Solutions

1
Solutions

• Solutions
• Chemistry I

2
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

3
Dissolution of Solid Solute

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

4
Click Here
5
Solution and Concentration

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

6
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
7
Two Other Concentration Units
MOLALITY, m
by mass
grams solute grams solution
by mass
8
Calculating Concentrations

• Dissolve 62.1 g (1.00 mol) of ethylene glycol in
  250. g of H2O. Calculate molality and by mass
  of ethylene glycol.

Ethylene Glycol
9
Calculating Concentrations
Dissolve 62.1 g (1.00 mol) of ethylene glycol in
250. g of H2O. Calculate m of ethylene glycol
(by mass).

• Calculate molality

Calculate weight
10
Try this molality problem

• 25.0 g of NaCl is dissolved in 5000. mL of water.
  Find the molality (m) of the resulting solution.

m mol solute / kg solvent 25 g NaCl 1
mol NaCl 58.5 g NaCl
0.427 mol NaCl
Since the density of water is 1 g/mL, 5000 mL
5000 g, which is 5 kg
0.427 mol NaCl 5 kg water
0.0854 m salt water
11
Dilution

• When a solution is diluted, solvent is added to
  lower its concentration. A stock solution refers
  to a more concentrated solution. When water is
  added to a stock solution, its concentration
  lowers.
• The amount of solute remains constant before and
  after the dilution
• moles BEFORE moles AFTER
• M1V1 M2V2

Click here for dilution movie
12

• Making a Dilute Solution

13
Making a dilute solution example
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.
14
Answer

• M1V1 M2V2
• (0.500M)V1 (0.348M)(0.250L)
• V1 (0.348M)(0.250L)
• V1 0.174L or 174 mL
• Answer Obtain 174 mL from the 0.500M stock
  solution. Place this into a 250. mL flask. Add
  water to the stock solution (76.0 mL).

0.500M
15
Miscible solutions

• Miscible means the two liquids dissolve into
  each other. Ex. Water ethanol
• Immiscible the two liquids do not dissolve into
  each other. Ex. Water and oil

Immiscible liquids
16
Strong electrolytes

• A strong electrolyte is a solute that will
  dissociate completely into ions in an aqueous
  solution.

17
Weak electrolytes

• A weak electrolyte partially dissociates. Most
  solute particles stay intact and only a few
  dissociate into ions. Acetic acid, as shown on
  the left, is an example of a weak electrolyte.

18
Nonelectrolytes

• A non-electrolyte is a solute that does not
  dissociate into ions. The solute particle stays
  intact. An example is sugar. Most covalent
  solutes are non-electrolytes

19
Like dissolves like. Polar solutes will dissolve
in polar solvents. Nonpolar solutes will dissolve
in nonpolar solvents.

• Ethylene glycol has polar properties so it will
  dissolve in water.
• Oil, a nonpolar substance, will not dissolve in
  water, a polar substance.

20
3 Stages of Solution Process

• Separation of Solute
• must overcome inter-molecular forces (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)

21
Dissolution at the molecular level?

• Consider the dissolution of NaOH in H2O

22
Heat of solution

• the heat evolved or absorbed when one mole of a
  substance is dissolved in a large volume of a
  solvent.
• ?H q
• Remember q (mass)(specific heat)(?T)

mol
23
Heat of Solution

• If 0.860 g copper(II) bromide is dissolved in
  100.0 mL of water, the temperature changes from
  23.10C to 23.41C. What is the heat of solution
  for copper(II) bromide? Assume the specific heat
  of water, 4.184 J/gC.
• The reaction is
• CuBr2 (s) ? Cu2 (aq) 2Br (aq)
• Was this an endothermic or exothermic reaction?

24
Answer

• To determine the amount of heat absorbed (q), use
  the heat equation. Remember that the heat energy
  is absorbed by the entire solution that includes
  the solvent as well as the solute. Recall that
  the density of water is 1.00 g/mL, so 100.0 mL of
  water is also 100.0 g water.
• q (mass)(specific heat)(T)
• q (100.860 g)(4.184 J/gC)(23.41 C 23.10
  C)
• q (100.860)(4.184)(0.31)
• q 130.8 J
• Since the temperature increases (exothermic), the
  ?H value will be negative.
• mol CuBr2    0.860 g x 1 mol
• 
  223.35 g  
• ?H    (130.8 J)
• 0.00385 mol  
• 33974 J/mol x 1 kJ /1000 J    33.974 kJ/mol
• ? Hsoln 3.4 x 104 J/mol or 34kJ/mol (2 sig
  figs)

  0.00385 mol so
   33974 J/mol
25
Heat of solution Ex. 2

• What is the heat of solution and the dissolution
  reaction if when 1.893 g lithium fluoride
  dissolves in 250.0 mL water, the temperature
  changes from 23.69 C to 19.59 C?
• The reaction is LiF (s) ? Li1(aq) F-1(aq)
• Is this an endothermic or exothermic reaction?

26
Answer

• The moles of lithium fluoride that dissolved are
• 1.893 g LiF x 1 mol
• 25.94 g  
• The heat evolved is determined from
• q (mass solution)(specific heat water)(T)
• q (251.9 g)(4.184 J/gC)(23.69 C 19.59 C)
• q (251.9)(4.184)(4.10)
• q 4321 J
• The heat of solution will be positive because the
  temperature of the solution decreased
  (endothermic).
• ?H    q/ mol    (4321 J)/ 0.07298 mol   
  59,211 J/mol  
•   5.92 x 104 J/mol OR
• 5.92 x 104 J/mol x 1 kJ
• 1000 J  

  0.07298 mol LiF
   59.2 kJ/mol
27
Rate of solution

• Rate of solution is how quickly a solute
  dissolves in a solvent. Factors determining the
  rate of solution are
• Surface Area When a solute dissolves, only the
  surface of the solute comes in contact with the
  solvent. Therefore, the more the surface area of
  the solute, the faster it dissolves.
• Stirring When you're dealing with solid and
  liquid solutes, stirring brings fresh parts of
  the solvent into contact with the solute and
  particles are forced to connect.
• Temperature Increasing the temperature also
  generally increases the amount of solute the
  solvent can hold (solid and liquid solutes but
  not gas solutes).

28
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.
29
The effect of pressure on gas solubility.
30
Solubility Curves

• 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.

31
Solubility curves

• Each point on any curve represents a saturated
  solution. A saturated solution cannot dissolve
  any more solute at that temperature.
• An unsaturated solution will dissolve more solute
  at a certain temperature until it becomes
  saturated.
• The solubility of KNO3 is 30 g/100 g of water at
  20 Co. How many more grams of solute must be
  added to keep the solution saturated at 50 Co.
• How many grams of Ce2(SO4)3 will precipitate out
  of solution when it is cooled from 70 Co to 10
  Co? (assume 100 g of solvent)
• What mass of KClO3 is required to saturate 50 g
  of water at 70 Co?

32
Definitions

• SUPERSATURATED SOLUTIONS contain more solute than
  is possible to be dissolved
• Supersaturated solutions are unstable. The
  supersaturation is only temporary, and usually
  accomplished in one of two ways
• Warm the solvent so that it will dissolve more,
  then cool the solution
• Evaporate some of the solvent carefully so that
  the solute does not solidify and come out of
  solution.

Click here to watch movie on supersaturated
solutions
33
Temperature the Solubility of GasesThe
solubility of gases DECREASES at higher
temperatures since the Intermolecular forces can
not be formed between the gas and the solvent.
34
Pressure and 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.
• Therefore, bubbles of CO2 escape from solution.

35
Colligative Properties

• Dissolving solute in pure liquid will change all
  physical properties of liquid, Density, Vapor
  Pressure, Boiling Point, Freezing Point
• 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.

36
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.
• The magnitude of the boiling-point elevation
  is proportional to the number of solute particles
  dissolved in the solvent.

Elevation of Boiling point occurs in the beaker
on the right.
37
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
• i Vant Hoff factor

38
vant Hoff factor (i)

• The vant Hoff factor is a whole number integer
  that has the value of 1, 2, 3, etc. It is never
  zero or a negative number.
• This number represents the number of particles
  based on the dissociation of a solute in water.
• The vant Hoff factor for any covalent solute is
  always 1. Covalent compounds are always nonmetals
  only.
• The vant Hoff factor for any ionic solute is
  equal to the number of ions in that compound.
  Ionic compounds always involve 1 metal.

39
vant Hoff factor examples

• Determine the number of particles that are
  represented in each of the following substances
• C6H12O6
• NaCl
• Al2(SO4)3
• Ca(C2H3O2)2
• CCl4
• CaCl2
• Al(NO3)3
• C7H6O2

1
2
5
3
1
3
4
1
40
Note the vant Hoff factor is 2
x 2
41
(No Transcript)
42
Ex. 2 Bioling pt. elevation

• What is the boiling point elevation when 11.4 g
  of ammonia (NH3) is dissolved in 200g of water?
  Kb for water is 0.52 Co/m.

43
Solution

• First find moles of solute 11.4 g
• Find molality of solution
• Determine the kg of solvent 200 g x 1kg
• Molality 0.669 mol/0.2000 kg 3.35 mol/kg
• ?T Kf x m x i
• ?T 0.52 m/Co x 3.35 m x 1
• ?T 3.37 Co so the boiling pt of this solution
  will be 103.37 Co

0.669 mol
17.04 g/mol
0.2000 kg
1000 g
44
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.
45
Change in Freezing Point
Ethylene glycol/water solution
Pure water

• The freezing point of a solution is LOWER than
  that of the pure solvent

46
x 1
47
(No Transcript)
48
Freezing pt. depression problem.

• How many grams of pyrozole (C3H4N2) must be added
  to 451 g of benzene to lower the freezing point
  by 5.00 Co? Kf for benzene is 5.12 Co/m.

49
Answer

• First analyze what you are given
• Benzene is the solvent. So,
• 451 g x 1 kg 0.451 kg
• ?T 5.00 Co
• Kf 5.12 Co/m
• i 1 since solute is covalent
• Substitute your givens into the equation
• ?T Kf x i x m
• 5.00 Co 5.12 Co/m x 1x m
• m 0.977mol/kg
• Solve for mass using the equation for molality
• m mol/kg so 0.977 mol/kg x 0.451 kg 0.954 mol

1000 g
50
Answer

• From the previous slide mol 0.954 mol
• Remember mol grams/molar mass. To solve for
  mass in grams multiply mol and molar mass.
• 0.954 mol x 68.09 g/mol 65.0 grams