solutions –
TRANSCRIPT
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SOLUTIONS –
• homogeneous dispersion of two or more substances in an ionic or molecular scale
Na+
O H
H
Cl- H O
H
CHO
CH OH
CH2OH
O H
H
Very small particles, too small to be seenBy the unaided eye
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Kinds of Mixture Particle Size
suspension
colloid
solution
10-4 cm. in diameter
10-5 – 10-6 cm. in diameter
10-7 cm. in diameter
Always suspended in the medium (too small to be pulled down by gravity)Shows Tyndall Effect : reflect light to produce a visible beam of light
Always suspended in the medium ( too small to be pulled down by gravity)
Settles upon standing
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Kinds of Solutions
Solid state Alloys ( Coin, brass)
Liquid state Sea water, alcohol in waterDissolved Oxygen (DO)
Gaseous state Air
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Parts of a Solution
solution
solute
solvent
Dissolved substance
Dissolving medium
is
is
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Parts of a solution
Physical state amount
Solute Dissolved substance
Solid, gas Less
Solvent Dissolving medium
liquid more
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Nature of Dissolving Process:
Na+
Na+
Na+
Na+
Cl-
Cl-
Cl-
Cl-
H OH Cl
-H2O
H2O
H2O
H2O
solvated iondissolving rate
crystallizing rate
salt
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Nature of the Dissolving Process
Na+1
Na+1 Na+1
Cl-1Na+1
Cl-1
Cl-1Cl-1
Cl-1 Na+1
H2O
H2O
H2O
H2O
Dissolving rate
Crystallizing rate
saltSolvated ion
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Nature of Dissolving Process:
• The nature of the dissolving process is• a 2 way process1. Process of dissolution2. Process of crystallization
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Which, at one point in time the 2 processes will attain equilibrium
• At equilibrium
• Rate of dissolution == rate of crystallization
• Solution is called saturated solution
• Concentration of the solution is constant
• corresponds to the solubility of the solute in question .•
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Solubility –
• maximum amount of solute that will be dissolved by a given amount of solvent producing a stable system, under a specified temperature
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Factors affecting SolubilityNature of solute and solvent :
• Like Dissolves Like
IMFAsolute &solvent
IMFAsolute
IMFAsolvent
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Temperature
Heat of solution
Solid in liquid
Gas in liquid
T increases solubility increases
Endothermic
T increases solubility decreases
exothermic
exothermicT increases solubility decreases
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Pressure
• Pressure –has little or no effect on the solubility of solid in liquid, liquid in liquid
• Gas in liquid : Pressure increases, solubility increases
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When can a gas become soluble in water ?
gas
liquid
For the gas to become soluble in the liquid , it must comeinto contact with the liquid: IMFA forming ( exothermic heatflow )
Applied Pressure: Pressure increases, solubility of gas in water increases
And this will be effectedby applying pressure to the gas so that IMFA is formed between the gas and water.
A process which involvesIMFA forming results for heatTo flow out to the sorrounding.
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Factors affecting Rate of Dissolution
1. Temperature : As temperature increases, rate of dissolving increases.
2. Stirring increases rate of dissolving3. Surface area – As surface area increases, rate
of dissolving increases
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Concentration = is the amount of solute present in a given amount of solvent producing a solution
Described
qualitatively quantitatively
dilute concentrated
saturated
unsaturated supersaturated
% Molarity (M)
Normality (N)
Molality (m) Mole fraction (X)
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1. Dilute- contains a relatively low amount of solute2. Concentrated – contains a relatively high amount of solute 3. Saturated – contains the maximum amount of solute that
can be dissolved by a measured amount of solvent (solubility equivalent )
4. Unsaturated – one which contains solute concentration lower than the concentration in the saturated solution.
5. Supersaturated – one which contain solute concentration higher than the concentration in the saturated Solution.
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Dilute concentrated
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46.5 g NaAc/100 g H2O25 C
solubility
80 g NaAC/100 g H20 50 C
all 80 g solutedissolves cool to
25 C
all 80 g NaAc is inwater as a solutionindefinitely
called
supersaturatedsolution
which canbe destroyed by
seeding agitation
Saturated supersaturated
If amount of NaAc is < 46.5 g in 100 g H2OAt 25 oC
Unsaturated solution
Saturated solutionWith undissolvedsolute
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A B C
At the start of Dissolving
Solute amount in Solvent is zero
During dissolving
Solute amount in solvent Has increased, but still lessThan the dissolving rate
After dissolving
Solute amount in solvent Has increased so that dissolvingRate is equal to the crystallizing rate
At what point during the dissolving rate is the saturated solution ?C
Not a solution Unsaturated solution Saturated solution
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QUANTITATIVE METHODS OF EXPRESSING CONCENTRATION OF SOLUTION
SOLUTION
SOLUTE SOLVENT/WATER
%
Molarity
Normality
molality
Mole fraction
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PerCent = Part Quantity/Total Quantity X 100
1. Percent by mass = grams ofsolute/grams of solution X 100
2. Percent by volume = volume of solute/volume of solution X 100
3. Percent by mass-volume = grams of solute/volume of solution X 100
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A 0.50 liter bottle of wine contains 60 ml ethanol. What is the % v/v ethanol in the solution ?
SOLUTION
SOLUTE SOLVENT
0.50 liter
60 ml
volume solute % = ------------------- X 100 volume solution
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How many grams of KCl are required to prepare 250 grams of an aqueous solution that is 10.0 % KCl by mass
SOLUTION
SOLUTE SOLVENT
?
250 grams
means
10 g KCl = 100 g solution
Conversion factor
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Molarity (M)
SOLUTION
SOLUTE SOLVENT/WATER
MOLE
Wt./MW
LITER
OR
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What is the M of a 10 % HCl solution of a density of 1.2 g /ml
SOLUTION
SOLUTE SOLVENT
10 g HCl = 100 g solution
10g HCl = 100 g soluton
means
M = wt/mwt/liter
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NORMALITY (N)
SOLUTION
SOLUTE SOLVENT/WATERNumber of equivalents ( (wt/MW ) X F )
Number of milliequivalents( (wt/MW) X F X 1000)
LITER
milliliter
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Mole fraction (X)
SOLUTION
SOLUTE SOLVENT/WATER
mole Mole Add to
Mole total
Mole solute/mole total
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Molality (m)
SOLUTION
SOLUTE SOLVENT/WATER
MOLE
Wt./MW
OR KG
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Calculate the molality of a solution of 2.34 g acetic acid, HC2H3O2 in 35.0 g water
SOLUTION
SOLUTE SOLVENT
2.34 g
HC2H3O2
35.0 g
MWT = 60.0 g/mole wt/ MWT solute m = ------------------- Kg solvent
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DILUTION
• = process of adding water to a solution of known concentration to obtain a new solution of different concentration
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6M HCl300 mL.
+ 300 ml water3M HCl600 mL
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solute solute
add water
What happens to the amount of solute upon dilution ?
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1.8 moles 1.8 moles
6M HCl300 mL.
+ 300 ml water3M HCl600 mL
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IN DILUTION
• :• Amount of solute in original solution = amount of solute in the prepared solution•
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Calculate the M of
55.0 g NaCl in 125 ml of solution
SOLUTIONso
SOLUTE SOLVENT
55.0 g 125 ml
M = wt/mwt/liter
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Theory Acid Base
ArrheniusOne which contains a hydrogen which will be yielded as a H+1 in water
One which contains a hydroxide which will be yielded as OH-! In water
Bronstead Lowry Proton donor Proton acceptor
Lewis E’ pair acceptor E’ pair donor
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H Cl H+1
+ Cl
It has H +1: has a potential to donate
Proton donor
Bronsted acid H+1
E pair acceptor
Lewis acid
Arrhenius acid
+ 2e' H
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Na OH OH-1
Na+1
+
it has OH w/c is negative; potential proton acceptor
Bronsted baseOH
-1Na
+1+
Has several e’ pairs; potential e’ pair donor
Lewis base
Arrhenius base
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HCl H2O H+ Cl
-
+
AcidConjugate base
What had become of the acid after donating a proton
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NH3 NH4+
H2O
OH-
BASE CONJUGATE ACID
What had become of the base after donating2 electrons
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HCl H2O H+ Cl
-
+
acid Conjugate base
base Conjugate acid
A strong acid has a weak conjugate base
SAWeak
SB
At equilibrium weak are favored
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solute Solvent (water)
solution
electrolytes
Strong :HCl H
+Cl
-+
NaOH Na+
OH-+
Weak
HC2H3O2 H+ C2H3O2
-+
Mg( OH) 2 Mg+2
2 OH-
+
Aside as being a solvent , something important is happening to water
Acid, base, salt
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H2O dissociates (into ions) to an extremely small but very important degree.
Kw = 1 X 10 -14 ( 0.00000000000001)
T= 250 C
H2O H+
OH-
+
Arithmetically of the value
1x10 -7 M 1x10 -7 M
WATER IS A NON ELECTROLYTE
Ion product constant :
Kw = (H+) (OH-)
Called
aciditybasicity
neutral
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solution
• solute • Solvent (water)electrolyte Non electrolyte
HCl H+1Cl +
NaOH OH-Na
+1+
acid
base
HOHH+OH
-
+
Interplay of these ions is
ACIDITY AND OR BASICITY
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Possible interplays are
H+ OH
-
H+ OH
-
H+ OH
-
<
>
=
Acidic
basic
neutral
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When an acid is added to water , the H+ concentration of the resulting solution is determined solely by the acid
base OH-
base
HCl Cl-
H+
0.01 M
H2O H+ OH
-
H+
solution
=
0.01
0.01
NaOH Na+
OH-
+
0.02 M
H2O H+
OH-
+
OH-
solution
=
0.02 M
0.02 M
Small amount
acidic
Small amount
basic
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Colligative Properties- dependent only on the ratio of the number of solute particle to the number of solvent particles and not on
the nature of the solute.
• Vapor pressure lowering• Boiling point elevation• Freezinf point Depression• Osmotic Pressure
Does not matter wether solute is an acid or base or salt
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Vapor pressure
P air
Temperature : Boiling point
solvent
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Vapor pressure
P air
Solute occupies some areas of the liquid and interfers with the evaporation
less
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Vapor pressure
P air
TBoiling point Elevation
more