Chapter 11 Properties of Solutions. Solution Composition Molarity (M) = moles solute / Liters...

Post on 26-Mar-2015

230 views 0 download

Transcript of Chapter 11 Properties of Solutions. Solution Composition Molarity (M) = moles solute / Liters...

Chapter 11

Properties of Solutions

Solution Composition

• Molarity (M) = moles solute / Liters solution

• Mass Percent – no units– (mass solute / mass solution) x 100

• Mole Fraction – no units – decimal number

χ a= (mol a / mol total)

• Molality (m) = moles solute / kg solvent – Does not change with temp

• Example - #25 p. 548

A solution is made by dissolving 25g of NaCl in enough water to make 1.00L of solution. Calculate the mass percent, molarity, molality, and mole fraction of the sodium chloride. Assume the density of the solution is 1.00g/mL

Solubility

• Likes dissolve likes– Polar dissolves polar– Nonpolar dissolves nonpolar

• For solids – Increased Temp usually increases solubility

• Gases– Increased Temp decreases solubility– Increased Pressure increases solubility

• Henry’s Law

Henry’s Law

• The amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution

• P=kC

• P is the pressure above the gas above the solution

• k is constant for a solution

• C is the concentration of the dissolved gas

Implications of Gas Solubility

From: http://tellus.ssec.wisc.edu/outreach/teach/ideas/kotoski/Minifact_Sheets/Minifact4_Dissolved_Oxygen.pdf

Vapor Pressure of Solutions

• A nonvolatile solute added to a solvent ALWAYS lowers the vapor pressure of the solution

• Fewer molecules to enter the gas phase

• Follows Raoult’s Law

• Called Ideal solutions

Raoult’s Law

• Psolution = χsolvent * Psolvent

• Where

– Psoluiton = Vapor Pressure of Solution

– χsolvent = mole fraction of solvent

– Psolvent = Vapor Pressure of the solvent

• Since the mole fraction is always less than one the pressure of the solution is lowered

• Example - #45 p. 549

A solution is prepared by mixing 50.0 grams of glucose C6H12O6 with 600.0 g of water. What is the vapor pressure of the solution at 25ºC? The vapor pressure of water at 25ºC is 23.8 torr.

More on Raoult’s Law

• Nonelectrolyes (covalent compounds) have the expected change you calculate

• Electrolytes (Ionic compounds) will see a bigger change in vapor pressure

• Ex. Ethanol lowers vapor pressure of water as expected– NaCl doubles the lowering of the vapor

pressure. MgCl2 triples

Deviations From Ideal

• When the solute and the solvent have similar polarity the vapor pressure of the solution is lower – Attraction b/t particles– Negative deviation

• When the solute and the solvent have dissimilar polarity the vapor pressure of the solution is higher– Positive deviation

Effect of Solute on Solution

• What effect will the lowering of the vapor pressure have on the solution?

• Boiling point will go up

• Freezing point will go down

Colligative Properties

• Properties that depend on the number of particles in solution

• Vapor Pressure Depression

• Boiling Point Elevation

• Freezing Point Depression

• Osmotic Pressure

Boiling Point Elevation

• ΔTb = Change in the boiling point• Kb = Boiling point elevation constant

Units of (ºC*kg/mol)– Constant for a given substance

• Data found in table 11.5 on page 532

– If ΔTb for salt water is 2.2ºC then the new B.P. is 102.2

The van’t Hoff Factor

• i = The van’t Hoff factor– Tells you how many moles of particles a

solute would form if 1 mole dissolves– All covalent compounds have an i of 1

• They do not ionize

– i for ionic compounds equals the # of ions• NaCl = 2• MgCl2 = 3• AlCl3 = 4

Freezing Point Depression

• ΔTf = Change in the freezing point

• Kf = Freezing point depression constant Units of (ºC*kg/mol)– If ΔTf for salt water is 2.2ºC then the new F.P.

is -2.2ºC

Supercooling

From: http://wiki.one-school.net/index.php/Changes_in_the_state_of_matter

http://www.chemprofessor.com/colligative.htm

Example• What is the freezing point of water if 0.25

mol. of glucose is dissolved in 500mL of water? Kf=1.86 ºC*kg/mol?

Example• What is the new boiling point of water when

100. grams of NaCl is dissolved in 2.0L of water? Kb=0.51 ºC*kg/mol

Example• Ethylene Glycol (C2H6O2 62.08 g/mol) is

placed in radiators to prevent the water from boiling. How many moles of EG must be added to 10.0L of water to make the water boil at 110.00 ºC? Kb=0.51 ºC*kg/mol

Osmosis

• The flow of solvent molecules into a solution through a semipermeable membrane

• High concentration of solvent to low concentration of solvent

Osmotic Pressure • The Pressure required to stop osmosis• Pressure required to cause reverse

osmosis• Π=iMRT• Π = osmotic pressure (atm)• i = van’t Hoff factor• M = Molarity• R = .08206 L*atm/mol*K• T = Temp in K

Homework

• P.548 #’s 26,28abd,38,46,57,63,70,72