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Solutions. Soluble : solute dissolves in solvent Insoluble: solute will not dissolve in solvent Solubility : amount of solid solute that dissolves in given quantity of solvent at specific temperature/ pressure Miscible : two liquids dissolve in one another/no separation of layers

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slide3
Soluble: solute dissolves in solvent

Insoluble: solute will not dissolve in solvent

Solubility: amount of solid solute that dissolves in given quantity of solvent at specific temperature/ pressure

Miscible: two liquids dissolve in one another/no separation of layers

Immiscible: two liquids do not dissolve in one another/separation of layers

slide4

http://college.hmco.com/chemistry/shared/media/animations/supersaturatedsodiumacet.htmlhttp://college.hmco.com/chemistry/shared/media/animations/supersaturatedsodiumacet.html

slide5

http://college.hmco.com/chemistry/shared/media/animations/dissolutionofsolidinliquid.htmlhttp://college.hmco.com/chemistry/shared/media/animations/dissolutionofsolidinliquid.html

like dissolves like
“Like” dissolves “like”
  • Polar solvents dissolve ionic/polar compounds
    • Most ionic compounds dissolve in water
      • Exception: ions w/greater attraction for each other than charged water molecules don’t dissolve
  • Nonpolar solvents dissolve nonpolar compounds
    • Won’t dissolve in water
      • Not attracted to water molecules (sucrose polar, so dissolve in water)
slide7
Collisions of solvent/solute’s surface particles causes solvation of solute particles

Polar water/ions attract each other

Solvated solute particles collide increasingly with remaining crystal and rejoin crystal

Solvation rate > crystallization rate = continuing solvation

When solvation rate = crystallization rate = dynamicequilibrium

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.html

Solvation

http://college.hmco.com/chemistry/shared/media/animations/solutionequilibrium.html

factors that affect solubility
Factors that affect solubility
  • Nature of solvent/solute (like dissolves like)
  • Temperature (generally, ↑T  ↑ solubility)
  • Pressure (only gases in liquids)
  • Rate of solution: measure of how fast substance dissolves
particles
Particles
  • Size
    • Larger solute molecules = lowered solubility
      • More difficult to surround w/solvent molecules
    • Increasing total surface area of solute particle increases dissolving (only takes place at surface of each particle)
  • Contact of solute w/solvent increases rate
    • Shaking/stirring (agitation) brings solvent in contact w/surface of solute
  • Saturation
    • Less solute in solution allows faster dissolving to take place
    • As solution approaches saturation, dissolving slows
temperature
Liquids/solid solutes

↑T ↑rate solute dissolves

Most solid solutes more soluble in warmer solutions

Increases KE of water molecules

Increases frequency/ force of collisions with crystal surfaces

Gases: reverse is true

↑T ↓solubility/rate of solution (cold soda-more carbonation than warm)

Temperature
pressure
Pressure
  • Solids/liquid solutes
    • ∆P has practically no effect on solubility
  • Gaseous solutes
    • Increasing P increases solubility
    • Decreasing P decreases solubility
      • Gas has more space to escape into
henry s law
Henry’s Law
  • At given T, solubility of gas (solute) in solution directly proportional to partial pressure of that gas above solution
    • As P of gas above liquid increases, solubility of gas increases
    • As P of gas decreases, solubility of gas decreases
  • S1/P1= S2/P2
    • S: gas solubility in liquid
    • P: Pgas above liquid
slide13
Example: The concentration of dissolved oxygen is 0.44g / 100 mL solution. The partial pressure of oxygen is 150 mm Hg. What is the predicted concentration if the partial pressure for oxygen is 56 mm Hg?

Solution:

P1 = 150 mm Hg

C1 = 0.44 g O2 /100 mL solution (C is being used as concentration instead of solubility)

P2 = 56 mm Hg

C2 = ?

C2 = 0.15 g O2 /100 mL solution

slide14
Solubility = grams of solute/100 grams of solvent

If 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 23oC, how much will dissolve in 1.0 L of water at 1.0 atm of pressure and the same temperature?

0.85/4.0 = S2/1.0 S2 = 0.21 g/L

If 0.55 g of a gas dissolves in 1.0 L of water at 20.0 kPa of pressure, how much will dissolve at 110.0 kPa of pressure?

homework
Homework:

Read 15.1, pp. 452-461

Q pg. 484, #52, 53, 64, 66, 67

concentration of a solution qualitative description

Concentration of a solution-qualitative description

Measure of how much solute is dissolved in a specific amount of solvent or solution

percent composition by mass
Percent Composition(by mass)
  • To calculate percent by mass of solute in solution, we need
    • Mass of solute in solution
    • Mass of solution
  • Calculate percent by mass
  • Concentration expressed as percent is ratio of measured amount of solute to measured amount of solution
example
Example
  • In order to maintain a sodium chloride concentration similar to ocean water, an aquarium must contain 3.6 g NaCl per 100.0 g of water. What is the percent by mass of NaCl in the solution?

3.6 g NaCl

103.6 g NaCl + H2O x 100 = 3.5%

  • What is the percent by mass of NaHCO3 in a solution containing 20.0 g NaHCO3 dissolved in 600 mL H2O?
using percent by volume both solute and solvent are liquids
Using percent by volume-both solute and solvent are liquids
  • Solute volume : solution volume as percent
  • Volumes of liquids not always additive
    • Sometimes volumes change when two liquids are mixed together
    • For example, mixing 70 ml of isopropyl alcohol and 30 ml of water will not give you exactly 100 ml of solution
    • Ethanol/water molecules interact differently with each other than they do with themselves
examples
Examples
  • Rubbing alcohol is 70% isopropyl alcohol. 70 volumes of isopropyl alcohol are dissolved in every 100 volumes of solution. So 30 volumes of water are in every 100 volumes of the rubbing alcohol.
  • What is % by volume of ethanol in a solution that contains 35 mL of ethanol dissolved in 115 mL of water? (23%)
  • If you have 100.0 mL of a 30.0% aqueous solution of ethanol, what volumes of ethanol and water are in the solution? (30 mL ethanol, 70 mL water)
  • What is % by volume of isopropyl alcohol in a solution that contains 24 mL of isopropyl alcohol in 1.1 L of water? (2%)
more examples
More Examples
  • What is % concentration of a solution that you made by taking 5.85 g of NaCl and diluting to 100 mL with H20?
    • 5.85 g/100 mL x 100 = 5.85% W/V solution of NaCl
  • What is % concentration of a solution that you made by taking 40 g of CaCl2 and diluting to 500 mL with H20?
    • 40 g/500 mL x 100 = 8%
  • How would you make 250 mL of a 8.5% NaCl solution?
    • x/250 mL = 8.5% x = 21.3 g
    • weigh out 21.3 g NaCl and dilute to 250 mL with H20.
  • How much 0.85% NaCl may be made from 2.55 g NaCl?
    • 2.55 g/x = .85% x = 300 mL
a solution is made up of two parts solute and solvent
A solution is made up of two partssolute and solvent
  • Powdered ice tea mix-4 scoops powder for every 2 quarts H2O
    • Normal recipe for ice tea mix looks like:

4 scoops of powderTo make ice tea of "normal" strength = -----------------------                                                         2 quarts of water

  • Make ice tea 2x strong as normal-use 8 scoops of powder w/2 quarts of water, or 4 scoops of powder w/1 quart water:
  • To make ice tea 2x "normal" strength =

8 scoops of powder 4 scoops of powder

---------------------------- or ----------------------------- 2 quarts of water          1 quart of water

molarity
Molarity
  • Need more exact strength of solution when doing quantitative analysis in lab 
    • Unit for molarity is M and read as "molar" (3 M = three molar)
    • "Moles" measures quantity of material
    • "molarity" measures concentration of that material
  • Concentration of substance: measure of how much solute is dissolved in specific amount of solvent or solution
  • Concentration of substance in solution: number of moles of that substance moles per unit volume of solution
example of molarity
Example of molarity
  • Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into some water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure that when everything was well-mixed, there was exactly 1.00 liter of solution.
  • What would be the molarity of this solution?
  • The answer is 1.00 mol/L. Notice that both the units of mol and L remain. Neither cancels.
  • A replacement for mol/L is often used. It is a capital M. So if you write 1.00 M for the answer, then that is correct.
example 2
Example #2
  • Suppose you had 2.00 moles of solute dissolved into 1.00 L of solution. What's the molarity?
  • The answer is 2.00 M.
  • Notice that no mention of a specific substance is mentioned at all. The molarity would be the same. It doesn't matter if it is sucrose, sodium chloride or any other substance. One mole of anything contains 6.022 x 1023 units.
example 3
Example #3
  • What is the molarity when 0.75 mol is dissolved in 2.50 L of solution?
  • The answer is 0.300 M.
example 4
Example #4
  • Now, let's change from using moles to grams. This is much more common. After all, chemists use balances to weigh things and balances give grams, NOT moles.
  • Example #4 - Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution?
  • The solution to this problem involves two steps which will eventually be merged into one equation.
    • Step One: convert grams to moles.
    • Step Two: divide moles by liters to get molarity.
  • 58.44 grams/mol is the molecular weight of NaCl.
  • Dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol.
  • Then, dividing 1.00 mol by 2.00 L gives 0.500 mol/L (or 0.500 M). Sometimes, a book will write out the word "molar," as in 0.500-molar.
example 5
Example #5
  • Calculate the molarity of 25.0 grams of KBr dissolved in 750.0 mL.
  • Note change from mL to L in the answer.
example 6
Example #6
  • 80.0 grams of glucose (C6H12O6, mol. wt = 180. g/mol) is dissolved in enough water to make 1.00 L of solution. What is its molarity?
example 7
Example #7
  • How many moles of CaCl2 would be used in the making of 5.00 x 102 cm3 of a 5.0M solution? Notice volume is given in cm3.  Since there are 1000 cm3 in 1 liter, 500 cm3 must be equal to 0.500 liters.
    • 5.00 x 102cm3 x 1 L/1000 cm3 = 0.500 L
    • 5 M = x mol/0.500 L = 2.5 mol CaCl2
  • How many grams of CaCl2 in above question?
    • Molar mass = 111 g/mol 2.5 mol x 111 g/mol = 280 g CaCl2
example 8
Example #8
  • Calculate the molarity of 20 g of NaCl added to water to make 1 L of solution
  • Molar mass of NaCl
    • 22.99 + 35.45 = 58.44 g/mol
    • 20 g/1mol/58.44 g = 0.342 mol
  • M = 0.342 mol/1L = 0.342 M
example 9
Example #9
  • Calculate the molarity when 0.25 g of Pb(NO3)2 in enough water to make 500 mL of solution
  • Molar mass of Pb(NO3)2 =207.2+28.01+96.00 = 331.2 g/mol
  • 0.25 g x 1mol/331.2 g = 7.5 x 10-4 mol
  • 500 ml x 1 L/1000 ml = 0.500 L
  • M = 7.5 x 10-4 mol/0.500 L=1.5 x 10-3 M
slide34
1) Calculate the molarity when 75.0 grams of MgCl2 is dissolved in 500.0 mL of solution.
  • 2) 100.0 grams of sucrose (C12H22O11, mol. wt. = 342.3 g/mol) is dissolved in 1.50 L of solution. What is the molarity?
  • 3) 49.8 grams of KI is dissolved in enough water to make 1.00 L of solution. What is the molarity?
slide35
When 2.00 grams of KMnO4 (molec. wt = 158.0 g/mol) is dissolved into 100.0 mL of solution, what molarity results?
  • How many grams of KMnO4 are needed to make 500.0 mL of a 0.200 M solution?
try some more
Try Some More!
  • 1) 10.0 g of acetic acid (CH3COOH) is dissolved in 500.0 mL of solution. What molarity results?
  • 2) How many mL of solution will result when 15.0 g of H2SO4 is dissolved to make a 0.200 M solution?
now try this
Now try this:
  • If you put a teaspoon of sugar, C12H22O11, or about 5.00 g, into a cup of water (about 250 ml) what would be the molar concentration?
  • Sugar has a molecular weight of 342 g/mole so 5.00 g is 1.46 x 10-2 moles.
and this one
And this one…
  • If I dissolve 2.38 g of vanillin, C8H8O3, which is the essential ingredient in the vanilla you put into cookies and cakes, into 930 ml of ethyl alcohol, what would be the molar concentration?
  • Notice that we can use any liquid for the solution. Water is just the most common solvent you're likely to encounter.
  • Vanillin has a molecular weight of 152 g/mole, so 2.38 grams is 1.57 x 10-2 moles. The resulting solution would be:
making dilutions
Making Dilutions
  • Concentrated solution: large amount of solute
  • Dilute solution: small amount of solute
  • To make dilutions
    • Increase solvent/solute remains constant
    • M1 V1 = M2V2
  • 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M solution. How many mL of 0.800 M can you make?
    • (1.50 mol/L) (53.4 mL) = (0.800 mol/L) (x)
    • 100. mL-don’t have to convert to L because given in mL
more examples of dilutions
More Examples of Dilutions
  • 100.0 mL of 2.550 M KBr solution is on hand. You need 0.5500 M. What is the final volume of solution which results?
    • (2.500 mol/L) (100.0 mL) = (0.5500 mol/L) (x)
    • x = 4545. mL
    • How much more water is added? 4545 - 100.0 = 4445 mL
  • How do you prepare 100ml of 0.40M MgSO4 from a stock solution of 2.0M MgSO4? How much more water needs to be added?
    • M1 = 2.0M MgSO4         V1 = unknown M2 = 0.40M MgSO4       V2 = 100mL
    • 2.0 M x = (0.40 M) (100 L) x = 20 mL
      • Add 80 mL distilled water to 20 mL of 0.40M MgSO4solution
and still more
And still more…
  • How do you prepare 100cm3 of 0.1M CoCl2 from a 0.52M solution of CoCl2?
    • Add 19.2cm3 of 0.52M CoCl2 solution to a graduate. Add distilled water to make the total volume 100cm3.
  • How many cubic centimeters of 0.69M Ba(NO3)2 are needed to prepare 200ml of 0.25M solution?
    • Add 72.5cm3 of 0.69M Ba(NO3)2 solution to a graduate. Add distilled water to make the total volume 200cm3.
slide42
Given a standard solution of 2M NH4Br, describe the preparation of 500ml of 0.15M ammonium bromide solution.
    • Add 37.5ml of 2M NH4Br solution to a graduate. Add distilled water to make the total volume 500ml.
  • Many solutions are prepared in the laboratory from purchased concentrated solutions.  What volume of concentrated 17.8 M stock sulfuric acid solution would a laboratory technician need to make 2.00 L of 0.200 M solution by dilution of the original, concentrated stock solution? (22 mL)
molality m
Molality, m
  • # moles of solute dissolved in exactly one kilogram of solvent
  • Need
    • Moles of solute present in solution
    • Mass of solvent (in kilograms) in solution
example44
Example
  • Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into exactly 1.00 liter water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure everything was well-mixed.
  • What would be the molality of this solution? Notice that my one liter of water weighs 1000 grams (density of water = 1.00 g / mL and 1000 mL of water in a liter). 1000 g is 1.00 kg, so:
  • The answer is 1.00 mol/kg. Notice that both the units of mol and kg remain. Neither cancels.
  • A replacement for mol/kg is often used. It is a lower-case m and is often in italics, m.
example45
Example
  • What is the molality when 0.75 mol is dissolved in 2.50 L of solvent?
  • The answer is 0.300 m.
example46
Example
  • Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 kg of pure water (the solvent). What would be the molality of the solution?
  • The solution to this problem involves two steps.
    • Step One: convert grams to moles.
    • Step Two: divide moles by kg of solvent to get molarity.
  • In the above problem, 58.44 grams/mol is the molecular weight of NaCl.
  • Dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol.
  • Then, dividing 1.00 mol by 2.00 kg gives 0.500 mol/kg (or 0.500 m).
calculate
Calculate:
  • Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water.
  • 80.0 grams of glucose (C6H12O6, mol. wt = 180. g/mol) is dissolved in 1.00 kg of water. Calculate the molality.
slide48
Calcuate the molality when 75.0 grams of MgCl2 is dissolved in 500.0 g of solvent.
  • 100.0 grams of sucrose (C12H22O11, mol. wt. = 342.3 g/mol) is dissolved in 1.50 L of water. What is the molality?
  • 49.8 grams of KI is dissolved in 1.00 kg of solvent. What is the molality?
slide49
Phrase "of solution" shows up in some of these problems

Molarity definition based on volume of solution

Temperature-dependent

Molality definition does not have volume in it

Temperature independent

Very useful concentration unit in colligative properties

mole fraction x of component in solution
Mole Fraction, X, of component in solution
  • Ratio of # moles of that component to total # moles of all components in solution
examples51
Examples
  • What is partial pressure of 0.50 moles of N in sample of gas that has total pressure of 6.34 atm/1.39 total moles of gas?
    • work out the mole fraction of nitrogen in the sample
      • XN2 = nN2/ntot
      • XN2 = 0.50/1.39
      • XN2 = 0.36
  • What are the mole fractions of HCl in an aqueous solution if there are 37.5 g of HCl?
    • nHCl = 37.5 g HCl x 1 mol/36.5 g HCl = 1.03 molHCl
    • nH2O = 62.5 g H2O x 1 mol/18.0 g H2O = 3.47 mol H2O
    • XHCl = nHCl/nHCl + nH2O = 1.03/1.03 + 3.47 = 0.229
    • XH2O = nH2O/nHCl + nH2O = 3.47/1.03 + 3.47 = 0.771
example52
Example
  • Determine the mole fraction concentration of Glucose in a solution in which 320 grams of glucose are dissolved in 4000 grams of water.
  • Determine the moles of glucose from molecular weight and given mass of glucose
    • 320 grams glucose X 1 mole / 180 grams = 2.0 moles Glucose
  • Convert grams of water to mols of water 4000 grams water X 1 mol / 18.0 grams = 222.2 mols water
  • Apply the definition for mole Fraction to determine the mole fraction concentration
    • Xglucose = moles Glucose / mols Glucose + mols water
  • Xglucose = 2.0 / 2.0 + 222.2 = 0.0089
normality
Normality
  • When comparing solutions on basis of concentration of specific ions or amount of charge that ions have
  • N = multiple of molarity of solution
      • CaCl2 has 2 moles of Cl-
        • For every mole of CaCl2 Ca2+ + 2 Cl-
        • For calcium chloride, N = 2
      • AlCl3 Al3+ + 3 Cl-
        • For aluminum chloride, N = 3
example54
Example
  • If a 0.5 M solution of BaI2 is totally dissociated into Ba2+ ions and I- ions, what is the molar concentration of each ion?
    • BaI2(s)   Ba2+(aq)  +  2 I1-(aq)
    • 0.5 M  therefore [Ba2+(aq)] = 0.5 M and the [I1-(aq)] = 1.0 M
homework55
Homework:

Read 15.2, pp. 462-470

Q pg. 470, #29, 32

Q pg. 484-485, #54, 55, 80, 82, 83a, 84, 85

solutes affect some of the physical properties of their solvents
Colligative properties

Depend only on # dissolved particles in solution and not their identity

Non-colligative properties

Depend on identity of dissolved species and solvent

Taste

Color

Viscosity

Surface tension

Solubility

Solutes affect some of the physical properties of their solvents
vapor pressure lowering
Pure solvent in closed container at constant T/P

Surface molecules gain enough energy to break free of liquid's intermolecular forces

Gaseous molecules coming into contact w/surface of liquid trapped by those intermolecular forces

When rate of escape = rate of capture, constant, equilibrium vapor pressure above pure liquid formed

http://college.hmco.com/chemistry/shared/media/animations/vapor_pressure_low_a.html

Vapor Pressure Lowering
slide58

http://college.hmco.com/chemistry/shared/media/animations/vapor_pressure_low_b.htmlhttp://college.hmco.com/chemistry/shared/media/animations/vapor_pressure_low_b.html

http://college.hmco.com/chemistry/shared/media/animations/vapor_pressure_low_c.html

raoult s law
Raoult’s Law
  • Amount of vapor pressure lowering proportional to amount of solute
    • Solutes that dissociate into particles have greater effect on vapor pressure than same concentration of nondissociating solute
  • Raoult's Law:  P = XP°   
    • P = VP of solution
    • P° = VP of pure solvent
    • X = mole fraction of solvent
examples60
Examples
  • If pure water has VP of 600 torr, calculate VP for 180g glucose/72g H2O
    • mol H2O = 72/18 = 4.0     
    • mol glucose = 180/180 = 1.0
    • P = XP° = (4/5)(600 torr) = 480 torr
  • If 90.0g water and 20.0g solute exert VP of 540 torr, calculate MW of solute. (P° = 600 torr)
    • X =P/P° = 540/600 = 0.90
    • X = mol H2O/total mol = (90/18)/(90/18 + 20/MW) = 5/(5 + 20/MW)
    • MW = 36
more examples61
More examples
  • Calculate VP for 68g (NH4) 2S & 180g H2O   (P° = 600 torr)
    • mol (NH4) 2S = 68/68 = 1.0    
    • mol H2O = 180/18 =10
    • But (NH4) 2S ionzes into 3 particles!
    • mol solute particles = 3.0
    • P = XP° = (10/13)(600 torr) = 462 torr
  • If solute volatile, P = XAPA° + XBPB°
  • Calculate VP for 172g hexane (MW=86) and 342g octane (MW=114).
    • PH° = 500 torr               
    • PO° = 400 torr
    • P = XHPH°  +  XOPO° = (2/5)(500 torr) + (3/5)(400 torr) = 200 torr + 240 torr = 440 torr
boiling point elevation
Boiling Point Elevation
  • Solute lowers solvent’s vapor pressure/affects solvent’s BP
    • Boiling point: T at which VPliquid = atmospheric pressure
  • Solution containing solute
    • Vapor pressure less than atmospheric pressure
    • Solution will not boil unless additional heat added
      • BP of solution made of liquid solvent with nonvolatile solute > BP of pure solvent
  • Boiling point elevation , ∆Tb
    • T difference between solution’s BP and pure solvent’s BP
    • Directly proportional to solution’s molality

http://college.hmco.com/chemistry/shared/media/animations/boilingpointelevation_a.html

http://college.hmco.com/chemistry/shared/media/animations/boilingpointelevation_b.html

http://college.hmco.com/chemistry/shared/media/animations/boilingpointelevation_c.html

slide64

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/propOfSoln/colligative.htmlhttp://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/propOfSoln/colligative.html

oC/m

examples65
Examples
  • Calculate BP point of solution of 0.0222 m glucose, a non-electrolyte in water. The Kb for water = 0.512 C/m. Tb0 = 100 C.
    • Tb = (0.512) (0.0222) = .0114
    • Tb= 100 + .0114 = 100.0114 C
  • If we used .05 molal solution of Na2SO4 what would be the BP?
    • Write the dissociation equation for Na2SO4
      • Na2SO4 + H2O  2 Na+ (aq) + SO4-2 (aq)
    • Calculate the i factor
      • i = # total moles after solution/# moles before solution
      • = sum coefficients on right/coefficients of solute on left = 2 + 1/1 = 3
    • Calculate BP of solution using BP elevation equation
      • Tb - Tb0 = i Kb m = (3)(0.512)(.05) = 0.0768
      • Tb = 100 + 0.0768 = 100.0768 C
slide66
What would be the boiling point of a solution of the same concentration (.0222m) of an electrolyte solute, Ca3 (PO4) 2. Kb = 0.512 and m = 0.0222. Assume complete dissociation.

Write the dissociation equation for Ca3(PO4) 2

Ca3(PO4) 2 + H2O  3Ca+2(aq) + 2PO4-3(aq)

Calculate the i factor

i = 3 + 2 / 1 = 5

Calculate BP of solvent in solution.

∆Tb = Tb - 100 = i Kb m

Tb - 100 = (5)(0.512)(.0222) = 0.0568

Tb = 100 + .0568 = 100.0568 C

slide67

http://college.hmco.com/chemistry/shared/media/animations/freezingpointdepression_a.htmlhttp://college.hmco.com/chemistry/shared/media/animations/freezingpointdepression_a.html

http://college.hmco.com/chemistry/shared/media/animations/freezingpointdepression_b.html

http://college.hmco.com/chemistry/shared/media/animations/freezingpointdepression_c.html

example68
Example
  • Calculate the freezing point of a solution of 5.00 g of diphenyl C12H10 and 7.50 g of naphthalene, C10H8 dissolved in 200.0 g of benzene (fp = 5.5 °C)
  • 1) The key to this problem is to calculate moles of each substance and add then together:
    • (5.00 g / 154.2 g mol¯1) + (7.50 g / 128.2) = 0.0909 mol
  • 2) Calculate the molality:
    • 0.0909 mol / 0.200 kg = 0.455 m
  • 3) Calculate the freezing point depression:
    • ΔT = i Kfm x = (1) (5.12 °C m¯1) (0.455 m) = 2.33 °C
    • The solution freezes at 5.5 - 2.33 = 3.17 °C
freezing point lowering
Freezing Point Lowering
  • Place two ice cubes in a container.
  • Pour salt on one, but not the other. Record your observations.

1. Which ice cube melted faster? Explain.

  • Moisten two cotton strings and place upon ice cubes. Pour salt on one of them. After one minute, pick up the loose ends of each string.

2. Why did the string freeze to the ice cube after salt was added, but did not freeze to the cube in which no salt was added?

3. Some people prefer to make their own ice cream. When doing so, they mix the cream in a container that rests within another container tht they fill with water, ice, and salt. Explain the purpose of that outer tub.

4. The freezing point of water decreases as salt is added until it reaches the limit of sodium chloride solubility at -21oC. What are the molarity and molality of the solution at this temperature?

homework73
Homework:

Read 18.3, pp. 471-475

Q pg. 475, #38, 39, 41

Q pp. 484-485, #57, 58, 86, 88

brownian motion
Brownian Motion
  • Random, vibrating motion of colloidal particles
  • Suspended particle constantly/ randomly bombarded from all sides by molecules of liquid
  • If particle very small, hits it takes from one side stronger than bumps from other side, causing it to jump
tyndall effect type of diffraction
Tyndall effect (type of diffraction)
  • Suspensions/colloids
    • Particles big enough to scatter light
    • Cloudy appearance of mixtures
    • Solutions always clear (particles too small to scatter light-smaller than wavelengths of visible light)
  • Examples
    • Sunbeams passing through window
    • Car headlight shining through fog
    • Exceptions: effect not be seen w/milk (colloid)
classification of dispersions pp 478 479
Classification of Dispersions (pp. 478-479)
  • Shake or stir contents, then observe over next 10 minutes.
  • Darken the room and shine a flashlight beam through the beakers.
  • Turn lights back on and continue to observe each container for signs of phase separation. On the basis of these two criteria, classify each on the chart (next slide).
solutions vs colloids and suspensions centrifuge
Solutions vs. Colloids and Suspensions (centrifuge)
  • Which of the mixtures were suspensions? Colloids? Solutions?
thixotropy
Thixotropy
  • Exhibited by certain gels (semisolid, jellylike colloids)
    • Thixotropic gel appears to be solid, maintains shape of its own until it is subjected to force or disturbance (shaking)
    • Then acts as sol (semifluid colloid) and flows freely
  • Thixotropic behavior is reversible
    • When allowed to stand undisturbed, sol slowly reverts to gel
  • Common examples: certain paints, printing inks, clays
water of hydration water of crystallization
Water of Hydration/water of crystallization
  • Some hydrated salts/hydrates form crystals w/definite amount of water molecules chemically combined as part of crystalline structure
    • Dot: attractive force between polar water molecules/+ metal ion
  • Heat overcomes attractive forces, water molecules released, becomes non-crystalline anhydrous substances

CuSO4 + 5H2O(s) CuSO4(s) + 5H2O(g)

Hydrated (white) anhydrous (blue)

slide84
Hygroscopic compounds: hydrated salts w/low vapor pressure, remove water from moist air to form hydrates

Desiccants absorb moisture from air but remain in solid state (drying agent)

Anhydrous silica gel common one

Deliquescents absorb moisture and form liquid sol

making gels canned heat
Making gels: “canned heat”
  • Dissolve 17 g calcium acetate in 50 mL of water.
  • Add 8 mL of ethyl or isopropyl alcohol.
  • A gel will form almost immediately.
  • Darken the room and ignite the “canned heat” with a long-stemmed match. It will produce a blue flame.
  • Place a watch glass over it to extinguish.
homework86
Homework:

Read 15.4, pp. 476-482

Q pg. 479, #46

Q pg. 484, #60, 63,

Test practice, pg. 487, all questions

Use link for quiz and submit as before.

http://www.glencoe.com/qe/science.php?qi=999