Chapter 14: Mixtures and Solutions
Types of Mixtures
Heterogeneous
Homogeneous
Simple Suspension
Colloid
Solution
Visible
Greater than 1 µm
Microscopic
1 µm down to 1 nm
Molecular / Ionic
1 nm down to 62 pm
Settles Out
Yes
No
No
Key Feature
Easily Separated
by Filtering
Tyndall Effect and
Brownian Motion
Clear
Particle Size
Soluble: solid, liquid, or gas solutes can be dissolved into a liquid solvent
Miscible: two liquids that are soluble in any proportions
Insoluble: substances that will not dissolve in a solvent
Immiscible: two liquids that separate upon standing
Saturation
Unsaturated: less than the maximum amount of solute is dissolved in the solvent
Saturated: the maximum amount of solute is dissolved in the solvent
Supersaturated: more than the usual maximum amount of solute is dissolved in the solvent
Example 1: A student dissolves 29 g of KNO3 in 50. g of water at 40°C. Is the solution
saturated or unsaturated?
First, use the solubility chart in Table G of the Reference tables to find the saturation
concentration = 64 g KNO3 / 100.g H2O at 40°C
50.g
For 50. g of H2O:
.
32g
Since the 29 g used is less than the 32 g required for saturation, the solution is unsaturated
Example 2: If the solution made in Example 1 sits overnight and cools to 30°C, is the
solution unsaturated, saturated, or supersaturated?
Check the solubility chart in Table G of the Reference tables to find the saturation
concentration = 47 g KNO3 / 100.g H2O at 30°C
50.g
For 50. g of H2O:
.
23.5g
Since the 29 g used is more than 23.5 g, the solution is supersaturated
Solution concentration
Volume – Volume: most often used to make solutions where the concentration is not critical
Percent by Volume:
100%
Example: If water is added to 35 mL sample of isopropyl alcohol to produce 50. mL of
solution, what is the volume percentage of alcohol in the solution?
Alcohol = 35 mL
%vol = Vsolute / Vsolution ) x 100%
Solution = 50. mL
= ( 35 mL / 50. mL ) x 100%
= 70. % isopropanol
Mass – Volume: used to make solutions dispensed by volume when concentration is critical
Molarity (M):
Example 1: What is the molar concentration of a solution made by dissolving
0.633 moles NaCl in enough water to form 0.250 L of solution?
0.633moles
M
2.53M
0.250L
Dilution Equation for Molarity: M1V1 = M2V2
Example 1A: If 250. mL of a 3.75 M solution of NaOH is diluted to 725 mL, what is the
new concentration?
M1 = 3.75 M
V1 = 250. mL
M2 = ?
V2 = 725 mL
M V
3.75M 250. mL
M V
M V ∴ M
1.29M
V
725mL
Mass – Mass: used to make solutions where the mole fraction of solute and solvent is critical
Percent by Mass (%mass):
100%
Example 1: What is the mass percent concentration of a saturated NaCl solution at 20°C?
Solute = 37 g NaCl
%mass = msolute / msolution ) x 100%
Solvent = 100. g H2O
= ( 37 g / 137 g ) x 100%
= 27 % NaCl
Parts Per Million (ppm):
Example 2: What is the concentration in ppm of a solution containing 5.0 x 10–4 moles of
lead (II) ions in 500 mL of water?
(Safe level in water is 15 ppb)
First, convert moles to mg:
207.2g
5.0 10– mole
0.104g 104mg
1mole
Next, even given the very large difference between the density of lead and of water,
assuming the solution to be pure water gives a percent error far below the number of
significant figures in use here, so:
Solute = 104 mg Pb2+
ppmPb = mgPb / kgsolution
Solvent = 500. g H2O
= 104 mg / 0.50 kg
= 210 ppm Pb2+
Mole Fraction (Xa):
Example 3: What is the mole fraction of a saturated NaCl solution at 20°C?
Solute
= 37 g NaCl
= 37g
Solvent = 100. g H2O = 100. g
n
n n .
0.633mole
.
0.633moles
0.633moles 5.55moles
5.55mole
0.102NaCl
Molality (m):
Example 4: What is the molal concentration of a solution made by dissolving
0.633 moles NaCl in 0.200 L of solution?
0.633moles
3.17
0.200kg
The solvation process
Hydration Spheres or Solvation Shells
from http://en.wikipedia.org/wiki/Solvation_shell
What would the Cl–(aq) ion look like?
How many hydration spheres do you think would exist
in a solution?
How would increasing temperature affect the number of
hydration spheres?
Factors affecting solvation
Agitation
Surface area
Temperature
Pressure (gases) and Henry’s Law:
Colligative Properties – depend only upon the number of solute particles in the solution
Vapor pressure lowering
Boiling point elevation
Freezing point depression
Osmotic pressure
The effect of a solute on colligative properties is best explained by a Boltzmann distribution
The addition of a solute lowers the mole fraction of solvent such that at the normal boiling
temperature fewer solvent molecules are available to move into the vapor phase which
requires a higher temperature to increase raise the vapor pressure back to one atmosphere
Electrolytes and nonelectrolytes
Nonelectrolytes – molecules that do not ionize in solution
Electrolytes – compounds that form ions in solution and add multiple particles
van’t Hoff factor: i = the number of discrete ions in a formula unit
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Calculating boiling point elevation: ∆Tb = Kb m i
Example: Find the boiling point of a saturated solution of NaCl
Table G of the Reference Tables: 40. g of NaCl dissolve in 0.100 kg of water at 100°C
1mol
0.6844mol
n 40. g
58.443g
0.6844mol
6.844
0.100kg
Kb for water = 0.512 °C m–1
i = 2 for NaCl [Na+(aq) and Cl–(aq)]
∆Tb = Kb m i
= ( 0.512 °C m–1 )( 6.844 m )( 2 )
= 7.01°C
New boiling point = normal boiling point + ∆Tb = 100°C + 7.01°C = 107.01°C
Calculating freezing point depression: ∆Tf = Kf m i
Example: Find the freezing point of a saturated solution of NaCl
Table G of the Reference Tables: 37 g of NaCl dissolve in 0.100 kg of water at 100°C
1mol
n 37g
0.6331mol
58.443g
0.6331mol
6.331
0.100kg
Kf for water = –1.86 °C m–1
i = 2 for NaCl [Na+(aq) and Cl–(aq)]
∆Tf = Kf m i
= ( –1.86 °C m–1 )( 6.331 m )( 2 )
= –23.6°C
New freezing point = normal freezing point + ∆Tf = 0°C – 23.6°C = –23.6°C