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?

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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

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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