I. PROPERTIES OF SOLUTIONS
• SOLUTION: - a homogeneous mixture
- looks the same throughout
• SOLUBILITY: the amount of solute (dissolved substance) that
dissolves in an of solvent (substance that does the dissolving) at a
specific temperature
• SATURATED: a solution containing the maximum amount of
dissolved solute
• UNSATURATED: a solution containing less than the maximum
amount of dissolved solute
• MISCIBLE: describes liquids that DO dissolve in each other
Ex. Water and vinegar
• IMMISCIBLE: describes liquids that DO NOT dissolve in each other
Ex. Water and oil
QUESTION:
Why would it be wrong to say that sugar
and water are miscible?
FACTORS AFFECTING SOLUBILITY:
1. Temperature:
- SOLIDS are MORE soluble in liquids at HIGHER temps.
- GASES are LESS soluble in liquids at HIGHER temps.
Why does warm soda go “flat?”
Soda “goes flat” when the CO2 gas escapes from the liquid;
At a warm temp. the CO2 gas is less soluble, so the gas
escapes the liquid faster
2. Particle size:
- SMALLER particles dissolve FASTER
3. Pressure:
- GASES are MORE soluble in liquids at HIGHER pressures
**4. Agitation:
- Particles dissolve faster when stirred, mixed, shaken, etc.
Solubility of compounds in 100g of H2O as a function of increasing temperature
Solubility of NaNO3
At 15°C?
85 g/ 100g H2O
Solubility of KClO3
at 70°C?
30 g/ 100g H2O
Use Figure 18.4 on pg. 504 to answer the following questions:
1.) What is the solubility of NH4Cl in water at 35°C? 40 g / 100 g H2O
2.) What is the solubility of NaNO3 in water at 10°C? 80 g / 100 g H2O
3.) What is the solubility of KBr in water at 65°C? 90 g / 100 g H2O
4.) What is the solubility of KNO3 in water at 70°C? 120 g / 100 g H O
2
5.) Which compound’s solubility is relatively unaffected by
increasing temperature? How do you know?
NaCl, b/c its solubility curve remains relatively flat
6.) Which compound’s solubility follows a trend opposite of the
others? How do you know?
Yb2(SO4)3 ; its solubility decreases with increasing temp.
• HENRY’S LAW:
S1 x P2 = S2 x P1
S = solubility P = pressure
Use to calculate:
The solubility of a gas in a
liquid at different pressures
Ex 1 : If the solubility of a gas in water is 0.77 g/L at 3.5 atm of
pressure, what is its solubility at 1.0 atm?
S1 = 0.77 g/L
S1 x P2 = S2 x P1
P1 = 3.5 atm
S2 = ?
P2 = 1.0 atm
S1 x P2 = S2
P1
0.77 g/L x 1.0 atm = S2
3.5 atm
0.22 g/L= S2
Note: Pressure decreased so
solubility decreased
Ex 2: If the solubility of CO2 gas in water is 1.87 g/L at 10.5 atm of
pressure, what must the pressure be in order for the solubility
to reach 2.56 g/L?
S1 = 1.87 g/L
S1 x P2 = S2 x P1
P1 = 10.5 atm
S2 = 2.56 g/L
P2 = ?
P2 = S2 x P1
S1
P2 = 2.56 g/L x 10.5 atm
1.87 g/L
Note: Pressure had to
P2 = 14.37 atm increase for the
solubility to increase
Page 507: 1, 2
1:
S1 = 0.16 g/L
S1 x P2 = S2 x P1
P1 = 104 kPa
S2 = ?
P2 = 288 kPa
S1 x P2 = S2
P1
(0.16 g/L) x (288 kPa) = S2
(104 kPa)
0.44 g/L= S2
Note: Pressure increased so
solubility increased
2:
S1 = 3.6 g/L
S1 x P2 = S2 x P1
P1 = 1.0 atm
S2 = 9.5 g/L
P2 = ?
P2 = S2 x P1
S1
P2 = (9.5 g/L) x (1.0 atm)
(3.6 g/L)
P2 = 2.64 atm
Note: Solubility increased b/c
pressure increased
Page 508: Section Review 18.1 (3, 4, 6)
3. temperature, particle size, and pressure
4. By using Henry’s Law : S1 x P2 = S2 x P1
6. a. Add more solvent
6. b. Add more solute
Practice: The solubility of CO2 in a can of soda is 4.2 g/L at 3.2 atm
of pressure. What is the solubility of the CO2 gas when the
can of soda is opened and the pressure drops to 1.0 atm?
What would happen to the solubility of the CO2 if the
temperature went up and WHY?
S1 = 4.2 g/L
S1 x P2 = S2 x P1
P1 = 3.2 atm
S2 = ?
P2 = 1.0 atm
S1 x P2 = S2
P1
(4.2 g/L) x (1.0 atm) = S2
(3.2 atm)
1.3 g/L= S2
Note: Pressure decreased so
solubility decreased
II. CONCENTRATIONS OF SOLUTIONS
• CONCENTRATION: a measure of the amount of solute dissolved in a
given amount of solvent (a general term; no dependence on temp.)
o DILUTE SOLUTION: contains a small amount of solute
o CONCENTRATED SOLUTION: contains a large amount of solute
• MOLARITY (M): the unit for concentration
M = mol
L
mol
M
L
Used to calculate:
mol = M x L
M = mol
L
L=
mol
M
Ex 1: What is the molarity of a solution containing 4.5 moles of KBr
in 3.0 L of solution?
M= ?
mol = 4.5 mol
L= 3.0 L
M = mol
L
M = 4.5 mol
3.0 L
= 1.5 M or 1.5 mol/ L
Ex 2: A saline solution contains 0.90 g of NaCl in exactly 100 mL of
solution. What is the molarity of the solution?
M= ? (this is what we solve for)
mol = ? (we get this from grams)
L= ? (we get this from mL)
0.90 g NaCl x
0.015 mol
0.100 L
1
mol NaCl =
________________
58.5
g NaCl
0.015 mol NaCl
(23.0) + (35.5)
100 mL = 0.100 L (move the decimal 3 places left)
M = mol
L
M = 0.015 mol
0.100 L
= 0.15 M (or mol/L)
Ex 3: a.) How many moles of CaCl2 is needed to make 2.0 L of a 0.50 M
solution?
b.) How many grams is this?
M= 0.50 M (mol/ L)
mol = ? (from moles, we can get the mass in grams)
L= 2.0 L
mol = M x L
mol
M
mol = (0.50 mol/L) x (2.0 L)
mol = 1.0 mol CaCl2
L
(40.1) + (2 x 35.5)
1.0 mol CaCl2 x
111.1 g CaCl2
________________
=
1
mol CaCl2
111.1 g CaCl2
Page 511:
#9: M= ?
mol = 0.70 mol
L= 0.250 L (250 mL = 0.250 L)
M = mol
L
#10:
M = (0.70 mol)
(0.250 L)
M = 2.8 mol/L
M= 0.425 mol/L
mol = ?
L= 0.335 L (335 mL = 0.335 L)
mol = M x L
mol = (0.425 mol/L) x (0.335 L)
mol = 0.14 mol NH4NO3
#11:
M= 2.0 mol/L
mol = ?
L= 0.250 L (250 mL = 0.250 L)
mol = M x L
mol = (2.0 mol/L) x (0.250 L)
mol = 0.50 mol CaCl2
0.50 mol CaCl2 x
111.1 g CaCl2
________________
1
mol CaCl2
=
55.6 g CaCl2
Page 515:
#21a:
M= ? (this is what we solve for)
mol = ? (we get this from 400 grams) 2.5 mol
L= 4.00 L
400 g CuSO4 x
M = mol
L
1
mol CuSO4 =
________________
159.6 g CuSO4
M = (2.5 mol)
(4.00 L)
2.5 mol CuSO4
M = 0.625 mol/L
Page 515:
#21b: M= ?
mol = 0.06 mol
L= 1.500 L (1500 mL = 1.500 L)
M = mol
L
M = (0.06 mol)
(1.500 L)
M = 0.04 mol/L
• DILUTION FORMULA: M x V = M x V
1
1
2
2
Use to calculate:
The molarity or volume
of a solution made
from another solution
*Volume can be in L or mL. You don’t have to convert!
Making Dilutions: moles of solution before dilution =
moles of solution after dilution
*There are no mole or gram conversions in these problems!!
Ex 1: To prepare 100.L of 0.400 M MgSO4, how many L of
0.200 M MgSO4 would you need?
M1 = 0.400 M
M1 x V1 = M2 x V2 (0.400 M) x (100. L) = V2
V1 = 100. L
M2 = 0.200 M
V2 = ?
(0.200 M)
M1 x V1 = V2
M2
200 L= V2
Ex 2: 500 mL of 2.5 M NaCl solution is diluted to 750 mL. What is
the new molarity of the solution?
M1 = 2.5 M
M1 x V1 = M2 x V2 (2.5 M) x (500 mL) = M2
V1 = 500 mL
M2 = ?
V2 = 750 mL
(750 mL)
M1 x V1 = M2
V2
1.7 M = M2
Pg. 513:
12:
M1 = 0.760 M
M1 x V1 = M2 x V2 (0.760 M) x (250.0 mL) = V2
V1 = 250.0 mL
M2 = 4.00 M
V2 = ?
(4.0 M)
M1 x V1 = V2
M2
47.5 mL = V2
13:
M1 = 0.20 M
M1 x V1 = M2 x V2 (0.20 M) x (250 mL) = V2
V1 = 250 mL
M2 = 1.0 M
V2 = ?
(1.0 M)
M1 x V1 = V2
M2
50 mL = V2
Pg. 514:
14:
?
=
10 mL
200 mL
% v/v = ?
Vol. solute (acetone) = 10 mL
Vol. solution = 200.0 mL
5%
x 100
• Percent by volume (%(v/v)):
% v/v = Volume solute
x 100
Volume solution
Use to calculate:
%v/v or vol. of solute
Ex 1: What is the percent by volume of a solution of rubbing
alcohol that contains 25.0 mL of alcohol in 85.0 mL of
solution?
% v/v = ?
25.0 mL
alcohol
85.0 mL
Vol. solute (alcohol) = 25.0 mL
Vol. solution = 85.0 mL
% v/v = 25.0 mL x 100
85.0 mL
= 29.4%
Ex 2: A bottle of hydrogen peroxide (H2O2) antiseptic is labeled 3.0%
H2O2 by volume. How many mL of H2O2 are in a 400 mL
bottle of this solution?
% v/v = 3.0 %
Vol. solute (H2O2) = ?
Vol. solution = 400 mL
12 mL H2O2
3.0 % =
?
400 mL
(400 mL ) x (3.0 %) = ?
100
400 mL
388 mL H2O
x 100
12 mL = vol. H2O2
IV. CALCULATIONS INVOLVING COLLIGATIVE PROPERTIES
Because volume depends on temperature and molarity depends on
Volume, we need a way of expressing solution concentration when
Temperature changes. To do that, we use molality:
• MOLALITY (m):
m = mol solute
kg solvent
Although we could use this
formula to solve for m, mol,
or kg, we will focus on using it
only to find molality (m)
Ex : Calculate the molality of a solution prepared by dissolving
7.5 mol LiCl in 2.1 kg of H2O.
Solute = LiCl
Solvent = H2O
m = ? (this is what we solve for)
mol = 7.5 mol
kg = 2.1 kg
m = mol
kg
m = 7.5 mol
2.1 kg
= 3.6 mol/kg
Ex : What is the molality of a solution containing 10.0 g NaCl
dissolved in 600. g of H2O?
Solute = NaCl
Solvent = H2O
m = ? (this is what we solve for)
mol = ? (we get this from g NaCl) 0.17 mol
kg =
? (we get this from g H2O)
10.0 g NaCl x
0.600 kg
1
mol NaCl =
________________
58.5
g NaCl
0.17 mol NaCl
600. g = 0.600 kg (move the decimal 3 places left)
m = mol
kg
m = 0.17 mol
0.600 kg
= 0.28 mol/kg
III. COLLIGATIVE PROPERTIES OF SOLUTIONS
• COLLIGATIVE PROPERTIES:
- depend only on the number of particles dissolved in a solution
NOT on what they are
→ Remember, only IONIC compounds (metal & nonmetal)
break apart in H2O, MOLECULAR do NOT
Ex. How many particles are released when each compound dissolves
in water?
NaCl 2
CaF2 3 MgO 2
LiBr 2 K2S 3
• BOILING POINT ELEVATION: the increase in temperature between
the boiling point of a solution & the boiling point of a pure solvent
Ex. Salt water boils ABOVE 100°C
• FREEZING POINT DEPRESSION: the decrease in temperature
between the freezing point of a solution & the freezing point of a
pure solvent
Ex. Salt water freezes BELOW 0°C
● BOILING POINT ELEVATION:
∆Tb = Kb x m x n
∆Tb : change in
Kb : Constant
boiling temp.
for the
solvent
m: molality n: # ions in
solution
M & _____),
NM
** If the solution contains an IONIC solid ( ______
you must multiply the molality of the solution by the number
of ions in solution
Ex. NaCl (aq) → Na+ (aq) + Cl- (aq) ____
2 ions, so n = 2
MgCl2 (aq) → Mg2+ (aq) + 2 Cl-(aq) ____
3 ions, so n = 3
Ex 1: What is the boiling point of an aqueous 1.50 m NaCl solution?
Kb for H2O = 0.512°C/m.
∆Tb = Kb x m x n
∆Tb = ?
Kb = 0.512 °C/m
m
= 1.50 m
n
= 2 (Na+ and Cl-)
∆Tb = (0.512 °C/m) x (1.50 m) x (2)
∆Tb = 1.5°C
New boiling pt = Original b.p. of solvent + ΔTb
New boiling pt = 100°C + 1.5°C = 101.5°C
Ex 2: What is the boiling point of a solution that contains 1.25 mol
CaCl2 in 1.4 kg of H2O? Kb for H2O = 0.512°C/m
∆Tb = Kb x m x n
∆Tb = ?
Kb = 0.512 °C/m
1.25 mol CaCl2 = 0.89 m
m = 1.4 kg H O
2
n
= 3 (Ca2+ and 2 Cl-)
∆Tb = (0.512 °C/m) x (0.89 m) x (3)
∆Tb = 1.4°C
New boiling pt = Original b.p. of solvent + ΔTb
New boiling pt = 100°C + 1.4°C = 101.4°C
● FREEZING POINT DEPRESSION:
∆Tf = Kf x m x n
∆Tf : change in
Kf : Constant m: molality
freezing temp.
for the
solvent
n: # ions in
solution
** These problems are solved the same way as boiling point elevation
problems except Kf is used and the ∆T tells you how much the
temperature goes down
Ex 1: What is the freezing point of a 0.20 m aqueous K2SO4
solution? Kf for H2O = 1.86°C/m.
∆Tf = Kf x m x n
∆Tf = ?
Kf =
1.86 °C/m
m
= 0.20 m
n
= 3 (2 K+ and 1 SO42-)
∆Tf = (1.86 °C/m) x (0.20 m) x (3)
∆Tf = 1.1°C
New freezing pt = Original f.p. of solvent - ΔTf
New freezing pt = 0.0°C - 1.1°C = - 1.1°C
Ex 2: An aqueous solution of glucose (C6H12O6) in water has a
concentration of 0.45 m. What is the freezing point of the
solution? Kf for H2O = 1.86°C/m.
∆Tf = Kf x m x n
∆Tf = ?
Kf =
1.86 °C/m
m
= 0.45m
n
= 1 (glucose is NOT ionic!)
∆Tf = (1.86 °C/m) x (0.45 m) x (1)
∆Tf = 0.8°C
New freezing pt = Original f.p. of solvent - ΔTf
New freezing pt = 0.0°C - 0.8°C = - 0.8°C
Ex 1: What is the freezing point of a solution in which 12.0 mol CCl4
is dissolved in 75.0 kg C6H6? Kf (°C/m) for C6H6 is 5.12 and the
freezing point is 5.5˚C.
∆Tf = Kf x m x n
∆Tf = ?
Kf =
5.12 °C/m
= 0.16 m
m
= 12.0 mol CCl4
75.0 kg C6H6
n
= 1 (CCl4 is not ionic so it doesn’t break apart in water)
∆Tf = (5.12 °C/m) x (0.16 m) x (1)
∆Tb = 0.8°C
New freezing pt = Original f.p. of solvent - ΔT
New freezing pt = 5.5°C - 0.8°C = 4.7°C
Pg. 528: 55.a.
What is the boiling point of a solution that contains:
0.50 mol glucose (molecular) in 1000 g of H2O?
(Kb (°C/m) for H2O is 0.512.)
∆T = K x m x n
b
b
∆Tb = ?
Kb = 0.512 °C/m
0.50 mol glucose = 0.50 m
m = 1.000 g H O
2
n
= 1 (b/c glucose is not ionic, its doesn’t break apart)
∆Tb = (0.512 °C/m) x (0.50 m) x (1)
∆Tb = 0.3°C
New boiling pt = Original b.p. of solvent + ΔT
New boiling pt = 100°C + 0.3°C = 100.3°C
Pg. 528: 55.a.
What is the boiling point of a solution that contains:
1.50 mol NaCl in 1000 g of H2O?
(Kb (°C/m) for H2O is 0.512.)
∆T = K x m x n
b
b
∆Tb = ?
Kb = 0.512 °C/m
1.50 mol NaCl
m = 1.000 g H O
2
n
= 1.50 m
= 2 (Na+ and Cl-)
∆Tb = (0.512 °C/m) x (1.50 m) x (2)
∆Tb = 1.5°C
New boiling pt = Original b.p. of solvent + ΔT
New boiling pt = 100°C + 1.5°C = 101.5°C
Ex : Calculate the molality of a solution prepared by dissolving
24.5 g NaCl in 2.5 kg of H2O.
Solute = NaCl
Solvent = H2O
m = ? (this is what we solve for)
mol = ? (we get this from grams NaCl) 0.42 mol
kg = 2.5 kg
24.5 g NaCl x
m = mol
kg
1
mol NaCl =
________________
58.5
g NaCl
m = 0.42 mol
2.5 kg
0.42 mol NaCl
= 0.17 mol/kg
Ex : How many moles of KBr must be dissolved in 2.0 kg of water to
form a 3.1 m solution?
Solute = KBr
Solvent = H2O
m = 3.1 mol/ kg
mol = ?
kg = 2.0 kg
mol = m x kg
mol = 3.1 mol x 2.0 kg
kg
mol
m
kg
= 6.2 mol
Ex : How many grams of LiF must be dissolved in 4.5 kg of water to
form a 2.8 m solution?
Solute = LiF
Solvent = H2O
m = 2.8 mol/ kg
mol = ? (from moles, we can find grams)
kg = 4.5 kg
mol = m x kg
mol = 2.8 mol x 4.5 kg = 12.6 mol
kg
mol
m
kg
25.9 g LiF = 326.3 g LiF
12.6 mol LiF x _____________
1 mol LiF