Solution Chemistry
Solutions
• A solution is a homogeneous mixture of two or
more substances in a single phase of matter.
• Examples of solutions include salt water, air and
alloys.
• The dissolving medium (water) in a solution is
called the solvent.
• The substance dissolved (salt) in a solution is
called the solute.
• The solute is generally designated as that
component of a solution that is of lesser quantity.
Examples
• If we had a mixture of 25 mL of ethanol
and 75 mL of water, the ethanol would
be the solute and water would be the
solvent.
• PRACTICE: Identify the solute and
solvent in a 1.00 M Sr(NO3)2(aq) solution.
Electrolytes
• Solutions can be characterized by their electrical
conductivity.
• Solutions that contain strong electrolytes will conduct
electricity, whereas solutions of weak or
nonelectrolytes will conduct little or no electricity.
• A strong electrolyte is a compound that dissociates
completely in solution and produce mobile ions, such
as NaCl.
• A weak electrolyte is a compound that only partially
dissociates (like vinegar, a weak acid), and a
nonelectrolyte will not dissociate (an example is
sugar).
•
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/conductivity.html
Solubility
• Solubility is defined as the amount of a
substance that can be dissolved in a given
quantity of solvent.
• Any substance whose solubility is less than
0.01 mol/L will be referred to as insoluble.
• We can predict whether a precipitate will
form when solutions are mixed if we know
the solubilities of different substances.
KISS (keep it simple solubility) Rules
1. All common compounds of Group I and
ammonium ions are soluble.
2. All nitrates, acetates, and chlorates are
soluble.
3 . All halogen compounds (other than F) with
metals are soluble, except those of Ag+, Hg22+,
and Pb2+. (Pb2+ halides are soluble in hot water.)
4. All sulfates are soluble, except those of
barium, strontium, calcium, lead, silver, and
mercury (I). The latter three are slightly
soluble.
* Except for rules above, other ions are generally
insoluble.
Solubility Table
What determines solubility?
Three factors will affect solubility:
#1: Nature of solute and solvent.
Remember the rule:
Like Dissolves Like
Polar solvents (partial + or – charges) will
easily dissolve charged particles or polar
molecules.
Nonpolar solvents (no charges, equal sharing
of e-) will dissolve nonpolar solutes.
Factors Affecting Solubility
#2: Temperature of solution. Generally warmer
solutions will hold more solute (except for
gases).
#3: Pressure (gases only) on solution will
increase solubility.
Sat. vs. Unsat. Solutions
• A solution that cant dissolve any more solute
is said to be saturated.
• Additional solute will not dissolve if added to
this solution.
• An unsaturated solution can still hold more
solute. It has not yet reached its capacity.
• A supersaturated solution can be made by
dissolving the solute under high temps and
then carefully cooling them. These are
unstable solutions and will suddenly
precipitate if provoked.
The Dissolving Process
Ion hydration
• Water is a polar solvent and
is attracted to polar solutes.
• Salt is polar.
• Water molecules surround
and isolate the surface ions.
The ions become hydrated
and move away from each
other in a process called
dissociation.
•
http://www.chem.iastate.edu/group/Greenbowe/sections/proj
ectfolder/flashfiles/thermochem/solutionSalt.html
Temp effects on
solubility
• This graph
represents the
solubility of NaCl,
NaNO3, and
KNO3 at different
temps.
• Notice that when
temp increases,
solubility increases.
Precipitation Reactions
• A solid that forms from solution is called a
precipitate.
• Precipitates form when a combination of
insoluble ions come together:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO 3(aq)
The insoluble solid AgCl will separate from
the solution and precipitate out.
Solution Concentrations
1. Molarity:
M = moles solute
Liters solution
(NOTE: solution is solute PLUS solvent)
2. molality:
m = moles solute
kg solvent
(used when temperature may affect
solution volume)
Molarity Calculations
Calculate the molarity of a solution prepared
with 35.2 grams of CO2 in 500. mL solution.
Step 1: convert 35.2 g of CO2 into moles:
35.2g CO2 x 1 mole CO2 = 0.800 mol CO2
44.01 g CO2
Step 2: divide moles by volume in liters
0.800 mol CO2 = 1.60 M
0.500 L
Molality calculations
• molality is the measure of the number of
moles of a solute per 1000g (1 Kg) of solvent.
m=
moles solute
1000 g (kg) solvent
• molality is represented by m.
Percent concentration
• Percent by volume:
% conc (vol) = volume of solute x 100%
volume of sol’n
• Percent by mass:
% conc (mass) = mass of solute x 100%
mass of sol’n
Colligative Properties
• Colligative comes from the Greek word
kolligativ meaning glued together.
• We use this term for the properties of
substances (solutes and solvents)
together.
• Colligative properties of solutions depend
only on the solvent and the concentration
of the solute, not its identity.
There are 4 colligative properties:
1)
2)
3)
4)
Vapor Pressure Lowering
Boiling Point elevation
Freezing Point Depression
Osmotic Pressure
Two of these properties will be investigated
in the next few slides….
What is a phase diagram?
Normal
melting point:
melting point
at one
atmosphere
(phase diagram for water)
critical point:
beyond this point
the vapor cannot
be liquified at any
pressure
Normal boiling
Point: boiling
point at one
atmosphere
triple point: T and P
at which all three
states coexist in
equilibrium
Vapor pressure and boiling point
• The boiling point of a substance is defined as
the temperature at which the vapor pressure of
a solution is equal to the atmospheric pressure.
vapor pressure: the pressure
of the vapor present at
equilibrium (saturated)
vapor pressure equals
atmospheric pressure,
solvent will boil
• Equilibrium vapor pressure animation:
•
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/vaporv3.swf
Boiling Point and Freezing Point
• Review the phase
diagram of a pure
substance.
• How will the phase
diagram of a solution
(freezing and boiling
points) differ from
those of a pure
solvent?
What happens to boiling point
when solute is added?
• The vapor pressure diagram for solutions shows
that equilibrium vapor pressure is lowered when a
solute is added.
• What will this do to the normal boiling point?
What happens to normal b.p. and f.p. in a solution?
ΔT= boiling point
elevation
ΔT= freezing point
depression
Molality revisited…
• Recall the units for Molarity:
M = moles solute
L of solution
• Molality is the measure of the number of
moles of a solute per 1000g (1 Kg) of solvent.
m=
moles solute
1000 g (kg) solvent
• Molality is best used to describe colligitive
properties and is represented by m.
What was that, again?
• The addition of a nonvolatile solute will require a
higher temperature in which to reach boiling point,
thus:
boiling point elevation:
ΔT =kbmsolute
where kb is a constant that depends on the solvent and
msolute is the molality of the solute
• The addition of a nonvolatile solute will also require a
lower temperature in which to reach freezing point,
thus:
freezing point depression:
ΔT =kfmsolute
where kf is a constant that depends on the solvent and
msolute is the molality of the solute
What are Kb and Kf for water?
• It has been found experimentally that 1 mole of
a nonvolatile solute particles will raise the
boiling temperatures of 1 kg of water by 0.52
C°.
• The same concentration of solute will lower the
freezing point of 1 kg of water by 1.86 C°.
• These two figures are the molal boiling point
constant (Kb) and the molal freezing point
constant (Kf) for dihydrogen oxide.
Reference Table
• The following table contains some examples of
the solvent-dependant constants Kb and Kf.
Solvent
Normal
boiling pt
(°C)
Water
100.0
Benzene
80.1
Ethanol
78.4
CCl4
76.8
Chloroform 61.2
Kb
Normal
(°C/m) freezing
pt (°C)
0.52
0.0
2.53
5.5
1.22
-114.6
5.02
-22.3
3.63
-63.5
Kf
(°C/m)
1.86
5.12
1.99
29.8
4.68
Ionic compounds and molality
• A 1m solution of sugar in water contains 1 mol of
solute particles per 1 kg of solvent. It does NOT
dissociate.
• A 1m solution of NaCl in water contains 2 mol of
solute particles (because NaCl is completely soluble,
it will dissociate in water into Na+ and Cl- ions) per 1
kg of solvent.
• How many mol of solute would a 1m calcium nitrate
[Ca(NO3)2] solution have per 1 kg of solvent?
That’s right!
3 mol because of the Ca2+ and the two NO3- ions.
Calculating ΔTb and ΔTf for ionic cmpds
• Boiling point elevation is:
ΔTb = Kb(m x #of particles)
(molality of ions)
(change in boiling point) (boiling point constant)
Freezing point depression is:
ΔTf = Kf(m x # of particles )
(molality of ions)
(change in freezing point) (freezing point constant)
Example Problem #1
If 55.0 grams of glucose (C6H12O6) are dissolved in 525
g of water, what will be the change in boiling and
freezing points of the resulting solution?
Step 1: Convert g of glucose to moles
55.0 g x (1 mol) = 0.305 mol
(180.18 g)
Step 2: Convert g of water to kg
525g  0.525kg
Step 3: Calculate m
0.305 mol /0.525 kg = 0.581 m
Continued…
• Step 4: Obtain molal Kb and Kf from reference
table.
• Step 5: Place values into equation
ΔTb = Kbm
ΔTb = (0.52°C/m)(0.581m) = 0.30 ºC
• This means that the boiling point will be
elevated by 0.30 °C. This solution will reach
boiling point at 100.30 °C.
• But what about the change in freezing point??
Example #2 (ionic cmpds)
Calculate the change in freezing point of 24.5g
nickel(II) bromide dissolved in 445 g of water.
(assume 100% dissociation)
Step 1: Convert g of NiBr2 into moles:
24.5g x (1 mol)
(218.49 g) = 0.112 mol
Step 2: Convert solvent to kg
445 g  0.445 kg
Step 3: Calculate m
0.112 mol/0.445 kg = 0.252m
Continued…
• We now have to take 0.252 and multiply by 3
because the dissociation of the ionic
compound makes 3 moles of ions (solute) per
kg of solvent:
NiBr2  Ni+2 + 2Br –
0.252 x 3 = 0.756m
• Step 4: Obtain molal Kf from table.
• Step 5: Place values into equation
ΔTf = (1.86°C/m)(0.756m) = 1.41 ºC
*Freezing point has been depressed to -1.41 °C.
• Coolants are used because it takes higher
temperatures to reach boiling point.
• Antifreeze causes fluids to need to reach
lower temperatures in order to freeze.
• This is also why salt is used on frozen
roads and walkways. The salt dissolves in
the snow and lowers the freezing point of
the meltwater. It now takes colder temps to
turn the water back into ice.
• A 10% salt solution freezes at 20 ºF (-6 ºC),
and a 20% solution freezes at 2 ºF (-16 ºC).
The End
Spectrophotometry:
An Analytical Tool
The process of light being absorbed by a solution
concentration 2
concentration 1
blank where Io = I
light
source
with sample
I < Io
detector
Io
I
b
Cell with
path length, b,
containing solution
As concentration
increased, less
light is
transmitted (more
light gets
absorbed).
Beer’s Law
A = abc
where a – molar absorptivity, b – path length,
and c – molar concentration
See the Beer’s Law Simulator
Analyze at what wavelength?
Scan visible wavelengths from 400 – 650 nm
(detector range) to produce an absorption
spectrum (A vs. l)
Crystal Violet Absorption Spectrum
1.4
Absorbance
1.2
1
0.8
0.6
lmax
0.4
0.2
0
200
250
300
350
400
450
500
wavelength, nm
550
600
650
phototube detector range
700
750
lmax - wavelength where maximum absorbance occurs
The BLANK
The blank contains all substances
except the analyte.
Is used to set the absorbance to zero:
Ablank = 0
This removes any absorption of light
due to these substances and the cell.
Therefore, all measured absorbance is
due to analyte.
The components of a Spec-20D
Light source - white light of constant intensity
slits
filter
occluder
Grating
Phototube
detects light &
measures intensity
slits
Sample
When blank is the sample
Io is determined
otherwise I is measured
Separates white light
into various colors
Rotating the grating
changes the wavelength
going through the sample
When calibration curves go bad!
• The linear Beer’s Law relationship starts to
show curvature at high concentrations
Absorbance
1
Calibration Curve
0.8
Non-linear
0.6
linear
curved
Linear (linear)
0.4
0.2
0
0
0.2
0.4
0.6
concentration
0.8
1
Calibrating the Spec 20
•
•
•
•
•
Plug in and turn on the Spec 20. It must warm up for 30
minutes before use.
Set the instrument to the proper wavelength by turning the
knob located on the right hand surface of the
spectrophotometer. The wavelength setting can be seen
through the window next to the knob.
Obtain a properly cleaned cuvette and fill it about 3/4 full of
the reference solution (water).
With no cuvette in the sample holder, close the cover and
rotate the zero light control knob(left front knob) to display a
reading of 0.0% transmittance. As long as this knob is not
moved, no other adjustments to this control are needed.
Place the reference solution cuvette in the sample holder,
close the cover, and rotate the light control knob (front right
knob) to display a reading of 100.0% transmittance. This
procedure must be repeated every time measurements are
taken at a new wavelength.
Making Measurements
•
•
•
•
Calibrate the instrument (see instructions on
last slide) at the wavelength you wish to
measure. You will use the solvent of your
sample solution as your reference (this will
usually be water).
Fill a properly cleaned cuvette 3/4 full of you
sample solution.
Place your sample cuvette in the sample
holder and close the cover.
Read either the absorbance or percent
transmittance as needed.
Chemistry of Ice Cream
When looking at ice cream at a microscopic
level it is found that ice cream is made up of
four phases: ice, air, fat and a concentrated
aqueous solution. It is the relative amount of
these phases and the interactions between
them that determines the properties of the ice
cream – whether soft and ‘whippy’ or hard.
electron micrographs of ice cream
a = air bubbles, c = ice crystals, f = fat and s = concentrated aqueous solution
ICE CREAM INGREDIENTS
Today's ice cream has the following composition :
a) At least10% milk fat by legal definition, and usually
between 10% and as high as 16% fat in some premium
ice creams.
b) between 9 and 12% milk solids (non-fat), the component
which contains the proteins and carbohydrates (lactose)
found in milk.
c) 12% to 16% sweeteners, usually a combination of
sucrose and glucose-based corn syrup sweeteners
d) 0.2% to 0.5% added stabilizers and emulsifiers
e) The rest, usually 55% to 64%, is water, which comes
from the milk. Air is also whipped into ice cream to add
volume and keep it from melting quickly.
ICE CREAM: FAT AND SUGAR
• The fat component in ice cream adds richness of flavor
and contributes to a smooth texture with creamy body.
The milk solids-not-fat component also contributes to the
flavor but more importantly improves the body and
texture of the ice cream by offering some "chew
resistance" and enhancing the ability of the ice cream to
hold its air.
• The sugars give ice cream its characteristic sweetness
and enhance the perception of various fruit flavors. In
addition, the sugars, including the lactose from the milk
components, contribute to a depressed freezing point so
that the ice cream has some unfrozen water associated
with it at very low temperatures typical of their serving
temperatures, -15o to -18oC. Without this unfrozen water,
the ice cream would be too hard to scoop.
ICE CREAM: STABILIZERS
• The stabilizers are compounds (usually cellulose or
bean gum) that are responsible for adding viscosity to
the unfrozen portion of the water and thus holding this
water so that it cannot migrate. This results in firmer
ice cream. Without the stabilizers, the ice cream would
become coarse with large ice crystals very quickly as
water migrates and refreezes.
• The smaller the ice crystals in the ice cream, the less
detectable they are to the tongue. Especially in the
distribution to supermarkets, the trunks of cars, and so
on, ice cream has many opportunities to warm up,
partially melt some of the ice, and then refreeze as the
temperature is once again lowered. This process is
known as heat shock and every time it happens, the
ice cream becomes more icy tasting. Stabilizers help
to prevent this.
ICE CREAM: EMULSIFIERS
• Colloids are mixtures of two substances that are
insoluble. Ice cream contains colloids since fat
does not dissolve in water. The emulsifiers are a
group of compounds in ice cream which aid in
blending the fats and water. Emulsifiers are
characterized by having a molecular structure
which allows part of the molecule to be readily
soluble in a polar compound such as water, and
another part of the molecule to be more readily
soluble in non-polar solvents such as fats. As a
result, emulsifiers reside at the interface
between fat and water.