JJ

2. STOICHIOMETRY IN
SOLUTIONS
 Identify the compound/element present
in the combined solution.
 Write the balanced net ionic equation.
 Calculate the moles of reactants.
 Calculate the moles of product(s).
 Convert to grams or other units.

3. 3
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
First write a balanced
equation.
Al(s) + HCl(aq)  AlCl3(aq) + H2(g)
2 6 2 3
Gram to Gram Conversions

4. 4
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
Al(s) + HCl(aq)  AlCl3(aq) + H2(g)
2 6 2 3
Now let’s get organized.
Write the information
below the substances.
3.45 g ? grams
Gram to Gram Conversions

5. 5
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
Al(s) + HCl(aq)  AlCl3(aq) + H2(g)
2 6 2 3
3.45 g ? grams
Let’s work the problem.
= g AlCl3
3.45 g Al
Al
g
27.0
Al
mol
We must always convert to moles.
Now use the molar ratio.
Al
mol
2
AlCl
mol
2 3
Now use the molar mass to convert
to grams.
3
3
AlCl
mol
AlCl
g
133.3
17.0
Units match
gram to gram conversions

6. 6
Molarity
Molarity is a term used to express concentration. The units of molarity are
moles per liter (It is abbreviated as a capital M)
When working problems, it
is a good idea to change M
into its units.
mL
1000
moles
Liter
moles
M 


7. 7
A solution is prepared by dissolving 3.73 grams of AlCl3 in
water to form 200.0 mL solution. A 10.0 mL portion of the
solution is then used to prepare 100.0 mL of solution.
Determine the molarity of the final solution.
What type of
problem(s) is
this?
Molarity
followed by
dilution.
Solutions

8. 8
A solution is prepared by dissolving 3.73 grams of AlCl3 in
water to form 200.0 mL solution. A 10.0 mL portion of the
solution is then used to prepare 100.0 mL of solution.
Determine the molarity of the final solution.
1st:
= mol
L
3.73 g
g
133.4
mol
200.0 x 10-3 L
0.140
2nd: M1V1 = M2V2
(0.140 M)(10.0 mL) = (? M)(100.0 mL)
0.0140 M = M2
molar mass of AlCl3
dilution formula
final concentration
Solutions

9. 9
Solution Stoichiometry:
Determine how many mL of 0.102 M NaOH solution are needed to
neutralize 35.0 mL of 0.125 M H2SO4 solution.
First write a balanced
Equation.
____NaOH + ____H2SO4  ____H2O + ____Na2SO4
2 1 2 1

10. 10
Solution Stoichiometry:
Determine how many mL of 0.102 M NaOH solution is needed to neutralize
35.0 mL of 0.125 M H2SO4 solution.
Now, let’s get organized. Place
numerical Information and
accompanying UNITS below each
compound.
____NaOH + ____H2SO4  ____H2O + ____Na2SO4
2 1 2 1
0.102 M
L
mol
? mL
35.0 mL
mL
1000
mol
0.125
L
mol
0.125

Since 1 L = 1000 mL, we can use
this to save on the number of conversions
Our Goal

11. 11
Determine how many mL of 0.102 M NaOH solution is needed to neutralize
35.0 mL of 0.125 M H2SO4 solution.
Now let’s get to work
converting.
____NaOH + ____H2SO4  ____H2O + ____Na2SO4
2 1 2 1
0.102 M
L
mol
? mL
35.0 mL
mL
1000
mol
0.125
L
mol
0.125

= mL NaOH
H2SO4
35.0 mL
H2SO4
0.125 mol
1000 mL
H2SO4
NaOH
2 mol
1 mol
H2SO4
1000 mL NaOH
0.102 mol NaOH
85.8
Units Match
Solution Stoichiometry:

12. 12
What volume of 0.40 M HCl solution is needed to
completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
1st write out
a balanced chemical
equation
Solution Stoichiometry

13. 13
What volume of 0.40 M HCl solution is needed to
completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
2HCl(aq) + Ba(OH)2(aq)  2H2O(l) + BaCl2
0.40 M 47.1 mL
0.75 M
? mL
= mL HCl
Ba(OH)2
47.1 mL
2
2
Ba(OH)
Ba(OH)
mL
1000
0.75mol
1 mol
Ba(OH)2
HCl
2 mol
0.40 mol
HCl
HCl
1000 mL
176
Units match
Solution Stoichiometry

14. 14
Solution Stochiometry Problem:
A chemist performed a titration to standardize a barium hydroxide solution.
If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of
the barium hydroxide solution, what was the concentration of the barium
hydroxide solution in moles per liter (M)?
First write a balanced
chemical reaction.
____HCl(aq) + ____Ba(OH)2(aq)  ____H2O(l) + ____BaCl2(aq)
2 1 2 1
23.28 mL
0.135 mol
L
25.00 mL
? mol
L

15. 15
48.0 mL of Ca(OH)2 solution was titrated with 19.2
mL of 0.385 M HNO3. Determine the molarity of
the Ca(OH)2 solution.
We must first
write a balanced
equation.
Solution Stochiometry Problem:

16. 16
48.0 mL of Ca(OH)2 solution was titrated with 19.2
mL of 0.385 M HNO3. Determine the molarity of
the Ca(OH)2 solution.
Ca(OH)2(aq) + HNO3(aq)  H2O(l) + Ca(NO3)2(aq)
2 2
48.0 mL 19.2 mL
0.385 M
L
mol
0.385

= mol(Ca(OH)2)
L (Ca(OH)2)
19.2 mL
HNO3
3
3
HNO
HNO
mL
1000
mol
0.385
3
2
HNO
2mol
Ca(OH)
1mol
48.0 x 10-3L
? M
units match!
0.0770
Solution Stochiometry Problem:

17. 17
Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15 mol 0.10 mol ? moles
Two starting
amounts?
Where do we
start?
Hide
one

18. 18
Calculate the molarity of a solution prepared by dissolving 25.6 grams of
Al(NO3)3 in 455 mL of solution.
L
mol
0.264
L
10
x
455
g
213
mole
g
25.6
3
-

After you have
worked the
problem, click here
to see
setup answer
Try this problem (then check your answer):

20.  Polar liquids tend to dissolve readily in polar solvents.
 Pairs of liquids that mix in all proportions are said to
be miscible and liquid that do not mix are immiscible.
 Hydrogen bonding interactions between solute and
solvent may lead to high solubility.
 The solubility of alcohol decreases as the number of
carbons increases.
 like dissolves like
 Non polar substances are soluble in non polar
solvents.
Nature of the solute
and solvent

21. Temperature
 The solubility of most solid solutes in water
increases as the temperature of the solution
increases.
 In contrast, the solubility of gases in water
decreases with increasing temperature.

22. Pressure
 The solubilities of solids and liquids are
not appreciably affected by pressure.
 The solubility of a gas in any solvent
increases as the pressure of the gas
over the solvent increases.
 E.g. Carbonated Beverages.

23. ASSESSMENT
How does the following factors affect the solubility of a
solution:
a. Nature of Solute and Solvent
b. Temperature (solid, liquid, gas)
c. Pressure (solid, liquid, gas)

25. Colligative Properties
 Solution properties that depend on the
number of solute particles present.
 Among colligative properties are
Vapor pressure lowering
Boiling point elevation
Melting point depression
Osmotic pressure

26. Vapor Pressure
As solute molecules are
added to a solution,
the solvent become
less volatile.
Solute-solvent
interactions contribute
to this effect.

27. Vapor Pressure
Therefore, the vapor
pressure of a solution
is lower than that of
the pure solvent.

28. Raoult’s Law
PA = XAPA
where
• XA is the mole fraction of compound A
• PA is the normal vapor pressure of A at that
temperature
NOTE: This is one of those times when you
want to make sure you have the vapor
pressure of the solvent.

29. SAMPLE EXERCISE 13.8 Calculation of Vapor-
Pressure Lowering
Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a
density of 1.26 g/mL at 25°C. Calculate the vapor pressure at
25°C of a solution made by adding 50.0 mL of glycerin to
500.0 mL of water. The vapor pressure of pure water at 25°C is
23.8 torr.
PRACTICE EXERCISE
The vapor pressure of pure water at 110°C is 1070 torr. A
solution of ethylene glycol and water has a vapor pressure of
1.00 atm at 110°C. Assuming that Raoult’s law is obeyed, what
is the mole fraction of ethylene glycol in the solution?

30. Boiling Point Elevation and
Freezing Point Depression
Solute-solvent
interactions also
cause solutions to
have higher boiling
points and lower
freezing points than
the pure solvent.

31. Boiling Point Elevation
The change in boiling
point is proportional to
the molality of the
solution:
Tb = Kb  m
where Kb is the molal
boiling point elevation
constant, a property of
the solvent.
Tb is added to the normal boiling
point of the solvent.

32. Freezing Point Depression
 The change in freezing
point can be found
similarly:
Tf = Kf  m
 Here Kf is the molal
freezing point
depression constant of
the solvent.
Tf is subtracted from the normal freezing
point of the solvent.

33. Boiling Point Elevation and
Freezing Point Depression
In both equations,
T does not
depend on what
the solute is, but
only on how many
particles are
dissolved.
Tb = Kb  m
Tf = Kf  m

34. Colligative Properties of
Electrolytes
Because these properties depend on the number of
particles dissolved, solutions of electrolytes (which
dissociate in solution) show greater changes than
those of nonelectrolytes.
e.g. NaCl dissociates to form 2 ion particles; its limiting
van’t Hoff factor is 2.

35. Colligative Properties of
Electrolytes
However, a 1 M solution of NaCl does not show
twice the change in freezing point that a 1 M
solution of methanol does.
It doesn’t act like there are really 2 particles.

36. Osmosis
 Semipermeable membranes allow some
particles to pass through while blocking
others.
 In biological systems, most
semipermeable membranes (such as cell
walls) allow water to pass through, but
block solutes.

37. Osmosis
In osmosis, there is
net movement of
solvent from the area
of higher solvent
concentration (lower
solute concentration)
to the are of lower
solvent
concentration
(higher solute
concentration).
Water tries to equalize the concentration on both sides until
pressure is too high.

38. Osmotic Pressure
 The pressure required to stop osmosis,
known as osmotic pressure, , is
n
V
 = ( )RT = MRT
where M is the molarity of the solution
If the osmotic pressure is the same on both sides of a membrane
(i.e., the concentrations are the same), the solutions are isotonic.

39. Osmosis in Blood Cells
 If the solute
concentration outside
the cell is greater than
that inside the cell, the
solution is hypertonic.
 Water will flow out of
the cell, and crenation
results.

40. Osmosis in Cells
 If the solute
concentration outside
the cell is less than
that inside the cell, the
solution is hypotonic.
 Water will flow into the
cell, and hemolysis
results.

42. Molar Mass from
Colligative Properties
We can use the
effects of a colligative
property such as
osmotic pressure to
determine the molar
mass of a compound.