1. SOLUTIONS
Unit 9

2. Overview Types of Mixtures Concentration Dilution
Suspension
Colloid
Solution
Concentration
Mass percent
Mole fraction
Molality
Molarity
Dilution
Solution Stoichiometry
Solubility
“Like dissolves like”
Factors affecting
Henry’s Law
Solubility product (intro)
Common ion effect (intro)
Saturation
Electrolytes
Colligative Properties
Vapor Pressure Lowering
Freezing Point Depression
Boiling Point Elevation
Osmotic Pressure

3. Vocabulary Solute – substance being dissolved
Solvent – the dissolving medium

4. Types of Mixtures
Colloid – heterogeneous mixture with intermediate sized solute particles that do not settle out of the mixture
Suspension – heterogeneous mixture with large solute particles that can settle out of the mixture
Solution – homogeneous mixture of two or more components

5. Colloids
Heterogeneous mixture with intermediate sized solute particles that do not settle out of the mixture
When filtered, particles do not separate out
Particles do not separate when left to stand
Colloid particles make up the dispersed phase
Example
When large soil particles settle out of muddy water, the water is still cloudy
Emulsion and foam are used to classify colloids
Emulsifying agents help keep colloid particles dispersed

6. Examples of Colloids Class of Colloid Phases Example Sol
Solid dispersed in liquid
Paints, mud
Gel
Solid network extending throughout liquid
Gelatin, jell-o
Liquid emulsion
Liquid dispersed in liquid
Milk, mayonnaise
Foam
Gas dispersed in liquid
Shaving cream, whipped cream
Solid aerosol
Solid dispersed in gas
Smoke, auto exhaust
Liquid aerosol
Liquid dispersed in gas
Fog, mist, clouds, aerosol spray
Solid emulsion
Liquid dispersed in solid
Cheese, butter
Solid foam
Gas dispersed in solid
Marshmallow, styrofoam
Solid sol
Solid dispersed in solid
Ruby glass

7. Tyndall Effect
Colloids scatter light, making a beam visible. Solutions do not scatter light.

8. Suspension
Heterogeneous mixture with large solute particles that can settle out of the mixture
Particles are so large that they settle out unless the mixture is constantly stirred
Particles can be separated from the mixture through filtration
Example
Muddy water

9. Solution Solvation: interactions between solute and solvent
In a solution…
The solute can’t be filtered out
The solute always stays mixed
Particles are always in motion
A solution will have different properties than the solvent
Solvation: interactions between solute and solvent
Hydration: when the solvent is water

10. Other Solution Examples
Solute
Solvent
Example
Gas
Air
Liquid
Soda
Alcohol in water
Solid
Salt in water
Hydrogen in palladium
Mercury in silver
Silver in gold

11. Solution or Colloid?

12. Comparison Colloid Suspension Solution Heterogeneous Homogeneous
Particle size = nm, dispersed; can be aggregates or small molecules
Particle size = over 1000 nm, suspended; can be large particles or aggregates
Particle size = nm, can be atoms, ions, molecules
Do not separate on standing
Particles settle out
Cannot be separated by filtration
Can be separated by filtration
Scatter light (Tyndall effect)
May scatter light but are not transparent
Do not scatter light

13. Concentration Concentrated solution – strong solution
Dilute solution – “Watered-down” solution
Low concentrations
Pollutants often found in air and water are typically found at very low concentrations. Two common units are used to express these trace amounts.
Parts per million (ppm) = (volume solute/volume solution) × 106
Parts per billion (ppb) = (volume solute/volume solution) × 109
Example. One cm3 of SO2 in one m3 of air would be expressed as 1 ppm or 1000 ppb.

14. Solution Concentration
Mass percent: the ratio of mass units of solute to mass units of solution, expressed as a percent
Mole Fraction (X): the ratio of moles of solute to total moles of solution

15. Solution Concentration
Molality (m): moles of solute per kilogram of solvent

16. Assumptions
Assume that solutions with water as the solvent have the density of pure water (1 mL = 1 gram)
1 ml of water = 1 gram of water
1000 ml of water = 1 liter = 1000 grams

17. Solution Concentration
Molarity (M): the ratio of moles of solute to liters of solution
Recognizes that compounds have different formula weights
A 1 M solution of glucose contains the same number of molecules as 1 M ethanol.
[ ] - special symbol which means molar (mol/L )
M =
moles solute mol
liters of solution L =

18. Molarity Examples
Calculate the molarity of a 2.0 L solution that contains 10 moles of NaOH MNaOH = 10 molNaOH / 2.0 L = 5.0 M
What’s the molarity of a solution that has g HCl in 2.0 liters?
First, you need the FM of HCl.
MassHCl = g/mol
Next, find the number of moles.
molesHCl = gHCl / g/mol
= 0.50 mol
Finally, divide by the volume.
MHCl = mol / 2.0 L
= M

19. Solution Preparation Solutions are typically prepared by:
Dissolving the proper amount of solute and diluting to volume.
Dilution of a concentrated solution.

20. Dissolving Solute in Proper Volume
How do you prepare ml of a M solution of sodium chloride.
First, you need to know how many moles of NaCl are in ml of a 0.5 M solution.
M = mol/liters and mL = liters
0.5 M = (X mol)/( liters)
X = mols NaCl
Next, we need to know how many grams of NaCl to weigh out.
mol NaCl ÷ mol = grams
Finally, you’re ready to make the solution.
Weigh out exactly grams of dry, pure NaCl and transfer it to a volumetric flask.
Fill the flask about 1/3 of the way with pure water and gently swirl until the salt dissolves.
Now, dilute exactly to the mark, cap and mix.

21. Diluting Solutions to Proper Concentration
Once you have a solution, it can be diluted by adding more solvent. This is also important for materials only available as solutions
M1V1 = M2V2
1 = initial 2 = final
Any volume or concentration unit can be used as long as you use the same units on both sides of the equation.

22. Diluting Solutions to Proper Concentration
What is the concentration of a solution produced by diluting ml of 1.5 MNaOH to liters?
M1V1 = M2V2
M1 = 1.5 M M2 = ???
V1 = ml V2 = 2000 ml
(1.5)(100.0) = (M2 )(2000)
M2 = M

23. Solution Stoichiometry
Extension of earlier stoichiometry problems.
First step is to determine the number of moles based on solution concentration and volume.
Final step is to convert back to volume or concentration as required by the problem.
You still need a balanced equation and must use the coefficients for working the problem.

24. Stoichiometry Example
Determine the volume of M HCl that must be added to completely react with 250 ml of 2.50 M NaOH
HCl(aq) + NaOH(aq) →NaCl(aq) + H2O (l)
We have 250 ml of a 2.50 M solution
2.50 mol/L = (molNaOH)(0.250 L)
molNaOH = 0.625
0.625 molNaOH = × 1 molHCl/1 molNaOH = molHCl

25. Stoichiometry Example (cont…)
Now we can determine what volume of our M HCl solution is required.
M = molHCl /literHCl
0.100 = (0.625 mol)/(X liters)
= L of HCl

26. Solubility of Solutions
“Like” dissolves “like”
Nonpolar solutes dissolve best in nonpolar solvents
Examples
Fats in benzene, steroids in hexane, waxes in toluene
Polar and ionic solutes dissolve best in polar solvents
Inorganic salts in water, sugars in small alcohols
Miscible – pairs of liquids that dissolve in one another
Immiscible – pairs of liquids that do not dissolve in one another

27. Factors Affecting Dissolution
Factors that increase the rate of dissolving a solute
Increasing surface area of solute
More contact surfaces for solvent to dissolve solute
Agitating (stirring/shaking) solution
Increase surfaces of solute exposed to solvent
Heating the solvent
Solvent molecules move faster
Average kinetic energy increases
At higher temperatures, collisions between solvent molecules and solute are more frequent

28. Henry’s Law
Solubility of a gas is directly proportional to the partial pressure of that gas on the surface of the liquid
In carbonated beverages, solubility of CO2 is increased by increasing the pressure
Carbonation escapes when the pressure is reduced to atmospheric pressure so CO2 escapes
Effervescence – the rapid escape of gas from a liquid in which it is dissolved
Sg = kPg
Sg = solubility of gas
K = Henry’s law constant (table p 536)
Pg = partial pressure of gas over solution

29. Effects on Solubility of Gases
To increase the solubility of gases
Increase pressure
Gas is forced into solution under pressure
Soda goes “flat” when exposed to atmospheric pressure
Decrease temperature
Increased temperature means increased kinetic energy
With increased kinetic energy, more gas molecules can escape from the liquid
Soda goes “flat” faster in warm temperatures than cold

30. Solubility Curves

31. Solubility Product
Salts are considered “soluble” if more than 1 gram can be dissolved in 100 mL of water.
Salts that are “slightly soluble” and “insoluble” still dissociate to some extent
Solubility product (Ksp) is the extent to which a salt dissociates in solution
Greater the value of Ksp, the more soluble the salt

32. Solubility Product
For the reaction: AaBb(s) ↔ aA(aq) + bB(aq) Solubility product: Ksp = [A]a[B]b [ ] = concentration Example: CaF2(s) ↔ Ca+2(aq) + 2F-(aq) Ksp = [Ca+2][F-]2

33. Ksp Values for Salts at 25°C
Name
Formula
Ksp
 Barium carbonate 
 BaCO3 
 2.6 x 10-9 
 Barium chromate 
 BaCrO4 
 1.2 x 10-10 
 Barium sulfate 
 BaSO4 
 1.1 x 10-10 
 Calcium carbonate 
 CaCO3 
 5.0 x 10-9 
 Calcium oxalate 
 CaC2O4 
 2.3 x 10-9 
 Calcium sulfate 
 CaSO4 
 7.1 x 10-5 
 Copper(I) iodide 
 CuI 
 1.3 x 10-12 
 Copper(II) iodate 
 Cu(IO3)2 
 6.9 x 10-8 
 Copper(II) sulfide 
 CuS 
 6.0 x 10-37 
 Iron(II) hydroxide 
 Fe(OH)2 
 4.9 x 10-17 
 Iron(II) sulfide 
 FeS 
 6.0 x 10-19 
 Iron(III) hydroxide 
 Fe(OH)3 
 2.6 x 10-39 
 Lead(II) bromide 
 PbBr2 
 6.6 x 10-6 
 Lead(II) chloride 
 PbCl2 
 1.2 x 10-5 
 Lead(II) iodate 
 Pb(IO3)2 
 3.7 x 10-13 
 Lead(II) iodide 
 PbI2 
 8.5 x 10-9 
 Lead(II) sulfate 
 PbSO4 
 1.8 x 10-8 
Name
Formula
Ksp
 Lead(II) bromide 
 PbBr2 
 6.6 x 10-6 
 Lead(II) chloride 
 PbCl2 
 1.2 x 10-5 
 Lead(II) iodate 
 Pb(IO3)2 
 3.7 x 10-13 
 Lead(II) iodide 
 PbI2 
 8.5 x 10-9 
 Lead(II) sulfate 
 PbSO4 
 1.8 x 10-8 
 Magnesium carbonate 
 MgCO3 
 6.8 x 10-6 
 Magnesium hydroxide 
 Mg(OH)2 
 5.6 x 10-12 
 Silver bromate 
 AgBrO3 
 5.3 x 10-5 
 Silver bromide 
 AgBr 
 5.4 x 10-13 
 Silver carbonate 
 Ag2CO3 
 8.5 x 10-12 
 Silver chloride 
 AgCl 
 1.8 x 10-10 
 Silver chromate 
 Ag2CrO4 
 1.1 x 10-12 
 Silver iodate 
 AgIO3 
 3.2 x 10-8 
 Silver iodide 
 AgI 
 8.5 x 10-17 
 Strontium carbonate 
 SrCO3 
 5.6 x 10-10 
 Strontium fluoride 
 SrF2 
 4.3 x 10-9 
 Strontium sulfate 
 SrSO4 
 3.4 x 10-7 
 Zinc sulfide 
 ZnS 
 2.0 x 10-25 

34. Common Ion Effect AgCl(s) is added to water
The Ag+ ions and the Cl- ions completely dissociate
NaCl(aq) is mixed with AgCl(aq)
The Na+ ions do not have an effect on the Ag+ or the Cl- ions
There are already Cl- ions in solution
The addition of a “common ion” can cause some of the salt to precipitate out of solution
The salt with the smaller Ksp value will precipitate first

35. Saturation of Solutions
Supersaturated solution – contains more solute than can dissolve in the solvent
Saturated solution – a solution that cannot dissolve any more solute under the given conditions
Unsaturated solution – a solution that is able to dissolve additional solute

36. Electrolytes Electrolyte – solution that conducts electricity
ionic compounds in polar solvents dissociate (break apart) in solution to make ions
may be strong (100% dissociation) or weak (less than 100%)
Strong Electrolyte – all or almost all of compound dissociates
Example: strong acids (H2SO4, HNO3, HClO4, HCl, HBr, HI)
Weak Electrolyte – Small amount of compound dissociates
Example – weak acids (HF, acetic acid)
Nonelectrolyte – solution that does not conduct electricity
solute is dispersed but does not dissociate
Example: sugar (dissolves but does not dissociate), organic acids (contain carboxyl groups)

37. Classifying Electrolytes
Strong Electrolyte
Weak Electrolyte
Nonelectrolyte
Ionic Compounds:
All substances
None
Covalent Compounds:
Strong acids
Weak acids
Weak bases
All other compounds

38. Concentration of Ions Strong electrolytes – ions completely dissociate
Relative concentration of ions depends on chemical formula
Example 1: In a solution of 0.1 M NaCl
Concentration of Na+ = 0.1 M
Concentraion of Cl - = 0.1 M
Example 2: In a solution of 0.1 M Na2SO4
Concentration of Na+ = 0.1 M × 2 = 0.2 M (account for 2 sodium ions)
Concentraion of SO4-2 = 0.1 M

39. Colligative Properties
Properties that depend on the concentration of solute particles but not on their identity
Vapor pressure lowering
Freezing point depression
Boiling point elevation
Osmotic Pressure

40. Vapor Pressure Lowering
Increased number of solute particles means fewer water molecules
Fewer molecules exist to escape from the liquid
Molecules have lower tendency to leave the liquid phase
When a solute is added to a solution, the vapor pressure of the solution decreases (Raoult’s Law)

41. Vapor Pressure Lowering
Raoult’s Law:
P = XP°
P = vapor pressure of the solution
P° = vapor pressure of the pure solvent
X = mole fraction of the solvent

42. Vapor Pressure Lowering (Example)
Calculate the vapor pressure of the solution when 235 grams of sugar (C12H22O11) are added to 650 mL of water at 40°C. The vapor pressure of water at 40°C is 55 mm Hg. 235 grams C12H22O11 = mol C12H22O mL H2O= 650 g H2O= mol H2O X= /( ) = P = XP° P = (0.981)(55mmHg) = 54.0 mmHg

43. Freezing Point Depression
When solute is added to a solution, freezing point decreases
Particles in first beaker can easily shift into solid phase from liquid phase at normal freezing temperature
Particles in second beaker are blocked by solute particles and cannot as easily move into the solid phase from the liquid phase so a colder temperature is needed to shift into the solid phase

44. Freezing Point Depression
∆tf = iKfm
∆tf = change in freezing temperature
i = van’t Hoff factor (number of particles into which the added solute dissociates)
Kf = molal freezing-point depression constant
m = molality

45. Boiling Point Elevation
When a solute is added to a solution, the boiling point of the substance increases
Fewer water molecules exist due to addition of solute particles
Fewer particles leave liquid phase
Higher temperature needed to increase kinetic energy and allow liquid particles to escape

46. Boiling Point Elevation
∆tb= iKbm
∆tb = change in boiling temperature
i = van’t Hoff factor (number of particles into which the added solute dissociates)
Kb = molal boiling-point elevation constant
m = molality

47. Osmotic Pressure
External pressure that must be applied to stop osmosis
Osmosis – movement of solvent through a semi-permeable membrane (allows passage of some particles but not others) from the side of lower solute concentration to the side of higher solute concentration
Semi-permeable membrane - allows passage of some particles but not others

48. Osmotic Pressure
Semi-permeable membrane allows passage of water molecules but not solute particles
Solute particles prevent water molecules from striking surface of semi-permeable membrane
On pure water side, water molecules free to hit surface

49. Osmotic Pressure (cont…)
Rate of water molecules leaving pure water side is greater than rate of water molecules leaving solution side
Height of solution rises until pressure exerted by the height of solution is large enough to forces water molecules back through the membrane at equal rate to which they enter the solution side

50. Osmotic Pressure П = RTi = MRTi λ = osmotic pressure (atm)
n = moles of solute
R = gas constant
T = temperature (K)
V = volume of solution
i = van’t Hoff factor (number of particles into which the added solute dissociates)
M = molarity of solution

51. Van’t Hoff Factor
Salt lowers the freezing point of water nearly twice as much as sugar does. Why is this?
Nonelectrolyte solutions – particles do not dissociate into ions
Strong electrolyte solutions – particles completely dissociate into ions
Sugar is a nonelectrolyte so sugar dissolves to produce only one particle in solution. Salt is a strong electrolyte so it dissociates completely to produce two ions in solution (Na+ and Cl-). Salt has twice as many particles to lower the freezing point as sugar.
This is accounted for with the van’t Hoff factor (i)
What is the van’t Hoff factor for Ba(NO3)2? _________

52. Forces of Attraction
A 0.1 m solution of KCl lowers the freezing point more than a 0.1 m solution of MgSO4. Why is that?
Forces of attraction (charges of ions) interfere with movements of aqueous ions
Only in very dilute solutions is the forces of attraction between ions negligible
Ions of higher charge attract each other more strongly and cluster together more acting as a single unit more than multiple particles
MgSO4 has ions of +2 and -2. Ions are more attracted to each other more than in KCl (ions are +1 and -1). MgSO4 particles cluster together more than KCl particles so they have less effect on colligative properties.