Solutions

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Solutions
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Definitions

• A solution is a homogeneous mixture
• A solute is dissolved in a solvent.
• solute is the substance being dissolved
• solvent is the liquid in which the solute is
  dissolved
• an aqueous solution has water as solvent
• A saturated solution is one where the
  concentration is at a maximum - no more solute is
  able to dissolve.
• A saturated solution represents an equilibrium
  the rate of dissolving is equal to the rate of
  crystallization. The salt continues to dissolve,
  but crystallizes at the same rate so that there
  appears to be nothing happening.

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Dissolution of Solid Solute

• What are the driving forces which cause solutes
  to dissolve to form solutions?
• 1. Covalent solutes dissolve by H-bonding to
  water or by LDF
• 2. Ionic solutes dissolve by dissociation into
  their ions.

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Sucrose or commonly known as sugar
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Solution and Concentration

• 4 ways of expressing concentration
• Molarity(M) moles solute / Liter solution
• Mass percent (mass solute / mass of solution)
  100
• Molality (m) - moles solute / Kg solvent
• Mole Fraction(?A) - moles solute / total moles
  solution

Note that molality is the only concentration
unit in which denominator contains only solvent
information rather than solution.
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Concentration

• (w/w)
• (w/v)
• (v/v)

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Concentration Mass Example

• 3.5 g of CoCl2 is dissolved in 100mL solution.
• Assuming the
• density of the solution is 1.0 g/mL, what is
  concentration of the solution in mass?

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Concentration Molarity Example

• If 0.435 g of KMnO4 is dissolved in enough water
  to give 250. mL of solution, what is the molarity
  of KMnO4?

As is almost always the case, the first step is
to convert the mass of material to moles.
0.435 g KMnO4 1 mol KMnO4 0.00275 mol
KMnO4 158.0 g KMnO4
Now that the number of moles of substance is
known, this can be combined with the volume of
solution which must be in liters to give the
molarity. Because 250. mL is equivalent to 0.250
L .
Molarity KMnO4 0.00275 mol KMnO4 0.0110
M 0.250 L solution
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Dilution

• When a solution is diluted, solvent is added to
  lower its concentration.
• The amount of solute remains constant before and
  after the dilution
• moles BEFORE moles AFTER
• C1V1 C2V2

A bottle of 0.500 M standard sucrose stock
solution is in the lab. Give precise
instructions to your assistant on how to use the
stock solution to prepare 250.0 mL of a 0.348 M
sucrose solution.
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3 Stages of Solution Process

• Separation of Solute
• must overcome IMF or ion-ion attractions in
  solute
• requires energy, ENDOTHERMIC ( DH)
• Separation of Solvent
• must overcome IMF of solvent particles
• requires energy, ENDOTHERMIC ( DH)
• Interaction of Solute Solvent
• attractive bonds form between solute particles
  and solvent particles
• Solvation or Hydration (where water
  solvent)
• releases energy, EXOTHERMIC (- DH)

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Dissolution at the molecular level?

• Consider the dissolution of NaOH in H2O

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Factors Affecting Solubility
1. Nature of Solute / Solvent. - Like dissolves
like (IMF) 2. Temperature - i) Solids/Liquids-
Solubility increases with Temperature Increase
K.E. increases motion and collision between
solute / solvent. ii) gas - Solubility decreases
with Temperature Increase K.E. result in gas
escaping to atmosphere. 3. Pressure Factor - i)
Solids/Liquids - Very little effect Solids and
Liquids are already close together, extra
pressure will not increase solubility. ii) gas
- Solubility increases with Pressure. Increase
pressure squeezes gas solute into solvent.
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Solubilities of Solids vs Temperature

• Solubilities of several ionic solid as a function
  of temperature. MOST salts have greater
  solubility in hot water.
• A few salts have negative heat of solution,
  (exothermic process) and they become less soluble
  with increasing temperature.

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Temperature the Solubility of GasesThe
solubility of gases DECREASES at higher
temperatures
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Henrys LawThe effect of partial pressure on
solubility of gases

• At pressure of few atmosphere or less, solubility
  of gas solute follows Henry Law which states that
  the amount of solute gas dissolved in solution is
  directly proportional to the amount of pressure
  above the solution.
• c k P
• c solubility of the gas (M)
• k Henrys Law Constant
• P partial pressure of gas
• Henrys Law Constants (25C), k
• N2 8.42 10-7 M/mmHg
• O2 1.66 10-6 M/mmHg
• CO2 4.4810-5 M/mmHg

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Henrys Law Soft Drinks

• Soft drinks contain carbonated water water
  with dissolved carbon dioxide gas.
• The drinks are bottled with a CO2 pressure
  greater than 1 atm.
• When the bottle is opened, the pressure of CO2
  decreases and the solubility of CO2 also
  decreases, according to Henrys Law.
• Therefore, bubbles of CO2 escape from solution.

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Henrys Law Application

• The solubility of pure N2 (g) at 25oC and 1.00
  atm pressure is 6.8 x 10-4 mol/L. What is the
  solubility of N2 under atmospheric conditions if
  the partial pressure of N2 is 0.78 atm?

Step 1 Use the first set of data to find k
for N2 at 25C Step 2 Use this constant to
find the solubility (concentration) when P is
0.78 atm
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Colligative Properties

• Dissolving solute in pure liquid will change all
  physical properties of liquid, Density, Vapor
  Pressure, Boiling Point, Freezing Point, Osmotic
  Pressure
• Colligative Properties are properties of a liquid
  that change when a solute is added.
• The magnitude of the change depends on the number
  of solute particles in the solution, NOT on the
  identity of the solute particles.

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Vapor Pressure Lowering for a Solution

• The diagram below shows how a phase diagram is
  affected by dissolving a solute in a solvent.
• The black curve represents the pure liquid and
  the blue curve represents the solution.
• Notice the changes in the freezing boiling
  points.

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Vapor Pressure Lowering

• The presence of a non-volatile solute means that
  fewer solvent particles are at the solutions
  surface, so less solvent evaporates!

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Application of Vapor Pressure Lowering

• Describe what is happening in the pictures below.
• Use the concept of vapor pressure lowering to
  explain this phenomenon.

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Raoults LawDescribes vapor pressure lowering
mathematically.

• The lowering of the vapour pressure when a
  non-volatile solute is dissolved in a volatile
  solvent (A) can be described by Raoults Law
• PA cAPA
• PA vapour pressure of solvent A above the
  solution
• cA mole fraction of the solvent A in the
  solution
• PA vapour pressure of pure solvent A

only the solvent (A) contributes to the vapour
pressure of the solution
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What is the vapor pressure of water above a
sucrose (MW342.3 g/mol) solution prepared by
dissolving 158.0 g of sucrose in 641.6 g of water
at 25 ºC? The vapor pressure of pure water at 25
ºC is 23.76 mmHg.
mol sucrose (158.0 g)/(342.3 g/mol) 0.462 mol
mol water (641.6 g)/(18 g/mol) 35.6 mol
Psoln Xwater P?water (0.987)(23.76 mm Hg)

23.5 mm Hg
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Mixtures of Volatile LiquidsBoth liquids
evaporate contribute to the vapor pressure
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Raoults Law Mixing Two Volatile Liquids

• Since BOTH liquids are volatile and contribute to
  the vapour, the total vapor pressure can be
  represented using Daltons Law
• PT PA PB
• The vapor pressure from each component follows
  Raoults Law
• PT cAPA cBPB
• Also, cA cB 1 (since there are 2
  components)

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Benzene and Toluene

• Consider a two solvent (volatile) system
• The vapor pressure from each component follows
  Raoult's Law.
• Benzene - Toluene mixture
• Recall that with only two components, cBz
  cTol 1
• Benzene when ?Bz 1, PBz PBz 384 torr
  when ?Bz 0 , PBz 0
• Toluene when ?Tol 1, PTol PTol 133 torr
  when ?Tol 0, PBz 0

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Normal Boiling Process

• Extension of vapor pressure concept
• Normal Boiling Point BP of Substance _at_ 1atm
• When solute is added, BP gt Normal BP
• Boiling point is elevated when solute inhibits
  solvent from escaping.

Elevation of B. pt. Express by Boiling point
Elevation equation
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Boiling Point Elevation

• ?Tb (Tb -Tb) i m kb
• Where, ?Tb BP. Elevation
• Tb BP of solvent in solution
• Tb BP of pure solvent
• m molality , kb BP Constant

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Freezing Point Depression

• Normal Freezing Point FP of Substance _at_ 1atm
• When solute is added, FP lt Normal FP
• FP is depressed when solute inhibits solvent from
  crystallizing.

When solution freezes the solid form is almost
always pure. Solute particles does not fit
into the crystal lattice of the solvent because
of the differences in size. The solute
essentially remains in solution and blocks other
solvent from fitting into the crystal lattice
during the freezing process.
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Freezing Point Depression
?Tf i m kf Where, ?Tf FP
depression i vant Hoff Factor m
molality , kf FP Constant Generally
freezing point depression is used to determine
the molar mass of an unknown substance. Derive
an equation to find molar mass from the equation
above.

• Phase Diagram and the lowering of the freezing
  point.

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Osmotic pressure

• Osmosis is the spontaneous movement of water
  across a semi-permeable membrane from an area of
  low solute concentration to an area of high
  solute concentration
• Osmotic Pressure - The Pressure that must be
  applied to stop osmosis

P i CRT where P osmotic pressure
i vant Hoff factor C molarity
R ideal gas constant T
Kelvin temperature
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Osmosis and Blood Cells
(a) A cell placed in an isotonic solution. The
net movement of water in and out of the cell is
zero because the concentration of solutes inside
and outside the cell is the same. (b) In a
hypertonic solution, the concentration of solutes
outside the cell is greater than that inside.
There is a net flow of water out of the cell,
causing the cell to dehydrate, shrink, and
perhaps die. (c) In a hypotonic solution, the
concentration of solutes outside of the cell is
less than that inside. There is a net flow of
water into the cell, causing the cell to swell
and perhaps to burst.