SolutionsSolutions
Various Types of Solutions
ExampleState of Solution State of Solute
State of Solvent
Air, natural gas Gas Gas Gas
Vodka, antifreeze Liquid Liquid Liquid
Brass Solid Solid Solid
Carbonated water (soda) Liquid Gas Liquid
Seawater, sugar solution Liquid Solid Liquid
Hydrogen in platinum Solid Gas Solid
Solution Composition
AA
moles of soluteMolarity ( ) =
liters of solution
mass of soluteMass (weight) percent = 100%
mass of solution
molesMole fraction ( ) =
total moles of solution
moles of soluteMolality ( ) =
kilogram of s
M
molvent
Molarity
moles of soluteMolarity ( ) =
liters of solution M
Exercise
You have 1.00 mol of sugar in 125.0 mL of solution. Calculate the concentration in units of molarity.
Exercise
You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar?
7
Exercise
Consider separate solutions of NaOH and KCl made by dissolving 100.0 g of each solute in 250.0 mL of solution. Calculate the concentration of each solution in units of molarity.
Mass Percent
mass of soluteMass (weight) percent = 100%
mass of solution
Exercise
What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water?
Mole Fraction
AA
molesMole fraction ( ) =
total moles of solution
Exercise
A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the mole fraction of H3PO4. (Assume water has a density of 1.00 g/mL.)
Molality
moles of soluteMolality ( ) =
kilogram of solvent m
Exercise
A solution of phosphoric acid was made by dissolving 8.00 g of H3PO4 in 100.0 mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.)
Formation of a Liquid Solution
1. Separating the solute into its individual components (expanding the solute).
2. Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent).
3. Allowing the solute and solvent to interact to form the solution.
Steps in the Dissolving Process
Steps in the Dissolving Process
• Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent.
• Step 3 usually releases energy.• Steps 1 and 2 are endothermic, and step 3 is often
exothermic.
Enthalpy (Heat) of Solution
• Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps:
ΔHsoln = ΔH1 + ΔH2 + ΔH3
• ΔHsoln may have a positive sign (energy absorbed) or a negative sign (energy released).
Enthalpy (Heat) of Solution
Concept Check
Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation, be sure to address how ΔH plays a role.
The Energy Terms for Various Types of Solutes and Solvents
H1 H2 H3 Hsoln Outcome
Polar solute, polar solvent Large Large Large, negative
Small Solution forms
Nonpolar solute, polar solvent Small Large Small Large, positive No solution forms
Nonpolar solute, nonpolar solvent
Small Small Small Small Solution forms
Polar solute, nonpolar solvent Large Small Small Large, positive No solution forms
In General
• One factor that favors a process is an increase in probability of the state when the solute and solvent are mixed.
• Processes that require large amounts of energy tend not to occur.
• Overall, remember that “like dissolves like”.
• Structural Effects: Polarity
• Pressure Effects: Henry’s law
• Temperature Effects: Affecting aqueous solutions
Pressure Effects
• Henry’s law: C = kP
C = concentration of dissolved gas
k = constant
P = partial pressure of gas solute above the solution
• Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.
A Gaseous Solute
Temperature Effects (for Aqueous Solutions)
• Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature.
• Predicting temperature dependence of solubility is very difficult.
• Solubility of a gas in solvent typically decreases with increasing temperature.
The Solubilities of Several Solids as a Function of Temperature
The Solubilities of Several Gases in Water
An Aqueous Solution and Pure Water in a Closed Environment
Vapor Pressures of Solutions
• Nonvolatile solute lowers the vapor pressure of a solvent.
• Raoult’s Law:
Psoln = observed vapor pressure of solution
solv = mole fraction of solvent
= vapor pressure of pure solvent
soln solv solv = P P
solvP
A Solution Obeying Raoult’s Law
Nonideal Solutions
• Liquid-liquid solutions where both components are volatile.
• Modified Raoult’s Law:
• Nonideal solutions behave ideally as the mole fractions approach 0 and 1.
Total A A B B = + P P P
Vapor Pressure for a Solution of Two Volatile Liquids
33
Summary of the Behavior of Various Types of Solutions
Interactive Forces Between Solute (A) and Solvent (B)
ParticlesHsoln
T for Solution
Formation
Deviation from
Raoult’s Law
Example
A A, B B A B Zero ZeroNone (ideal
solution)Benzene-toluene
A A, B B < A BNegative
(exothermic)Positive Negative
Acetone-water
A A, B B > A BPositive
(endothermic)Negative Positive
Ethanol-hexane
Concept Check
For each of the following solutions, would you expect it to be relatively ideal (with respect to Raoult’s Law), show a positive deviation, or show a negative deviation?
a) Hexane (C6H14) and chloroform (CHCl3)
b) Ethyl alcohol (C2H5OH) and water
c) Hexane (C6H14) and octane (C8H18)
• Depend only on the number, not on the identity, of the solute particles in an ideal solution: Boiling-point elevation Freezing-point depression Osmotic pressure
Colligative Properties
• Nonvolatile solute elevates the boiling point of the solvent.
• ΔT = Kbmsolute
ΔT = boiling-point elevation
Kb = molal boiling-point elevation constant
msolute= molality of solute
Boiling-Point Elevation
• When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent.
• ΔT = Kfmsolute
ΔT = freezing-point depression
Kf = molal freezing-point depression constant
msolute= molality of solute
Freezing-Point Depression
Changes in Boiling Point and Freezing Point of Water
Exercise
A solution was prepared by dissolving 25.00 g glucose in 200.0 g water. The molar mass of glucose is 180.16 g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution.
Exercise
You take 20.0 g of a sucrose (C12H22O11) and NaCl mixture and dissolve it in 1.0 L of water. The freezing point of this solution is found to be -0.426°C. Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.
Exercise
A plant cell has a natural concentration of 0.25 m. You immerse it in an aqueous solution with a freezing point of –0.246°C. Will the cell explode, shrivel, or do nothing?
• Osmosis – flow of solvent into the solution through a semipermeable membrane.
• = MRT
= osmotic pressure (atm)
M = molarity of the solution
R = gas law constant
T = temperature (Kelvin)
Exercise
When 33.4 mg of a compound is dissolved in 10.0 mL of water at 25°C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound.
• The relationship between the moles of solute dissolved and the moles of particles in solution is usually expressed as:
van’t Hoff Factor, i
moles of particles in solution =
moles of solute dissolvedi
Ion Pairing
• At a given instant a small percentage of the sodium and chloride ions are paired and thus count as a single particle.
• The expected value for i can be determined for a salt by noting the number of ions per formula unit (assuming complete dissociation and that ion pairing does not occur). NaCl i = 2 KNO3 i = 2
Na3PO4 i = 4
Examples
• Ion pairing is most important in concentrated solutions.
• As the solution becomes more dilute, the ions are farther apart and less ion pairing occurs.
• Ion pairing occurs to some extent in all electrolyte solutions.
• Ion pairing is most important for highly charged ions.
Ion Pairing
Modified Equations
= T imK
= iMRT
• A suspension of tiny particles in some medium.
• Tyndall effect – scattering of light by particles.
• Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm.
Types of Colloids
• Destruction of a colloid.• Usually accomplished either by heating or
by adding an electrolyte.
Coagulation