Chpt 11 - Solutions Concentrations Energy of solutions Solubility Colligative Properties Colloids HW: Chpt 11 - pg. 531-538, #s 12, 14, 20, 24, 29, 34,

Chpt 11 - Solutions

• Concentrations

• Energy of solutions

• Solubility

• Colligative Properties

• Colloids

• HW: Chpt 11 - pg. 531-538, #s 12, 14, 20, 24, 29, 34, 41, 42, 44, 46, 52, 66, 72, 77, 81 Due Mon Dec. 3

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

Solvent is majority component. Solute is minority component, usually the substance dissolved in the solvent (liquid).

Solution composition

MolarityYou have 1.00 mol of sugar in 125.0 mL

of solution. Calculate the concentration in units of molarity.

8.00 M

You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar?

0.200 L

Molarity (M) example

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.

10.0 M NaOH 5.37 M KCl

Mass percent (%)

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?

6.6%

Mole fraction (A)

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.)

0.0145

Molality (m)

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.)

0.816 m

Solution Formation Schematic

Solution Formation Process

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.

Solution Formation Energies• 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 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).

Exo vs. Endo Hsoln

Demo NH4NO3 and NaOH examples

Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation,

be sure to address how ΔH plays a role.

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

Solubility Factors

• Structural Effects: Polarity (like dissolves like)

• Pressure Effects: Henry’s law (for dissolved gases)

• Temperature Effects: Affecting aqueous solutions

Pressure effects

• Henry’s law: C = kPC = concentration of dissolved

gask = constantP = 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.

Gas solubility in liquid

Soda pop’s carbonated water has the carbon dioxide forced into the solution under pressure. When the can is opened Patm is much lower than Pcan so CO2 leaves -> pop goes flat.

Temperature effects

• 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.

Temp solubility charts

Colligative Properties

• Depend only on the number, not on the identity, of the solute particles in an ideal solution: Vapor pressure loweringhttp://www.chem.purdue.edu/gchelp/solutions/colligv.html

Boiling-point elevation Freezing-point depression Osmotic pressure

Vapor Pressure of solutions

If the Pvap of the solvent (water) > Pvap of the solution, equilibrium is reached when the solvent evaporates and the solvent is absorbed by solution. It does this to lower the Pvap towards its equilibrium value.

Raoult’s Law

• Nonvolatile solute lowers the vapor pressure of a solvent.

• Raoult’s Law:

Psoln = observed vapor pressure of solution

solv = mole fraction of solvent

Posolv = vapor pressure of pure

solvent

Raoult’s Law - ideal solution

Ideal solution occurs with a nonvolatile solute in solution

Also the vapor pressure is then proportional to the mole fraction of the solvent using (total moles of ions of solute) in the solvent

Boiling Point elevation

• 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 particles

Freezing Point depression

• 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 particles

Phase Diagram of solutions

Boiling Point - Freezing Point explanation

http://en.wikipedia.org/wiki/Boiling-point_elevation

http://en.wikipedia.org/wiki/Freezing-point_depression

Boiling Pt Elev Problem

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. Kb =

0.51oC.kg/mol

100.35 °C

Osmotic Pressure

• Osmosis – flow of solvent into the solution through a semipermeable membrane. (Kidney dialysis uses this Principle).

= MRT = osmotic pressure (atm)M = molarity of the solutionR = gas law constantT = temperature (Kelvin)

Osmotic Pressure graphic

Osmotic Pressure Problem

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.

Strategy: need Temp in K, Pressure in atm, use R with atm unit to get molarity. Then use vol get moles, then mass get molar mass.

111 g/mol

Colloids• Intermediate mixture - a heterogeneous

mixture with particle size between a suspension and a solution

• 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

Tyndall Effect graphic

Freezing Pt problem

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. Kf = 1.86oC.kg/mol

Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture.

72.8% sucrose and 27.2% sodium chloride;mole fraction of the sucrose is 0.313

Derivation of Colligative Properties

• Specific derivation of the partial derivatives and derivation for colligative properties are found on the website.

http://www.chem.arizona.edu/~salzmanr/480a/480ants/colprop/colprop.html