the oil drop is stabilized in water.

Removal of Colloidal Particles. . . . . . . • Colloid particles are too small to be separated by

physical means (e.g. filtration).• Colloid particles are coagulated (enlarged) until they

can be removed by filtration.• Methods of coagulation:

– heating (colloid particles move and are attracted to each other when they collide);

– adding an electrolyte (neutralize the surface charges on the colloid particles).

– Dialysis: using a semipermeable membranes separate ions from colloidal particles