Water and the Properties of
Solutions
Homework
• The first 8 objectives are covered well in your text. You are responsible
for making notes:
1.
2.
3.
4.
5.
6.
7.
8.
Distinguish between heterogeneous and homogeneous mixtures. (13-1).
Distinguish between electrolytes and nonelectrolytes. (13-1)
Compare the properties of suspensions, colloids, and solutions. (13-1)
List and explain three factors that influence the rate of dissolving of a solid
in a liquid. (13-2)
Explain terms related to solutions and solubility. (13-2)
Explain how solutions form and use principles of polarity to determine
whether a compound will be miscible or immiscible in a particular solvent.
(13-2)
List the three interactions that contribute to the heat of solution, and
explain what causes dissolving to be exothermic or endothermic. (13-2)
Compare the effects of temperature and pressure on solubility. (13-2)
Water
• is an unusual substance, primarily
because of polarity and hydrogen
bonding.
• exists as a liquid at room temperature. It’s
melting and boiling points are very high
compared to substances of similar
molecular weight
Substance
CH4
C3H8
C4H10
O2
CO2
N2
H2O
Molar Mass
Boiling Point
16.04 g/mol
- 164 ºC
44.11 g/mol
- 42.1 ºC
58.14 g/mol
- 0.5 ºC
32.00 g/mol
- 183 ºC
44.01 g/mol
- 78.5 ºC
28.02 g/mol
- 196 ºC
18.02 g/mol
+ 100 ºC
Water
• occurs in three phases normally on Earth
(solid, liquid, vapour).
Water
• solid is less dense
than the liquid.
• ice floats.
Water
• has a high specific heat capacity.
• has a very high heat of fusion and
vapourization.
• this allows water to store and move vast
quantities of heat on the planet.
Water
• has a high surface
tension.
• this leads to capillary
action which allows
vascular plants to grow
to great heights.
Water
• is the universal solvent
• everything dissolves in water to some
degree.
• can generalize “Like dissolves Like”
Solubility
• Soluble in water
– polar solutes
• sugars
• alcohols
– ionic solutes
• salts
– low molecular weight, non-polar molecules
• O2 N 2
CO2
• Not soluble in water
– most hydrocarbons
• fats
• oils
• gasoline
Solutions
• Solutions are homogeneous mixtures of two or
more pure substances.
• In a solution, the solute is dispersed uniformly
throughout the solvent.
Solutions
The intermolecular
forces between solute
and solvent particles
must be strong enough
to compete with those
between solute particles
and those between
solvent particles.
How Does a Solution Form?
As a solution forms, the solvent pulls solute
particles apart and surrounds, or solvates, them.
How Does a Solution Form
If an ionic salt is
soluble in water, it is
because the iondipole interactions are
strong enough to
overcome the lattice
energy of the salt
crystal.
Types of Solutions
• Saturated
 Solvent holds as much
solute as is possible at
that temperature.
 Dissolved solute is in
dynamic equilibrium
with solid solute
particles.
Types of Solutions
• Unsaturated
 Less than the
maximum amount of
solute for that
temperature is
dissolved in the
solvent.
Types of Solutions
• Supersaturated
 Solvent holds more solute than is normally
possible at that temperature.
 These solutions are unstable; crystallization can
usually be stimulated by adding a “seed crystal” or
scratching the side of the flask.
Factors Affecting Solubility
• Chemists use the axiom
“like dissolves like”:
 Polar substances tend to
dissolve in polar solvents.
 Nonpolar substances tend
to dissolve in nonpolar
solvents.
Factors Affecting Solubility
The more similar the
intermolecular
attractions, the more
likely one substance
is to be soluble in
another.
Factors Affecting Solubility
Glucose (which has
hydrogen bonding) is
very soluble in water,
while cyclohexane
(which only has
dispersion forces) is
not.
Factors Affecting Solubility
• Vitamin A is soluble in nonpolar compounds (like
fats).
• Vitamin C is soluble in water.
Gases in Solution
• In general, the
solubility of gases in
water increases with
increasing mass.
• Larger molecules
have stronger
dispersion forces.
Gases in Solution
• The solubility of
liquids and solids
does not change
appreciably with
pressure.
• The solubility of a gas
in a liquid is directly
proportional to its
pressure.
Henry’s Law
Sg = kPg
where
• Sg is the solubility of
the gas;
• k is the Henry’s law
constant for that gas in
that solvent;
• Pg is the partial
pressure of the gas
above the liquid.
Temperature
Generally, the
solubility of solid
solutes in liquid
solvents increases
with increasing
temperature.
Temperature
• The opposite is true
of gases:
 Carbonated soft drinks
are more “bubbly” if
stored in the
refrigerator.
 Warm lakes have less
O2 dissolved in them
than cool lakes.
Concentration
• is a measure of the amount of solute, per
unit of solution or solvent.
• is central to quantitative chemistry, since
most chemical reactions happen in
solution.
Molar Concentration
• is the most common measure of
concentration used in the lab.
• is a measure of the number of moles of
solute, per litre of solution (mol/L, mol∙L-1).
• is also called Molarity (M).
• for instance, 0.12 mol/L ≡ 0.12 M
Examples of Molarity
• Calculate the molar concentration when 0.256 mol of
CuSO4 is dissolved in 1.1 L of solution.
• molar concentration =
•
•
c
=
=
=
moles of solute
volume of solution
n
v
0.256 mol
1.1 L
0.23 mol/L (2 s.d.)
Examples of Molarity
•
Calculate the molar concentration when 28.2 g of KNO3 is dissolved in 750. mL of
solution.
•
•
molar mass:
•
moles of substance = 28.2 g
101.11 g/mol
•
•
volume must be reported in L:
750. mL x
•
•
molar concentration:
•
KNO3
1xK
1xN
3xO
c =
=
n
V
= 0.372 mol/L
=
=
=
1 x 39.10 g/mol
1 x 14.01 g/mol
3 x 16.00 g/mol
101.11 g/mol
0.279 mol
1L
1000 mL
=
= 0.750 L
0.279 mol
0.750 L
Examples of Molarity
• Calculate the volume of solution made with 0.455 mol of
sodium phosphate in a 2.00 M solution.
•
c
=
•
V
=
•
=
•
=
n
V
n
c
0.455 mol
2.00 mol/L
0.228 L
Molarity Examples
• How many moles are present in 4.50 L of a
0.025 mol/L solution of magnesium nitrate?
•
•
•
•
•
c
=
n
=
=
=
n
V
cV
(0.025 mol/L)(4.50 L)
0.11 mol
Molarity Problems
• text, pages 941 & 942
• questions 332, 333, 335, 336, 346
Molal Concentration
• is used in industry and for measurements
of colligative properties (see Ch. 14); it is
useful because it does not depend on the
final volume of solution.
• is a measure of the number of moles of
solute, per kilogram of solvent (mol/kg,
mol∙kg-1).
• is also called Molality (M).
• for instance, 0.12 mol/kg ≡ 0.12 M
Molal Examples
• Calculate the molal concentration when 2.56 mol of
CuSO4 is dissolved in 12.0 kg of water.
• molal concentration
•
•
•
M
= moles of solute
mass of solvent
=
n
ms
•
•
=
2.56 mol
12.0 kg
•
=
0.213 mol/kg
Molal Examples
•
Calculate the molality when 19 g of Mg(NO3)2 is dissolved in 1200 g of
water.
•
•
•
•
•
Mg(NO3)2
1 x Mg
2xN
6xO
•
•
1200 g x
•
•
M
=
•
•
M
=
•
M
= 0.11 mol/kg
=
=
=
1 kg
1000 g
1 x 24.30 g/mol
2 x 14.01 g/mol
6 x 16.00 g/mol
148.32 g/mol
=
n
ms
m
msM
=
1.2 kg
n
=
19 g
(1.2 kg)(148.32 g/mol)
m
M
Molality Examples
•
Calculate the mass of solvent used with 61.2 g of sodium phosphate in a 2.00 M solution.
•
sodium phosphate = Na3PO4
•
•
•
•
3 x Na
1xP
4xO
•
•
M
=
n
ms
•
•
ms
=
n
•
•
•
•
ms
=
•
ms
ms
M
m
MM
=
61.2 g
(2.00 mol/kg)(163.94 g/mol)
= 0.187 kg
=
=
=
n =
3 x 22.99 g/mol
1 x 30.97 g/mol
4 x 16.00 g/mol
163.94 g/mol
m
M
Molality Examples
• What mass of lithium hydroxide is present in a 0.025 mol/kg solution
with 360. kg of water?
• lithium hydroxide = LiOH
•
•
•
•
• M=
n
•
ms
1 x Li
1xO
1xH
=
=
=
n =
m
M
• M
•
• m
=
•
•
•
= (0.025 mol/kg)(360. kg)(23.95 g/mol)
m
msM
= M msM
= 220 g
1 x 6.94 g/mol
1 x 16.00 g/mol
1 x 1.01 g/mol
23.95 g/mol
Molality Problems
• text, pages 941 & 942
• questions 338, 339, 340, 350, 352
Other Measures of Concentration
• Parts per million ("ppm") denotes one particle of a given
substance for every 999,999 other particles. This is roughly
equivalent to one drop of ink in a 150 litre (40 gallon) drum
of water, or 1 g of solute in 1000 kg of solution.
• Parts per billion ("ppb") denotes one particle of a given
substance for every 999,999,999 other particles. This is
roughly equivalent to one drop of ink in a lane of a public
swimming pool, or 1 gram of solute in 106 kg of solution
(roughly Riversdale pool).
• Parts per trillion ("ppt") denotes one particle of a given
substance for every 999,999,999,999 other particles. This is
roughly equivalent to one drop of ink in a shipping canal lock
full of water , or 1 g of solute in Blackstrap lake.
Dilution
• many substances, especially acids, are received
in a concentrated form (hydrochloric acid is 12.4
mol/L). This is called the stock solution.
• to use the chemicals in the lab they are usually
diluted to a concentration much less than they
are received.
• the problem is to decide how much of the stock
solution you need to make the solution you want.
Dilution
• For instance; you have a 12.4 mol/L stock
solution of HCl and you want to make
2.00 L of 0.100 mol/L solution.
• You can calculate the number of moles in
the solution:
•
•
•
n
= cv
= (0.100 mol/L)(2.00 L)
= 0.200 mol
Dilution
• 0.200 mol represents the number of moles of HCl you
need from the stock solution.
• Now you have the number of moles of stock solution and
the concentration of the stock solution, so you can
calculate the volume of concentrated HCl you need to
add to 2.00 L of water to make the dilute solution:
•
•
v
=
n
c
=
0.200 mol
12.4 mol/L
•
= 0.0161 L (1000 mL/L)
•
= 16.1 mL of concentrated HCl is added to
water to make 2.00 L of a 0.100
mol/L solution.
Dilution
• Fortunately, there is an easier way to do this. Since the
moles taken from the stock solution (ns) is the same
number of moles that goes into the dilute solution (nd)
•
• and
•
•
ns = nd
ns = csvs
nd = cdvd
• so the formula for dilution is:
•
csvs = cdvd
Dilution
• What volume of concentrated sulfuric acid
(18 M) is needed to make 500. mL of 2.00
M dilute solution?
• CsVs = CdVd
• (18 mol/L)Vd = (2.00 mol/L)(500. mL)
• Vd = 56 mL concentrated H2SO4 is
needed.
Dilution
• What is the concentration of dilute solution if
25.0 mL of glacial acetic acid (24.0 mol/L) is
added to make 2.00 L of dilute solution?
• CsVs = CdVd
• (24.0 mol/L)(25.0 mL) = (Cd)(2.00 L)(1000 mL/1 L)
• Cd = 0.300 mol/L acetic acid.
Dilution
• What volume of a 1.00 mol/L stock solution of
aluminum chloride is needed to make 500. mL
of 3.0 x 10-4 mol/L solution?
• CsVs = CdVd
• (1.00 mol/L)Vd = (3.0 x 10-4 mol/L)(500. mL)
• Vd = 0.15 mL AlCl3 stock solution is needed.
Serial Dilution
• Serial dilutions are an accurate method of making solutions of low
molar concentrations.
• Since measuring small volumes of solution is prone to error, a series
of dilutions are performed in order to gradually reduce the
concentration of the solution from that of the stock solution.
• Serial "ten-fold" dilutions are commonly used.
• To carry out a serial "ten-fold" dilution you would do the following:
•
•
•
•
•
Add 9 ml of distilled water to each test tube.
To the first test tube, add 1 ml of the stock solution.
Mix or vortex.
Now add 1 ml of this solution to the second test tube and mix.
Repeat until you have reached your desired concentration.
Ions in Aqueous Solution
• when a substance dissolves it is called a solvation reaction:
• sugar is a non-electrolyte:
C12H22O11 (s) → C12H22O11 (aq)
• soluble ionic compounds undergo dissociation in solution; they break up
into their constituent ions:
NaCl (s) → Na1+(aq) + Cl1-(aq)
(this represents both solvation and dissociation)
• Where
–
–
–
–
–
(s) means solid
(aq) means aqueous (dissolved in water)
(l) means liquid
(g) means gas
(ppt) means precipitate
Ions in Aqueous Solution
• to write a dissociation equation for any ionic
compound you must find out the identity of the
cation and anion, then write the equation, paying
attention to the stoichiometry:
Al2(SO4)3 (s) contains the Al3+ and SO42- ions.
• When it dissociates you get 2 Al3+ and 3 SO42ions:
Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO42-(aq)
Ions in Aqueous Solution
• this means that the concentration of the ions
may be different than the calculated
concentration of the substance in solution:
• If the aluminum sulfate is a 0.200 mol/L solution,
then
Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO42-(aq)
•
•
0.200 mol/L 2(0.200 mol/L) 3(0.200 mol/L)
= 0.400 mol/L = 0.600 mol/L
Ions in Aqueous Solution
• 0.35 mol/L NaOH
•
NaOH(s) → Na1+(aq) + OH1-(aq)
•
0.35 M
1(0.35 M) 1(0.35 M)
•
= 0.35 M
= 0.35 M
Ions in Aqueous Solution
• 1.12 mol/L (NH4)2CO3
• (NH4)2CO3 (s) → 2 NH41+(aq) + CO32-(aq)
• 1.12 M
2(1.12 M)
1(1.12 M)
•
= 2.24 M
=1.12 M
Ions in Aqueous Solution
• 0.056 mol/L V3(PO4)5
• V3(PO4)5 (s) → 3 V5+(aq) + 5 PO43-(aq)
• 0.056 M
3(0.056 M) 5(0.056 M)
•
= 0.17 M
= 0.28 M
Chemical Reactions in Solution
•
there are 4 indications that a chemical
reaction has occurred:
1.
2.
3.
4.
gas production
energy change
colour change
precipitate formed
Chemical Reactions in Solution
• a precipitate is a solid formed from the
reaction of two soluble ions in solution. For
example:
• Pb(NO3)2 (aq) + 2 KI(aq) → PbI2 (s) + 2 KNO3 (aq)
• this is the chemical reaction equation for
the combination of lead (II) nitrate and
potassium iodide.
Chemical Reactions in Solution
• If we recognize the fact that both ionic reactants
have dissociated and are in fact ions in solution
we would get the overall ionic equation:
• Pb2+(aq) + 2 NO31-(aq) + 2 K1+(aq) + 2 I1-(aq)
→
PbI2(s) + 2 K1+(aq) + 2 NO31-(aq)
• the lead (II) iodide does not break up into ions
because it is a solid at the bottom of the beaker.
Chemical Reactions in Solution
• participate in the chemical reaction and
are called spectator ions.
• If we eliminate the spectators we get the
net ionic equation:
•
Pb2+(aq) + 2 I1-(aq) → PbI2 (s)
• if there is no precipitate formed, no
chemical reaction occurs.
Chemical Reactions in Solution
• FeCl2 (aq) + K2S (aq) → FeS (s) + KCl (aq)
• Balance chemical reaction equation:
• FeCl2 (aq) + K2S (aq) → FeS (s) + 2 KCl (aq)
• Overall ionic equation:
• Fe2+(aq) + 2 Cl1-(aq) + 2 K1+(aq) + S2-(aq)
 FeS(s) + 2 K1+(aq) + 2 Cl1-(aq)
• Net Ionic Equation:
• Fe2+(aq) + S1-(aq) FeS(s)
• AlBr3 (aq) + NaOH (aq) → NaBr (aq) + Al(OH)3 (s)
• Balance chemical reaction equation:
• AlBr3 (aq) + 3 NaOH (aq) → 3 NaBr (aq) + Al(OH)3 (s)
• Overall ionic equation:
• Al3+(aq) + 3 Br1-(aq) + 3 Na1+(aq) + 3 OH1-(aq)
 Al(OH)3 (s) + 3 Na1+(aq) + 3 OH1-(aq)
• Net Ionic Equation:
• Al3+(aq) + 3 OH1-(aq)  Al(OH)3 (s)
• (NH4)3PO4 (aq) + Ca(NO3)2 (aq) → NH4NO3 (aq) + Ca3(PO4)2 (s)
• Balance chemical reaction equation:
• 2 (NH4)3PO4 (aq) + 3 Ca(NO3)2 (aq)
→ 6 NH4NO3 (aq) + Ca3(PO4)2 (s)
• Overall ionic equation:
• 6 NH41+(aq) + 2 PO43-(aq) + 3 Ca2+(aq) + 6 NO31-(aq)
 Ca3(PO4)2 (s)+ 6 NH41+(aq) + 6 NO31-(aq)
• Net Ionic Equation:
• 3 Ca2+(aq) + 2 PO43-(aq)  Ca3(PO4)2 (s)
• Aqueous solutions of silver nitrate and sodium carbonate
react to produce a precipitate of silver carbonate and
aqueous sodium nitrate.
• Write and balance chemical reaction equation:
• 2 AgNO3 (aq) + Na2CO3 (aq) → Ag2CO3 (s) + 2 NaNO3 (aq)
• Overall ionic equation:
• 2 Ag1+(aq) + 2 NO31-(aq) + 2 Na1+(aq) + CO32-(aq)
 Ag2CO3 (s) + 2 Na1+(aq) + 2 NO31-(aq)
• Net Ionic Equation:
• 2 Ag1+(aq) + CO32-(aq)  Ag2CO3 (s)
• Aqueous solutions of barium chloride and potassium
sulphate react to produce a precipitate of barium sulphate
and aqueous potassium chloride
• Write and balance chemical reaction equation:
• BaCl2 (aq) + K2SO4 (aq) → BaSO4 (s) + 2 KCl (aq)
• Overall ionic equation:
• Ba2+(aq) + 2 Cl1-(aq) + 2 K1+(aq) + SO42-(aq)
 BaSO4 (s) + 2 K1+(aq) + 2 Cl1-(aq)
• Net Ionic Equation:
• Ba2+(aq) + SO42-(aq)  BaSO4 (s)
Mixing Solutions
• What happens if you mix two ionic solutions ? Using what we
just learned you can calculate the concentration of the ions
in solution before mixing. Take these two solutions:
•
•
•
•
•
•
•
CaCl2 (aq) →
0.25 mol/L
Ca2+(aq) +
2 Cl1-(aq)
1(0.25 mol/L) 2(0.25 mol/L)
= 0.25 mol/L
= 0.50 mol/L
Mg3(PO4)2 (aq)
0.011 mol/L
→
3 Mg2+(aq)
+
2 PO43-(aq)
3(0.011 mol/L)
2(0.011 mol/L)
= 0.033 mol/L
= 0.022 mol/L
Mixing Solutions
• take 300. mL of the CaCl2 and 200. mL of
the Mg3(PO4)2 and we mix them.
• Assuming that no precipitate is formed,
what will the concentration of each ion in
solution be after mixing ?
Mixing Solutions
• we use the dilution formula:
CsVs = CdVd
• In this case, Cs and Vs is the original
concentration and volume of each ion (0.25
mol/L and 300. mL for the Ca2+, for instance)
and d2 is the total volume of the mixture (300.
mL + 200. mL). We are looking for Cd, the ion
concentration after mixing:
Mixing Solutions
• Cd =
CsVs
Vd
• For the Ca2+ ion the calculation is as follows:
• [Ca2+] = (0.25mol/L)(300.mL)
(300. mL + 200. mL)
•
= 0.15 mol/L
Mixing Solutions
• [Cl1-] =
•
(0.50 mol/L)(300. mL)
(300. mL + 200. mL)
= 0.30 mol/L
• [Mg2+] =
•
(0.033 mol/L)(200. mL)
(300. mL + 200. mL)
= 0.013 mol/L
• [PO43-] = (0.022 mol/L)(200. mL)
(300. mL + 200. mL)
•
= 0.0088 mol/L
Problems
1. 2.0 L of 0.40 mol/L MgSO4 mixed with
2.0 L of 0.080 mol/L KI
2. 200. mL of 0.600 mol/L AlBr3 mixed with
300. mL of 0.400 BaBr2
3. 20.0 mL of 0.50 mol/L FeCl3 mixed with
80.0 mL of 0.15 mol/L NH4Cl
Predicting Solubility
• substances dissolve in water when the attraction
of the solute particles to the water molecules
exceeds the attraction of the undissolved solute
for itself.
• if the solute particles are held together very
strongly, the attraction of the water molecules
may not be enough to cause it to dissolve. Such
solutes will be insoluble, or have a very low
solubility.
• we can’t tell from looking at a molecule whether
it will be soluble or insoluble in water. We can
use empirical evidence to make predictions.
Negative Ion
Positive Ion
Result
Any anion
Group 1 ions
Soluble
Any anion
Ammonium ion (NH41+)
Soluble
Nitrate (NO31-)
Any cation
Soluble
Acetate (CH3COO1-)
Ag1+
any other cation
.
Not soluble
Soluble
.
Chloride (Cl1-), Bromide
(Br1-), Iodide (I1-)
Ag1+, Pb2+, Cu1+, Hg22+ .
any other cation
Not soluble
Soluble
.
Sulfate (SO42-)
Ca2+, Sr2+, Ba2+, Ra2+, Ag1+, Pb2+
any other cation
Not soluble
Soluble
.
Sulfide (S2-)
Group 1 & 2 ions, NH41+
any other cation
Soluble
Not soluble
.
Hydroxide (OH1-),
Phosphate (PO43-),
Carbonate (CO32-)
Sulfite (SO32-)
Ammonium ion and group 1 ions
Soluble
.
any other cation
.
.
Not soluble
Predicting Solubility
• Na2S
– soluble, because Na is an alkali metal (group 1)
• NH4Cl
– soluble, because the NH41+ ion is soluble with any anion
• CuNO3
– soluble, because NO31- is soluble with any cation
• AgCH3COO
– insoluble
• Mg(CH3COO)2
– soluble
• PbCl4
– soluble; in this compound the Pb is 4+
• PbCl2
– insoluble; in this compound the Pb is 2+
Predicting Solubility
• Predict the solubility of the compounds in
question 2 on page 447 of the text.
Predicting the Results of Chemical
Reactions
• given any two ionic solutions you should be able to predict
the products and if a precipitation reaction occurs:
•
AgNO3 (aq) + NaCl(aq) → ???
• first, rewrite the reactants as separate ions in solution:
•
Ag1+(aq) + NO31-(aq) + Na1+(aq) + Cl1-(aq) →
• double replacement reaction; cation of the first compound
combines with the anion of the second, and vice versa:
•
•
Ag1+(aq) + Cl1-(aq) → AgCl
Na1+(aq) + NO31-(aq) → NaNO3
Predicting the Results of Chemical
Reactions
• use the solubility table to predict the solubility of the two products of
this reaction:
•
AgCl
insoluble
•
NaNO3
soluble
• write the chemical equation for this reaction:
•
AgNO3 (aq) + NaCl(aq) → AgCl(s) + NaNO3 (aq)
• write the overall ionic equation:
• Ag1+(aq) + NO31-(aq) + Na1+(aq) + Cl1-(aq) → AgCl(s) + Na1+(aq) + NO31-(aq)
• write the net ionic equation:
•
•
Ag1+(aq) + Cl1-(aq) → AgCl(s)
Predicting the Results of Chemical
Reactions
•
•
•
determine the identity of the possible products.
find out of the products are soluble or insoluble.
if no insoluble compound is formed, simply write No Reaction in the
product place of the chemical equation:
NaCl(aq) + NH4NO3 (aq) → No Reaction
•
if one (or both) of the possible products is insoluble, write a chemical
equation, an overall ionic equation and a net ionic equation.
•
•
•
•
•
NaBr and AgNO3
NiCl2 and KOH
CuSO4 and K2S
CoCl2 and (NH4)3PO4
Cu(NO3)2 and KI
NaBr and AgNO3
• Na1+
Br1Ag1+
• NaNO3
soluble
• AgBr
not soluble
NO31-
• NaBr(aq) + AgNO3 (aq) → NaNO3 (aq)+ AgBr(s)
• Na1+(aq) + Br1-(aq) + Ag1+(aq) + NO31-(aq) →
AgBr(s) + Na1+(aq) + NO31-(aq)
• Ag1+(aq) + Br1-(aq) → AgBr(s)
NiCl2 and KOH
• Ni2+
Cl1• Ni(OH)2
• KCl
K1+
not soluble
soluble
OH1-
• NiCl2 (aq) + 2 KOH(aq) → Ni(OH)2 (s) + 2 KCl (aq)
• Ni2+ (aq) + 2 Cl1-(aq) + 2 K1+ (aq) + 2 OH1-(aq) →
Ni(OH)2 (s) + 2 K1+ (aq) + 2 Cl1-(aq)
• Ni2+ (aq) + 2 OH1-(aq) → Ni(OH)2 (s)
CuSO4 and K2S
• Cu2+
SO42K1+
• CuS
not soluble
• K2SO4
soluble
S2-
• CuSO4(aq) + K2S (aq) → CuS(s) + K2SO4(aq)
• Cu2+ (aq) + SO42-(aq) + 2 K1+ (aq) + S2-(aq) →
CuS(s) + 2 K1+ (aq) + SO42-(aq)
• Cu2+ (aq) + S2-(aq) → CuS(s)
CoCl2 and (NH4)3PO4
• Co2+
Cl1• Co3(PO4)2
• NH4Cl
NH41+
not soluble
soluble
PO43-
• 3 CoCl2(aq) + 2 (NH4)3PO4(aq) →
Co3(PO4)2 (s) + 6 NH4Cl(aq)
• 3 Co2+ (aq) + 6 Cl1-(aq) + 6 NH41+ (aq) + 2 PO43-(aq) →
Co3(PO4)2(s) + 6 NH41+ (aq) + 6 Cl1-(aq)
• 3 Co2+ (aq) + 2 PO43-(aq) → Co3(PO4)2(s)
Cu(NO3)2 and KI
• Cu2+
• KNO3
• CuI2
NO31K1+
soluble
soluble
I1-
• Cu(NO3)2 (aq) + KI(aq) → No Reaction
Colligative Properties
• Changes in colligative properties
depend only on the number of solute
particles present, not on the identity of
the solute particles.
• Among colligative properties are
Vapor pressure lowering
Boiling point elevation
Melting point depression
Osmotic pressure
Vapor Pressure
Because of solutesolvent intermolecular
attraction, higher
concentrations of
nonvolatile solutes
make it harder for
solvent to escape to
the vapor phase.
Vapor Pressure
Therefore, the vapor
pressure of a solution
is lower than that of
the pure solvent.
Boiling Point Elevation and
Freezing Point Depression
Nonvolatile solutesolvent interactions
also cause solutions
to have higher boiling
points and lower
freezing points than
the pure solvent.
Boiling Point Elevation
The change in boiling
point is proportional to
the molality of the
solution:
Tb = Kb  m
Tb is added to the normal
boiling point of the solvent.
where Kb is the molal
boiling point elevation
constant, a property of
the solvent.
Freezing Point Depression
• The change in freezing
point can be found
similarly:
Tf = Kf  m
• Here Kf is the molal
freezing point
depression constant of
the solvent.
Tf is subtracted from the normal
freezing point of the solvent.
Boiling Point Elevation and
Freezing Point Depression
Note that in both
T
=
K

m
b
b
equations, T does
not depend on what
the solute is, but only
on how many
particles are
dissolved.
Tf = Kf  m
Colligative Properties of
Electrolytes
Since these properties depend on the number of
particles dissolved, solutions of electrolytes (which
dissociate in solution) should show greater changes
than those of nonelectrolytes.
Colligative Properties of
Electrolytes
• For Instance:
Al2(SO4)3 (s) → 2 Al3+(aq) + 3 SO42-(aq)
0.200 mol/L 2(0.200 mol/L) 3(0.200 mol/L)
= 0.400 mol/L
= 0.600 mol/L
• Total ion concentration
= 0.400 mol/L + 0.600 mol/L
= 1.000 mol/L
van’t Hoff Factor
One mole of NaCl in
water does not really
give rise to two moles
of ions.
van’t Hoff Factor
Some Na+ and Cl−
reassociate for a
short time, so the true
concentration of
particles is somewhat
less than two times
the concentration of
NaCl.
Osmosis
• Some substances form semipermeable
membranes, allowing some smaller
particles to pass through, but blocking
other larger particles.
• In biological systems, most
semipermeable membranes allow water
to pass through, but solutes are not free
to do so.
Osmosis
In osmosis, there is net movement of solvent from
the area of higher solvent concentration (lower
solute concentration) to the are of lower solvent
concentration (higher solute concentration).
Osmotic Pressure
• The pressure required to stop osmosis,
known as osmotic pressure, , is
=(
n
)
RT = MRT
V
where M is the molarity of the solution
If the osmotic pressure is the same on both sides
of a membrane (i.e., the concentrations are the
same), the solutions are isotonic.
Osmosis in Blood Cells
• If the solute
concentration outside
the cell is greater than
that inside the cell, the
solution is hypertonic.
• Water will flow out of
the cell, and crenation
results.
Osmosis in Cells
• If the solute
concentration outside
the cell is less than
that inside the cell, the
solution is hypotonic.
• Water will flow into the
cell, and hemolysis
results.
Molar Mass from
Colligative Properties
• We can use the effects of a colligative
property such as freezing point depression
to determine the molar mass of a
compound.
• Since Tf = Kf  M
• Then M = Tf ÷ Kf
• If we know the mass of solvent we can
calculate the number of moles.
• If we know the mass of the solute we can
calculate the molar mass.
Colloids:
Suspensions of particles larger than
individual ions or molecules, but too small to
be settled out by gravity.
Tyndall Effect
• Colloidal suspensions
can scatter rays of light.
• This phenomenon is
known as the Tyndall
effect.
Colloids in Biological Systems
Some molecules have
a polar, hydrophilic
(water-loving) end and
a nonpolar,
hydrophobic (waterhating) end.
Colloids in Biological Systems
Sodium stearate
is one example of
such a molecule.
Colloids in Biological Systems
These molecules can
aid in the
emulsification of fats
and oils in aqueous
solutions.