1. Solutions Ch. 15 (p )

2. Solutions Sugar in water Air Dental fillings Saline
Solution: A homogeneous mixture of two
or more components in a single physical state
Sugar in water
Air
Dental fillings
Saline

3. Characteristics of a Solution
The solute can’t be filtered out.
The solute always stays mixed.
Particles are always in motion.
Volumes may not be additive.
A solution will have different properties than the solvent.
A solution consists of two components:
solvent - component in the greater extent
solute - component in the lesser extent
(There may be more than one.)

4. Physical states of solutions
Solutions can be made that exist in any of the three states. (See Fig. 15-3, p. 503)
Solid solutions
dental fillings, 14K gold, sterling silver
Liquid solutions
saline, liquor, antifreeze, vinegar
Gas solutions
air, anesthesia gases

5. Solution Definitions Alloy: solution of two or more metals
Soluble: able to dissolve in another substance
Insoluble: not able to dissolve in another substance
Miscible: pairs of liquids that mix in any amount (Ex. ethanol and water)
Immiscible: pairs of liquids that do not mix in any amount (Ex. oil and water)
Aqueous: solutions with water as the solvent

6. Amount of Solute in Solution
Concentration: the amount of solute per quantity of solvent or solution
There are many systems - we will cover four.
Molarity (M)
Mass percent
Mole fraction (X)
Molality (m)

7. Example: Molarity MNaOH = 10 mol / 2.0 L
Calculate the molarity of a 2.0 L solution that contains 10 moles of NaOH.
MNaOH = 10 mol / 2.0 L
= 5.0 M
moles solute mol
liters of solution L
Molarity (M) = =

8. Use the same units for both
Mass Percent
Mass Solute
Total Mass
Mass % =
x 100
Use the same units for both
If a ham contained 5 grams of fat in 200 g
of ham, what is the mass percent?
5 g / 200g * 100 = 2.5 %
On the label, it would say 97.5 % fat free.

9. Molality moles solute mol kilograms of solvent kg Molality (m) = =
Example: Calculate the molality of a 10.0 g of KCl in 80.0 grams of water.

10. Mole fraction Mole fraction
The moles of solute, expressed as a fraction of the total number of moles in the solution.
XA =
Because the units in the numerator and denominator are the same, mole fraction is a unitless quantity.
The sum of all components must equal one. nA
nA + nB + nC

11. Example: Mole Fraction
What is the mole fraction of sulfur dioxide (SO2) in an industrial exhaust gas containing g of sulfur dioxide dissolved in g of carbon dioxide.
XCO2 = 1 – XSO2 = 1 – = .9446

12. Saturation
When a solution contains as much solute as it can at a given temperature.
Unsaturated Can still dissolve more.
Saturated Have dissolved all you can.
Supersaturated Temporarily have dissolved
too much.
Precipitate Excess solute that falls out
of solution.

13. Properties of Aqueous Solutions
There are two general classes of solutes.
Electrolytes
ionic compounds in polar solvents
dissociate in solution to make ions
conduct electricity
may be strong (100% dissociation) or weak (less than 100%)
Nonelectrolytes
do not conduct electricity
solute is dispersed but does not dissociate

14. Colligative Properties
“Bulk” properties that change when you add a solute to make a solution.
Based on how much you add but not
what the solute is.
Effect of electrolytes is based on number of ions produced.
Colligative properties
vapor pressure lowering
freezing point depression
boiling point elevation
osmotic pressure

15. Vapor Pressure Lowering
Volatile: has a measurable vapor pressure
The introduction of a nonvolatile solute will reduce the vapor pressure of the solvent in the resulting solution.
The vapor pressure of a nonvolatile component is essentially zero.
It does not contribute to the vapor pressure of the solution.
However, the solution’s vapor pressure is dependent on the solute mole fraction.

16. Demo: Vapor Pressure Lowering
Water will end up in the ‘salt’ solution because it’s
vapor pressure is lower than the pure water.

17. Vapor Pressure (A Dynamic Process)
liquid

18. Boiling Point Elevation
When you add a nonvolatile solute to a solvent, the boiling point goes up. This is because the vapor pressure has been lowered.
DTb = Kb m
The boiling point will continue to be elevated as you add more solute until you reach saturation.
Examples Cooking pasta in salt water
Antifreeze

19. Example: Boiling point elevation
Determine the boiling point for a m aqueous solution of sucrose.
Kb = oC/m for water.
DTb = oC/m (0.222 m)
= oC
Tb = oC oC = oC

20. Phase Diagram (Boiling Point Elevation)
Diagram:

21. Freezing Point Depression
When you add a solute to a solvent, the freezing point goes down.
DTf = Kf m
The more you add, the lower it gets.
This will only work until you reach saturation.
Examples “Salting” roads in winter
Making ice cream

22. Example: Freezing Point Depression
Determine the freezing point for a m aqueous solution of sucrose.
Kf = 1.86 oC/m for water.
DTf = oC/m (0.222 m)
= oC
Tf = 0.00 oC oC = oC

23. Example Constants (p. 522/526)
Normal Kb Normal Kf
Solvent Tb, oC oC/m Tf, oC oC/m
Water
Benzene
Camphor
Ethanol

24. Phase Diagram (Freezing Point Depression)
Diagram:

25. Electrolytes and Colligative Properties
Ionic substances have a greater effect per mole than covalent.
1 mol/kg of water for glucose = 1 molal
1 mol/kg of water for NaCl = 2 molal ions
1 mol/kg of water for CaCl2 = 3 molal ions
Effects are based on the number of particles!

26. Van’t Hoff Factor Van’t Hoff factor (i): number of ions created.
Glucose (nonelectrolyte)—i = 1
NaCl (Na+ + Cl-)—i = 2
CaCl2(Ca Cl-)—i = 3
Equations become:
Tb = iKbm
Tf = iKfm

27. Example: Electrolytes
Determine the boiling point for a 1.35 m aqueous solution of MgCl2.
i = 3 (Mg Cl-)
Kb = oC/m for water.
DTb = (3)0.512 oC/m (1.35 m)
= 2.07oC
Tb = oC oC = oC

28. Determining Molar Mass
EXAMPLE: A 10.0 gram sample of an unknown nonvolatile, nonelectrolyte compound is dissolved in 100 g of water. The boiling point of the solution is elevated to ◦C. What is the molar mass of the unknown sample? (Kb for water is 0.52 ◦C/m)
Given:
munk=10.0 g
msolv=100 g = .100 kg
Tb = ◦C
i=1
Kb=0.52 ◦C/m

29. Determining Molar Mass (page 2)
Given:
munk=10.0 g
msolv=100 g = .100 kg
Tb = ◦C
Kb = 0.52 ◦C/m
Equation: Tb = Kbm
m= Tb/Kb = 0.433/0.52 = 0.83 m (Step 1)
By Definition: m = mol solute/kg solvent
Rearranging: mol solute = m x kg solvent (Step 2)
= 0.83 m * .100 kg = mol
Conclusion: 10 g/0.083 mol = 120 g/mole (Step 3)

30. How Solutions Form
When an ionic solid is placed in a polar solvent, the outer ions are exposed to the polar molecules.
Water will pull the ions from the solid and surround them - solvate them.
Solvation of ions is an exothermic process which helps overcome the lattice energy that holds the crystal together. Hydration is this same process—with water as the solvent. (Solvation is entropically favorable, as well.)

31. Solution of solids

32. Factors Affecting Solubility
Nature of Solvent and Solute
2) Temperature
3) Pressure

33. Nature of Solvent & Solute
“Like dissolves like.”
Materials with similar polarity are soluble in each other. Dissimilar ones are not
Ionic substances are not soluble in nonpolar solvents like hexane.
A large amount of energy is need to separate the ions.
A nonpolar solvent can’t solvate ions so there is no solvation energy to offset the lattice energy.

34. Pressure and solubility of gases
Henry’s Law
At constant temperature, the solubility of a gas is directly proportional to the pressure of the gas above the solution.
cg = kpgas
This law is accurate to
within 1-3% for slightly
soluble gases and
pressures up to one
atmosphere.
Pressure (atm)
(g/100g water)
Solubility
0.010
0.005
0.000 O2 N2 He

35. Solubilities of solids
Ionic substances are not soluble in nonpolar solvents like hexane.
A large amount of energy is need to separate the ions.
A nonpolar solvent can’t solvate ions so there is no solvation energy to offset the lattice energy.
Predicting the solubility of ionic solids in water is difficult because a number of competing factors are involved.

36. Solution of solids
While covalent compounds do not dissociate, they are solvated in solution.

37. Saturated Solutions
At saturation, the solute is in dynamic equilibrium. The concentration is constant.
Solute species are
constantly in
motion, moving
in and out of
solution.

38. Why do some curves decrease with increasing temperature and some increase?