1. Physical Properties of Solutions
Chapter 13
Solution Stoichiometry

2. A solution is a homogenous mixture of 2 or more substances
The solute is(are) the substance(s) present in the smaller amount(s)
The solvent is the substance present in the larger amount
12.1

3. An electrolyte is a substance that, when dissolved in water, results in a solution that can conduct electricity.
A nonelectrolyte is a substance that, when dissolved, results in a solution that does not conduct electricity.
nonelectrolyte
weak electrolyte
strong electrolyte
4.1

4. A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.
An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature.
A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature.
Sodium acetate crystals rapidly form when a seed crystal is
added to a supersaturated solution of sodium acetate.
12.1

5. 1. The ease of solution depends on two factors
The solution Process
1. The ease of solution depends on two factors
a. the change in energy (enthalpy)
b. the change in disorder (Entropy)
2. The solution process is favored when there is a
decrease in enthalpy an increase in entropy

9. 3. Heat of solution Hsolution
a. if a solution gets hotter as a substance dissolves
- energy is being released
i. -Hsolution designates the release of energy
ii. The more negative the Hsolution, the more
favorable the solution process

10. 4. Main interactions that affect the solution process
strong solvent - solvent attractions does not favor solubility
weak solvent-solvent attractions favor solubility
weak solute - solute attractions favor solubility
5. three process
a. breaking apart the solute particles (endothermic)
b. breaking apart solvent solvent particles (endothermic)
c. interactions of solute and solvent particles (exothermic)

11. Three types of interactions in the solution process:
solvent-solvent interaction
solute-solute interaction
solvent-solute interaction
Requires heat
Requires heat
Gives off heat
DHsoln = DH1 + DH2 + DH3
12.2

12. “like dissolves like”
Two substances with similar intermolecular forces are likely to be soluble in each other.
non-polar molecules are soluble in non-polar solvents
CCl4 in C6H6
polar molecules are soluble in polar solvents
C2H5OH in H2O
ionic compounds are more soluble in polar solvents
NaCl in H2O or NH3 (l)
12.2

13. F. dissolution of solids in liquids
the process in which solvent molecules surround and
interact with solute ions or molecules is called solvation
When the solvent is water - it is called hydration
G Dissolution of liquids in liquids (miscibility)
1. Miscibility - ability of one liquid to dissolve
in another
miscible - mutually soluble in all proportions

14. Hydration is the process in which an ion is surrounded by water molecules arranged in a specific manner. d+ d-
H2O
4.1

15. H factors that effect solubility
1. Types of solvents and solute
a. General rule - like dissolves like-
polar dissolves polar -
nonpolar dissolves nonpolar
b. nonpolar mixed with polar will be immiscible
2. surface area - cause more collisions of the solute
and the solvent

16. a. in general - as temperature increase so does
solubility for solids
b. for gases - an increase in temperature causes a
decrease in solubility

17. Concentration Units
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
Percent by Mass
x 100%
mass of solute
mass of solute + mass of solvent
% by mass = =
x 100%
mass of solute
mass of solution
Mole Fraction (X)
XA =
moles of A
sum of moles of all components
12.3

18. Concentration Units Continued
Molarity (M)
M =
moles of solute
liters of solution
Molality (m)
m =
moles of solute
mass of solvent (kg)
12.3

19. You must first find the number of moles of H3PO4
What is the molarity of g of H3PO4 in enough water to make
2.00 L
M = moles
Liters
You must first find the number of moles of H3PO4
( 1 mole H3PO4)
(200.0 g H3PO4)
= moles
= M
(98.02 g H3PO4)
2.00 L

20. Molarity continued
How many liters can be prepared of a 2.0 M solution of KOH
if g of KOH is available
Remember Molarity is = moles
Liters
Change Molarity back into these units
2.0 M = 2.0 moles
Liters
This gives us units that we can work with

21. ( ) ( ) ( ) ) ( Change grams to moles 1 mole KOH 400.0 g KOH
= moles KOH
56.01 g KOH ( ) ) (
1 liter
7.142 Moles KOH
= 3.6 L
2.0 Moles KOH

22. First change Molar into mole/liter
How many gram of NaOH are required to make 5.00 L of a
2.5 Molar solution
First change Molar into mole/liter
2.5 Molar = 2.5 moles
liters ( )
40.0 g NaOH ( )
= 500 g
2.5 moles
liter
( 5.00 L)
1 mole NaOH

23. b. Colloids - particles that are intermediate in size.
i. Particles are on the order of 1 to 100 nm
ii. look cloudy
iii. Brownian motion
iv. Tyndall Effect
v. e.g. mayonnaise, gels, foams, smoke, fog, smog

24. Suspensions a. Suspensions - large particles that will settle
if not agitated
i. Particles are on the order of 100 nm,
which is about 1000 times bigger than atoms

25. Temperature and Solubility
Solid solubility and temperature
solubility decreases with increasing temperature
solubility increases with increasing temperature
12.4

26. Temperature and Solubility
Gas solubility and temperature
solubility usually decreases with increasing temperature
12.4

27. Pressure and Solubility of Gases
The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).
c is the concentration (M) of the dissolved gas
c = kP
P is the pressure of the gas over the solution
k is a constant (mol/L•atm) that depends only
on temperature
low P
high P
low c
high c
12.5

28. 4.5

29. Dilution is the procedure for preparing a less concentrated solution from a more concentrated solution.
Dilution
Add Solvent
Moles of solute
before dilution (i)
after dilution (f) =
MiVi
MfVf =
4.5

30. How would you prepare 60.0 mL of 0.2 M
HNO3 from a stock solution of 4.00 M HNO3?
MiVi = MfVf
Mi = 4.00
Mf = 0.200
Vf = 0.06 L
Vi = ? L
Vi =
MfVf Mi =
0.200 x 0.06
4.00
= L = 3 mL
3 mL of acid
+ 57 mL of water
= 60 mL of solution
4.5

31. Equivalence point – the point at which the reaction is complete
Titrations
In a titration a solution of accurately known concentration is added gradually to another solution of unknown concentration until the chemical reaction between the two solutions is complete.
Equivalence point – the point at which the reaction is complete
Indicator – substance that changes color at (or near) the
equivalence point
Slowly add base
to unknown acid
UNTIL
the indicator
changes color
4.7

33. What volume of a 1.420 M NaOH solution is
Required to titrate mL of a 4.50 M H2SO4
solution?
WRITE THE CHEMICAL EQUATION!
H2SO4 + 2NaOH H2O + Na2SO4 M
acid rx
coef. M
base
volume acid
moles acid
moles base
volume base
4.50 mol H2SO4
1000 mL soln x
2 mol NaOH
1 mol H2SO4 x
1000 ml soln
1.420 mol NaOH x
25.00 mL
= 158 mL
4.7

34. Fractional Distillation Apparatus
12.6

35. The Cleansing Action of Soap
12.8

36. Solution Stoichiometry
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
M = molarity =
moles of solute
liters of solution
What mass of KI is required to make 500. mL of
a 2.80 M KI solution?
M KI
M KI
volume KI
moles KI
grams KI
1 L
1000 mL x
2.80 mol KI
1 L soln x
166 g KI
1 mol KI x
500. mL
= 232 g KI
4.5