Role of Disorder in Solutions

• Disorder (Entropy) is a factor• Solutions mix to form maximum

disorder

Ch 13: Solutions

Two Ways to Form Solutions

1. Physical Dissolving (Solvation)• NaCl(s) Na+(aq) + Cl-(aq)• C12H22O11(s) C12H22O11(aq)

• Particles are surrounded by solvent molecules• Can evaporate water/solvent to get original

compound back

Types of Reactions

2. Chemical reaction• Ni(s) + 2HCl(aq) NiCl2(aq) + H2(g)

• Evaporating solvent gives the products

Solubility – Maximum amount of a solute that can dissolve in 100 mL of a solutionEx: NaCl 35.7 g/100mL– Saturated solution – Contains the max. amount of

solute with some undissolved solid,– Unsaturated – more solute will dissolve.

– Supersaturated – More than the max is dissolved by heating and slowly cooling.

Like Dissolves Like: Miscibility

• Polar dissolves polar (dipole-dipole Forces) and ionic (ion:dipole)– Water and Ammonia

• Non-Polar dissolves non-polar (London Forces)– Soap and grease

Would acetone (shown below) dissolve in water?

Acetone

:O: ||CH3CCH3

Using you knowledge of “like dissolves like”, explain the following trends in solubility.

Alcohol Solubility in H2O (mol/100 g H2O at 20oC)

CH3OH ∞

CH3CH2OH ∞

CH3CH2CH2OH ∞

CH3CH2CH2CH2OH 0.11

CH3CH2CH2CH2CH2OH 0.030

CH3CH2CH2CH2CH2CH2OH 0.0058

CH3CH2CH2CH2CH2CH2CH2OH 0.0008

Pressure Effects

• Solubilty of a gas increases with pressure of gas over the liquid (soda bottle)

• Henry’s Law

Sgas = kPgas

Henry’s Law

The Henry’s law constant for CO2 is 0.031 mol/L-atm.

a. Calculate the concentration of CO2 in a soda bottle pressurized to 4.00 atm of CO2.

b.After the bottle has been opened, the concentration drops to 9.3 X 10-6 M. Calculate the partial pressure of CO2 over the soda.

Temperature Effects

• Solubility of most solids increases with temperature

• Solubility of most gases decreases with temperature (warm soda)– Warm water is deoxygenated– Problem with thermal pollution of

lakes

Ways of Expressing Concentration

Mass % = mass of compound in soln X 100total mass of soln

Parts Per Million

ppm = mass of component X 106

total mass of soln

Concentration: Ex 1

13.5 g of C6H12O6 is dissolved in 0.100 kg of water. Calculate the mass percentage.

mass % = 13.5 g X 100 = 11.9%(100 g + 13.5 g)

Concentration: Ex 2

A 2.5 g sample of groundwater is found to contain 5.4 g of Zn2+. What is the concentration of the Zn2+ ion in ppm.

Concentration: Ex 2A 2.5 g sample of groundwater is found to contain

5.4 g of Zn2+. What is the concentration of the Zn2+ ion in ppm.

5.4 g | 1X10-6g = 5.4 X 10-6 g | 1 g

ppm = mass of component X 106

total mass of solnppm = 5.4 X 10-6 g X 106 = 2.2 ppm

2.5g

Concentration: Ex 3

Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water.

ANS: 2.91 %

Concentration: Ex 4

Bleach is 3.62 % NaOCl. What mass of NaOCl is contained in 2500 g of bleach?

ANS: 90.5 g NaOCl

Mole Fraction

Mole FractionX = moles of component

total moles of all components

What is the mole fraction of HCl if 36.5 grams is dissolved in 144 grams of water?

ANS: 0.111

Molality

Molality = moles of solutekilograms of solvent

Why not use Molarity?• Molarity varies with temperature• Total volume of a solution changes with

temperature (liquid expands)• Mass does not change with temperature

Molality: Ex 1

A solution is made by dissolving 4.35 grams of C6H12O6 in 25.0 mL of water. Calculate the molality of the glucose.

4.35 g 1 mol = 0.0241 mol180.2 g

Molality = 0.0241 mol = 0.964 m0.0250 kg

Molality: Ex 2

Calculate the molality of a solution made by dissolving 36.5g C10H8 in 425 grams of toluene (solvent).

ANS: 0.671 m

Molality: Ex 3

A solution of HCl contains 36 percent HCl by mass. Calculate the mole fraction and molality of HCl.

Pretend 100 g36 g HCl64 g H2O

36 g HCl 1 mol HCl = 0.99 mol HCl36.5 g HCl

64 g H2O 1 mol H2O = 3.6 mol HCl

18 g H2O

XHCl = 0.99 mol HCl = 0.22

0.99 mol + 3.6mol

Molality = 0.99 mol HCl = 15 m0.064 kg H2O

Molality: Ex 4

A commercial bleach solution contains 3.62 percent NaOCl by mass. Calculate the mole fraction and molality of NaOCl.

ANS: XNaOCl = 0.00900, 0.505 m

Molality: Ex 5

The density of a solution of 5.0 g of toluene (C7H8) and 225 g of benzene (C6H6) is 0.876 g/mL. Calculate the molality and molarity of the toluene.

Molality: Ex 5

The density of a solution of 5.0 g of toluene (C7H8) and 225 g of benzene (C6H6) is 0.876 g/mL. Calculate the molality and molarity of the solution.

Molality5.0 g 1 mol = 0.054 mol

92.0 g

m = 0.054 mol/ 0.225 kg = 0.24 m

Molarity

D = mass/VV = mass/DV = 230 g = 263 mL

0.876 g/ml

M = 0.054 mol = 0.21 M0.263 L

Molality: Ex 6

A solution containing equal masses of glycerol (C3H8O3) and water has a density of 1.10 g/mL. Calculate

a)molality (10.9 m)b)mole fraction (XC3H8O3 = 0.163)

c) molarity of glycerol in the solution (5.97 M)

Colligative Properties: Vapor Pressure Lowering

Non-volatile solutes lower the vapor pressure of the solvent

Raoult’s lawPA = XAPo

A

PA = Vapor pressure

XA = Mole fraction of solvent

PoA = Pressure of pure solvent

Raoult’s Law: Ex 1

What is the vapor pressure of a solution made by adding 50.0 mL of glycerin (C3H8O3) to 500.0 mL of water? The density of glycerin is 1.26 g/mL and the vapor pressure of pure water is 23.8 torr.

MassC3H8O3 = (50.0 mL)(1.26 g/mL ) = 63.0 g

MolesC3H8O3 = 63.0 g/92.1 g/mol = 0.684 mol

MolesH2O = 500.0 g/18 g/mol = 27.8 mol

XH2O = 27.8 mol = 0.976

(27.8 mol + 0.684 mol)

PA = XAPoA

PA = (0.976)(23.8 torr) = 23.2 torr

Raoult’s Law: Ex 2The vapor pressure of water at 110oC is 1070 torr. A

solution of ethylene glycol and water has a vapor pressure of 1 atm at 110oC. What is the mole fraction of ethylene glycol in the solution?

ANS: 0.290

Colligative Properties: Boiling Point Elevation

• Non-volatile solute raises the boiling point of a solution

• Shifts the phase diagram

• The pressure of the solution reaches atmospheric pressure at a higher temp.

Tb = iKbm

Colligative Properties: Freezing Point Depression

• Solutions freeze at a lower temperature than pure solvent

• Salt water freezes lower (-2oC) than distilled water (0oC)

Tf = iKfm

i = Van’t Hoff factorm = molality of the nonvolatile solute

• The more ions produced, the greater the freezing point depression or boiling point elevation

• C12H22O11 (i=1)

• NaCl = Produces two ions (i=2)• CaCl2 = Produces three ions (i=3)

Pure Solvent Solution

Boiling Point

Boiling Point

Freezing Point

Freezing Point

Colligative: Ex 1

Ethylene Glycol, C2H6O2, is used in antifreeze. What will be the freezing and boiling point of a 25.0 mass percent solution of ethylene glycol and water?

Pretend 100 grams of solution25 grams of C2H6O2

75 grams of H2O (0.075 kg)

25 grams of C2H6O2 = 0.403 moles

m = 0.403 moles = 5.37 m0.075 kg H2O

Tb = iKbm = (1)(0.52oC/m)(5.37 m) = 2.8oC

Tf = iKfm = (1)( 1.86oC/m)(5.37 m) =10.0oC

Boiling Point = 102.8oCFreezing Point = -10.0oC

ooo

Colligative: Ex 2

Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of C10H18O. Kf for CHCl3 is 4.68oC/m and the normal freezing point is -63.5 oC.

ANS: -65.6oC

Colligative: Ex 3

Rank the following aqueous solutions in order of their expected freezing points:

0.050 m CaCl2

0.15 m NaCl0.10 m HCl0.050 m HC2H3O2 (acetic acid)

0.10 m C12H22O11 (sugar)

0.050 m CaCl2 (0.15 m in particles)

0.15 m NaCl (0.30 m in particles)0.10 m HCl (0.20 m in particles)0.050 m HC2H3O2 (just above 0.05 m)

0.10 m C12H22O11 (0.10 m in particles)

Lowest FP Highest FPNaCl < HCl < CaCl2 < C12H22O11 < HC2H3O2

Colligative: Ex 4

Rank the following in order of the increase in boiling point that they will produce in 1 kg of water

1 mol Co(NO3)2

2 mol KCl3 mol C2H6O2 (a very, very weak electrolyte(acidic))

1 mol Co(NO3)2 (3 mol particles)

2 mol KCl (4 mol of particles)3 mol C2H6O2 (3+ mol of particles)

Lowest BP Highest BPCo(NO3)2 < C2H6O2 < KCl

Freezing Pt Depression: Ex 5

What would be the molality of salt water if it freezes at 0 oF? Kf = 1.86 oC/m.

ANS: 4.78 m

Colligative Properties: Osmotic Pressure

• Osmosis – movement of solvent from high concentration to low concentration

• semipermeable membrane – allows to passage of some particles but not others

Cucumber Skin cell afterin salt water soaking in a tub

• Note that solvent moves both ways

• Solute too large to pass through membrane

• Net movement is to try to dilute the side with solutes

• Osmotic Pressure () – pressure required to prevent osmosis

PV = inRT V = inRT = inRT V = iMRT

M = molarity

Osmotic Pressure: Ex 1

The average osmotic pressure of blood is 7.7 atm at 25oC. What concentration of glucose will be isotonic with blood?

(0.31 M)

Osmotic Pressure: Ex 2

What is the osmotic pressure at 20oC of a 0.0020 M sucrose, C12H22O11, solution? Express your answer both in atmosphere and in torr.

ANS; 0.048 atm, 37 torr

Molar Mass: Ex 1

A solution of an unknown nonelectrolyte was prepared by dissolving 0.250 g in 40.0 g of CCl4. The boiling point of the resulting solution was 0.357oC higher than that of the pure solvent. Kb for CCl4 is 5.02 oC/m. Calculate the molar mass of the unknown.

Tb = iKbm

m = Tb /iKb

m = (0.357oC)/(1 X 5.02 oC/m) = 0.0711 m

m = mol of solute kilograms of solvent

molsolute = (m)(kg of solvent)

molsolute = (0.0711 m)(0.0400 kg) = 0.00284 mol

Molar Mass = 0.250 g = 88.0 g/mol0.00284 mol

Molar Mass: Ex 2

Camphor, C10H16O, melts at 179.8oC and has a Kf of 40.0 oC/m. When 0.186 g of an unknown substance is dissolved in 22.01 g of liquid camphor, the freezing point is 176.7oC. What is the molar mass of the solute?

ANS: 110 g/mol

Molar Mass: Ex 3

A solution contains 3.50 mg of protein dissolved in water to form 5.00 mL of a solution. The osmotic pressure at 25oC was found to be 1.54 torr. Calculate the molar mass of the protein.

1.54 torr 1 atm = 0.00203 atm760 torr

= iMRTM = /iRTM = 0.00203 atm = 8.28X10-5 M

(1)(0.0821 L-atm/mol-K)(298)M = moles

litermoles = (M)(liters) = (8.28X10-5 M)(0.00500L)moles = 4.14 X 10-7 mol

Molar mass = 3.50 X 10-3 g = 8454 g/mol4.14 X 10-7 mol

Molar Mass: Ex 4

A 2.05 g sample of a plastic was dissolved in enough toluene to form 100 mL of solution. The osmotic pressure of this solution is 1.21 kPa at 25oC. Calculate the molar mass of the plastic. (1 atm = 101.325 kPa).

ANS: 42,000 g/mol

ColloidsSolutions/

HomogeneousColloids Suspensions/

Heterogeneous Mixture

Ions/Molecular size solute particles

Medium particles (10 to 2000 Å) Larger particles (like dirt in water)2 or more separate phases

Never separate Separates quickly

• Examples– Fog– Smoke– Whipped Cream– Milk

• Tyndall effect – scattering of light

Stabilization of Colloids

• Hydrophobic/hydrophilic imf– Biomolecules

Emulsifying agents– Soap– Sodium stearate(used to digest fats)

6. Larger Noble gases have greater London Forces (greater dipole: induced dipole forces)

16.a) Ion:Dipole b) Dipole: induced dipolec) Hydrogen bonding(weakest) b < c < a (strongest)

30.a) glucose (OH’s allow h-bonding)b) sodium propionate (has an ion)c) HCl (small and polar)

34.S = kP k = S/P = 1.38 X 10-3M/0.21 atmk =6.57 X 10-3 M/atm

Partial Pressure of O2 at the higher elevation:

650 torr = 0.855 atmPO2 = (0.855 atm)(0.21)

PO2 = 0.180 atm

S = kP S=(6.57 X 10-3 )(0.180 atm)S = 1.18 X 10-3 M

36a) 7.2% I2 b) 7.9 ppm Sr2+

38a) 0.0285 b) 5.66% c) 0.638 m40a) 0.125 M b) 0.140 M c) 0.630 M42a) 4.34 m b) 3.1 g S844a) 27.7% b) 0.0377 c) 2.18 m

d) 1.92 M46a) 0.0439 b) 0.498 m c) 0.417 M48a) 0.278 mol b) 6.25X10-5 mol c) 0.00329 mol50a) 21.8 g b) 7.7 g/112.3g c) 209 g

d) 11 mL of 6.0 M HCl52) 15 M NH3

62.a) 222 torr b) 150 g64. a) 0.75 b) 0.4766. 10% sucrose < 10% glucose < 10% NaNO3

68. 0.030 m phenol < KBr = 0.040 m glycerin70.a) -115.2, 78.8 b) -78, 72.4

c) -9.3, 102.672.-18 oC74.2.8 atm76.180 g/mol78. 380 g/mol

Write Net Ionic Equations for:MgCO3(s) + HNO3(aq)

H2SO4(aq) + 2KOH(aq)

NaHCO3(aq) + HCl(aq)

A solution of nickel(II)sulfate is stored in a zinc coated bucket.

a. Write the net ionic reaction that occurs.b.Suppose the nickel(II)sulfate was stored in a

copper bucket. Would this be a better choice?c. Write the net ionic reaction that occurs between

nickel(II)sulfate and barium nitrate.

1. Lots of ice, little water2. Stir and take temperature (~2 min)3. Add medicine cup of rock salt4. Stir and take temperature (~2 min)5. Clean Up

1. Rinse Foam cup, medicine cup, and thermometer2. Place in drying rack

A 80.5 g sample of ascorbic acid (C6H8O6) is dissolved in 210.0 grams of water. The resulting solution has a density of 1.22 g/mL.

a)Calculate the molarity of the solutionb)Calculate the molality of the solution.c)Calculate the freezing point of the solution.

(Assume ascorbic acid is a very weak electrolyte, Kf = 1.86 oC/m)

d)Would you expect the solution to have a higher or lower vapor pressure than pure water?

e)Calculate the vapor pressure of the solution if the vapor pressure of pure water is 17.5 torr.

Benzene (C6H6) freezes at 5.50 oC and has a Kf of 5.12 oC/m. 50.0 grams of an unknown solute is dissolved in 100.0 g of benzene. The resulting solution freezes at -8.72 oC.

a.Calculate the molality of the solution. (2.78 m)

b.Calculate the molar mass of the solute. (180)c.Why is benzene a good choice for this

experiment?d.Comment on the polarity of the unknown

molecule.e.What is the hybridization of the carbon

atoms in benzene?