Chpt 11 - Solutions Concentrations Energy of solutions Solubility Colligative Properties Colloids...
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Transcript of Chpt 11 - Solutions Concentrations Energy of solutions Solubility Colligative Properties Colloids...
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