chem 106, prof. t.l. heise 1 che 106: general chemistry u chapter thirteen copyright © tyna l....

Chem 106, Prof. T.L. Heise

11CHE 106: General Chemistry

CHAPTER THIRTEEN

Copyright © Tyna L. Heise 2001 - 2002

All Rights Reserved


22The Solution Process

• SolutionSolution - homogeneous mixture of two or more substances

• SolventSolvent - substance of greater amount in the homogeneous mixture (solution)

• SoluteSolute - compounds “dissolved” in the solvent

• ConcentrationConcentration - the amount of solute dissolved in a solvent. Expressed in molarity (M) Chap

13.1



• SolutionSolution -may be gases, liquids or solids

–Examples: air

ocean water

10K gold (alloy)

Chap 13.1



A solution is formed when one substance disperses uniformly through another.

All solutions, except gas mixtures, involve substances in a condensed state

Intermolecular forces are also going to operate between solute and solvent

- interactions are known as solvation

- when solvent is water, the interactions are known as hydration

Chap 13.1



Chap 13.1



Energy changes during solution formation is the sum of three energy changes:

H1separation of solute molecules

H2 = separation of solvent molecules

H3 = formation of solute-solvent interactions

H1 and H2 are endothermic because you are breaking or overcoming interactive forces

H3 is exothermic due to the creation of new forcesChap 13.1



Chap 13.1



Solution Formation depends on two factors

1) energy or enthalpy changes H

2) chaos or entropy changes S

Natural phenomenon's occur to satisfy two basic laws - energy content decreases

- disorder content increases

** Formation of solutions is favored by the increase in disorder that accompanies mixing

Chap 13.1



Solution Formation can occur during two basic processes 1) physical changes

2) chemical changes

Our focus is on physical changes, key to recognizing difference is examining whether you can get the salt BACK unchanged when reaction is done.

Chap 13.1


1010Saturated Solutions and Solubility

As a solid solute is dissolved, the concentration of dissolved particles increases, as does the chance of a collision between to dissolved particles. If two dissolved particles collide, the attractive forces could cause recrystallization

In any solutionSolute + solvent dissolve solution

crystallize

Chap 13.2



Saturated: in equilibrium, rate of dissolving equals rate of crystallization, concentration remains constant

Solubility: the amount of solute needed to form a saturated solution in a given temperature

Unsaturated: dissolved less the amount needed to form a saturated solution

Supersaturated: dissolving more than the amount needed Chap 13.2



Chap 13.2


1313Factors Affecting Solubility

Chap 13.3

•Solute-Solvent Interactions»the stronger the the attractions

between solute and solvent, the greater the solubility

»miscible and immiscible» like dissolves like



Chap 13.3

•Pressure Effects»the solubility of a gas in any

solvent is increased as the pressure of the gas over the solvent is increased



Chap 13.3

•Pressure Effects»relationship between pressure and

the solubility of a gas is expressed in terms of a simple equations known as Henry’s Law



Chap 13.3

Sample Exercise: Calculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO2

partial pressure of 3.0 x 10-4 atm.



Chap 13.3



1) Cg = kPg



Chap 13.3



1) Cg = kPg

Cg = x

k = 3.1 x 10-2 mol/L-atm(given page 480)

Pg = 3.0 x 10-4 atm



Chap 13.3



1) Cg = kPg

x = 3.1 x 10-2 mol/L-atm(3.0 x 10-4 atm)



Chap 13.3



1) Cg = kPg

x = 3.1 x 10-2 mol/L-atm(3.0 x 10-4 atm)

x = 9.3 x 10-6 mol/L



Chap 13.3

• Temperature Effects»solubility of most solid solutes

increases as the temperature of the solution does

»solubility of most gases decreases as temperature of the solution does

» look up solubility on a solubility table to find exact trend



Chap 13.3• Temperature Effects


2323Expressing Concentration

Chap 13.4

•Dilute - relatively small concentration of solute in solution

•Concentrated - relatively large concentration of solute in solution

•Ways to express numerically-•mass percentage•mole fraction•molarity•molality



Chap 13.4

•mass percentagemass of component *100

total mass of soln• ppm

mass of component *100,000 total mass of soln

• ppbmass of component *100,000,000

total mass of soln



Chap 13.4

Sample Exercise: Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water



Chap 13.4


mass percent = mass of component * 100total mass of soln



Chap 13.4

Sample Exercise: Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water mass percent = mass of component * 100total mass of soln

= 1.50 g NaCl *100 51.50 g soln



Chap 13.4



= 1.50 g NaCl *100 51.50 g soln

=2.91 %



Chap 13.4

Sample Exercise: a commercial bleaching solution contains 3.62 mass percent sodium hypochlorite, NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of bleaching solution?



Chap 13.4





Chap 13.4


mass percent = mass of component * 100total mass of soln 3.62

% = x * 100 2500 g soln 3.62 % * 2500g = x

100%



Chap 13.4


mass percent = mass of component * 100total mass of soln 3.62 % = x *

100 2500 g soln 3.62 % * 2500g = x 100%90.5 g NaOCl



Chap 13.4

•mole fraction mole fraction = moles of

component total moles of all components

• symbol X is often used to represent mole fraction

•molarity = moles of solute/L of solution molality = moles of solute/kg of solvent



Chap 13.4

Sample Exercise: What is the molality of a solution made by dissolving 36.5 g naphthalene, C10H8, in 428 g of toluene, C7H8?



Chap 13.4


1) molality = moles of solute kg of solvent



Chap 13.4



moles = 36.5 g C10H8 1 mol = 0.285 mol 128 g C10H8



Chap 13.4



moles = 0.285 mol kg = 428g 1 kg = 0.428 kg

1000 g



Chap 13.4


1) molality = moles of solute kg of solventmoles = 0.285 mol = 0.285 mol kg = 0.428 kg 0.428 kg

= 0.670 mol/kg



Chap 13.4

Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (a) the molality

(b) the mole fraction of NaOCl



Chap 13.4


mass % = mass of solute *100

mass of solution



Chap 13.4


mass % = mass of solute *100 mass of solution

3.62% = mass of solute * 100 100 g solution * when not

given an amount f solution, assume 100g.



Chap 13.4


mass of solute = 3.62 g molality = moles kg

3.62 g NaOCl 1 mol NaOCl = 0.0486 mol 74.44 g NaOCl



Chap 13.4


molality = moles = 0.0486 mol = 0.505 mol kg 0.09638 kg

kg *don’t forget we assumed 100 g of solution, so 3.62 g of NaOCl and 96.28 g water



Chap 13.4

Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (b) mole fraction of

NaOClMole fraction = moles of solute

total moles of components



Chap 13.4

Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (b) mole fraction of NaOCl

Mole fraction = moles of solute total moles of components 3.62 g NaOCl 1 mol NaOCl = 0.0486 mol

74.44 g NaOCl NaOCl 96.38 g H2O 1 mol H2O = 5.35 mol H2O 18 g H2O



Chap 13.4

Sample Exercise: A commercial bleach solution contains 3.62 mass percent NaOCl in water. Calculate (b) mole fraction of

NaOClMole fraction = moles of solute

total moles of components

0.0486 mol NaOCl = 0.00900 0.0486 mol NaOCl + 5.35 mol H2O


4747Colligative Properties

Chap 13.5

Certain physical properties of solutions differ from the pure solvent

- lowering freezing point- raising boiling point- reduction of vapor pressure- alteration of osmotic pressure



Chap 13.5

Vapor Pressure Vapor pressure is the pressure exerted by the vapor on the surface when closed flask achieves equilibrium



Chap 13.5

Boiling Point

Kb is the molal boiling point constant and is equal to 0.52°C/m

- this constant is related to the number of dissolved particles



Chap 13.5

Freezing Point

Kf is the molal boiling point constant and is equal to 1.87°C/m

- this constant is related to the number of dissolved particles



Chap 13.5

Sample Exercise: Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol, C10H18O, a fragrant substance found in the leaves of the eucalyptus tree.



Chap 13.5


molality = moles of solutekg of solvent



Chap 13.5

Sample Exercise: Calculate the freezing point of a solution containing 0.600 kg of CHCl3 and 42.0 g of eucalyptol, C10H18O, a fragrant substance found in the leaves of the eucalyptus tree. molality = moles of solute kg of solvent

42.0 g C10H18O 1 mol C10H18O = 0.273 mol 154 g C10H18O



Chap 13.5


molality = moles of solutekg of solvent = 0.273 mol0.600 kg = 0.455 m



Chap 13.5


Tf = Kfm = 4.68(0.455) = 2.13°C Normal Tf = -

63.5°C -2.13°C

New Tf = -65.6°C



Chap 13.4

Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water:

1 mol Co(NO3)2

2 mol KCl3 mol C2H6O2



Chap 13.4

Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 1 mol Co(NO3)2

Tf = Kfm = 1.87(1) = 1.87°C



Chap 13.4

Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 2 mol KCl

Tf = Kfm = 1.87(4) = 7.48°C



Chap 13.4

Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 3 mol C2H6O2

Tf = Kfm = 1.87(3) = 5.61°C



Chap 13.4

Sample Exercise: Which of the following solutes will produce the largest increase in boiling point upon addition of 1 kg of water: 1 mol Co(NO3)2 = -1.87 °C2 mol KCl = -7.48 °C

3 mol C2H6O2 = -5.61 °C



Chap 13.5

Osmotic PressureThe net movement of solvent is always toward the solution with the higher solute concentration.

The pressure needed to stop such movement is called osmotic pressure



Chap 13.5

Osmotic Pressure


6363Colloids

Chap 13.6

chem 106, prof. t.l. heise 1 che 106: general chemistry u chapter thirteen copyright © tyna l....

Documents

solution process chap

solubility chap

formation of solute

solvent interactions

mixing chap

hydration chap

solvent interactions

saturated solutions