ii iii i i. the nature of mixtures mixtures heterogeneous mixture – a non-uniform or unequal...
TRANSCRIPT
MixturesMixtures
Heterogeneous Mixture – a non-uniform or unequal blend of two or more pure substances• Suspension – mixture containing
particles that will settle out• Colloid – mixture containing particles
with a size of 1nm – 1000nm, and do not separate – stay suspended
MixturesMixtures
Tyndall Effect – Caused by dilute colloids, which appear to be homogeneous.• Is the scattering of light as it passes
though a dilute colloid
MixturesMixtures
Solutions - Solutions - homogeneous mixtures
Solvent Solvent - present in greater amount
Solute Solute - substance being dissolved
Homogeneous Mixtures
• A substance that dissolves in a solvent is soluble.
• Two liquids that are soluble in each other in any proportion are miscible.
• A substance that does not dissolve in a solvent is insoluble.
• Two liquids that can be mixed but separate shortly after are immiscible.
SolutionsSolutions
Solvation – Solvation – the process of dissolving – sugar dissolves in to water
solute particles are separated and pulled into solutionhttp://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.html
solute particles are surrounded by solvent particles
SolvationSolvation
StrongElectrolyte
Non-Electrolyte
solute exists asions only
- +
salt
- +
sugar
solute exists asmolecules
only
- +
acetic acid
WeakElectrolyte
solute exists asions and
molecules
SolvationSolvation
Molecular Molecular SolvationSolvation• molecules
stay intact
C6H12O6(s) C6H12O6(aq)
SolvationSolvation
DissociationDissociation• separation of an ionic solid into
aqueous ions• http://chemmovies.unl.edu/ChemAnime/NACL1D/NACL1D.html
NaCl(s) Na+(aq) + Cl–(aq)
SolvationSolvation
NONPOLAR
NONPOLAR
POLAR
POLAR
““Like Dissolves Like”Like Dissolves Like”““Like Dissolves Like”Like Dissolves Like”
SolvationSolvation
Heat of Solution – energy change that occurs during the formation of the solution.• Exothermic – solvation produces a
warm solution• Endothermic – solvation produces a
cold solution
Factors Affecting Solvation (solubility)
Factors Affecting Solvation (solubility)
Agitation – stirring or shaking – allows for more collisions (mixing) between solute and solvent
Surface Area – smaller pieces – more collisions
Temperature – in general solids dissolve faster in higher temps. – more collisions• Opposite for gases – colder the better
SolubilitySolubility
SolubilitySolubility• maximum grams of solute that will
dissolve in 100 g of solvent at a given temperature
• varies with temp• based on a saturated soln• Book Reference: p 969 R-3 and p 974
R-8• http://chemmovies.unl.edu/ChemAnime/SLBIL1D/SLBIL1D.html
Solubility RulesSolubility Rules
In general:
• Group I ions and Ammonium are soluble
• Acetates and Nitrates are soluble
• Cl, Br, I are soluble, except with Pb, Ag, Hg2+2
• Sulfates are soluble, except with Ba, Sr, Pb, Ca, Ag, Hg2
+2
• Carbonates, Hydroxides, oxides, sulfides, phosphates are INsoluble
SolubilitySolubility
SATURATED SOLUTION
no more solute dissolves
UNSATURATED SOLUTIONmore solute dissolves
SUPERSATURATED SOLUTION
becomes unstable, crystals form
concentration
SolubilitySolubility
Solubility CurveSolubility Curve• shows the
dependence of solubility on temperature
SolubilitySolubility
Solids are more soluble at...Solids are more soluble at...• high temperatures.
Gases are more soluble at...Gases are more soluble at...
• low temperatures &
• high pressures (Henry’s Law).
Types of Reactions in Aqueous Solutions
• When two solutions that contain ions as solutes are combined, the ions might react.
• If they react, it is always a double replacement reaction.
• Three products can form: precipitates, water, or gases.
SolubilitySolubility
Replacement Reactions • Double Replacement Reactions occur
when ions exchange between two compounds.
• This figure shows a generic double replacement equation.
SolubilitySolubility
SolubilitySolubility
• Aqueous solutions of sodium hydroxide and copper(II) chloride react to form the precipitate copper(II) hydroxide.
2NaOH(aq) + CuCl2(aq) → 2NaCl(aq) + Cu(OH)2(s)
• Ionic equations that show all of the particles in a solution as they actually exist are called complete ionic equations.
2Na+(aq) + 2OH–(aq) + Cu2+ (aq)+ 2Cl–(aq) → 2Na+
(aq) + 2Cl–(aq) + Cu(OH)2(s)
Types of Reactions in Aqueous Solutions (cont.)
• Ions that do not participate in a reaction are called spectator ions and are not usually written in ionic equations.
• Formulas that include only the particles that participate in reactions are called net ionic equations.
2OH–(aq) + Cu2+(aq) → Cu(OH)2(s)
A. Concentration is...
a measure of the amount of solute dissolved in a given quantity of solvent
A concentrated solution has a large amount of solute
A dilute solution has a small amount of solute• thus, only qualitative descriptions
A. ConcentrationA. Concentration
Describing Concentration Quantitatively
• % by mass
• % by volume
• Molarity
• Molality
• Mole Fraction
B. PercentB. Percent
100xsolution of mass
solute of massMassby Percent
100xsolution ofvolume
solute ofvolume Volumeby Percent
B. PercentB. Percent
What is the percent solution if 25 mL of CH3OH is diluted to 150 mL with water?
17% 4.8 g of NaCl are dissolved in 82.0 mL of
solution. What is the percent of the solution? 5.53% How many grams of salt are there in 52 mL of
a 6.3 % solution? 3.3g
C. MolarityC. Molarity
solvent of L
solute of moles(M)molarity
Volume of solvent only
1 kg water = 1 L water
L 1
mol0.25 0.25M
C. MolarityC. Molarity
Find the molarity of a solution containing 75 g of MgCl2 in 250 mL of water.
75 g MgCl2 1 mol MgCl2
95.21 g MgCl2
= 3.2 M MgCl2
0.25 L water
L
molM
C. MolarityC. Molarity
How many grams of NaCl are req’d to make a 1.54 M solution using 0.500 L of water?
0.500 L water 1.54 mol NaCl
1 L water
= 45.0 g NaCl
58.44 g NaCl
1 mol NaCl
L 1
mol1.54 1.54M
D. MolalityD. Molality
1 L water = 1 Kg water so … Molality (m) = moles of solute / Kg of solvent
Basically the same calculations, but different units
solvent ofKg
solute of moles(m)molality
2211 VMVM
E. DilutionE. Dilution
Preparation of a desired solution by adding water to a concentrate.
Moles of solute remain the same.
E. DilutionE. Dilution
What volume of 15.8M HNO3 is required to make 250 mL of a 6.0M solution?
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3
F. Preparing Solutions F. Preparing Solutions
500 mL of 1.54M NaCl
500 mLwater
45.0 gNaCl
• mass 45.0 g of NaCl• add water until total
volume is 500 mL• mass 45.0 g of NaCl• add 0.500 kg of water
500 mLmark
500 mLvolumetric
flask
1.54m NaCl in 0.500 kg of water
F. Preparing SolutionsF. Preparing Solutions
250 mL of 6.0M HNO3 by dilution
• measure 95 mL of 15.8M HNO3
95 mL of15.8M HNO3
water for
safety
250 mL mark
• combine with water until total volume is 250 mL
• Safety: “Do as you oughtta, add the acid to the watta!”
Solution Preparation LabSolution Preparation Lab Turn in one paper per team. Complete the following steps:
A) Show the necessary calculations.
B) Write out directions for preparing the solution.
C) Prepare the solution. For each of the following solutions:
1) 100.0 mL of 0.500M NaCl
2) 0.250m NaCl in 100.0 mL of water
3) 100.0 mL of 1.00M Red Solution from 12.1M concentrate.
A. DefinitionA. Definition
Colligative PropertyColligative Property
• property that depends on the
concentration of solute particles, not
their identity
A. ElectrolytesA. Electrolytes
Dissolved Solute particles disrupt the normal Intermolecular Forces of the Solvent
Molecules – Count as one dissolved solute particle• Example: CH3OH = 1 solute particle
Ionic Compounds – number of solute particles is equal to the total number of ions in the neutral formula• Example: AlCl3 = 4 solute particles • 1 Al and 3 Cl ions
B. TypesB. Types
Vapor Pressure Lowering• Solutions have a lower vapor pressure
than the original pure solvent• Because the solute particles are
attracted to solvent particles cause more IMF
• So more IMF = less solvent particles becoming vapors
B. TypesB. Types
Freezing Point DepressionFreezing Point Depression (tf)
• f.p. of a solution is lower than f.p. of the pure solvent
Boiling Point ElevationBoiling Point Elevation (tb)
• b.p. of a solution is higher than b.p. of the pure solvent
B. TypesB. Types
Freezing Point Depressionhttp://chemmovies.unl.edu/ChemAnime/SOLND/SOLND.html
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.html
E. CalculationsE. Calculations
t: change in temperature (°C)
k: constant based on the solvent (°C·kg/mol)
m: molality (m)
n or i: # of particles
t = k · m · n
E. CalculationsE. Calculations
# of Particles# of Particles
• Nonelectrolytes (covalent)• remain intact when dissolved • 1 particle
• Electrolytes (ionic)• dissociate into ions when dissolved• 2 or more particles
E. CalculationsE. Calculations
At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil?
m = 3.2mn = 1tb = kb · m · n
WORK:
m = 0.73mol ÷ 0.225kg
GIVEN:b.p. = ?tb = ?
kb = 3.56°C·kg/moltb = (3.56°C·kg/mol)(3.2m)(1)
tb = 11°C
b.p. = 181.8°C + 11°C
b.p. = 193°C
E. CalculationsE. Calculations
Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water.
m = 4.8m
n = 2
tf = kf · m · n
WORK:
m = 0.48mol ÷ 0.100kg
GIVEN:
f.p. = ?
tf = ?
kf = 1.86°C·kg/mol
tf = (1.86°C·kg/mol)(4.8m)(2)
tf = 18°C
f.p. = 0.00°C - 18°C
f.p. = -18°C