Liquids, Solids and Intermolecular Forces
Any substance can exist as a solid, liquid or gas under the proper conditions.
The phase in which matter exists at standard conditions depends on intermolecular
forces of attraction or the attractive forces between individual particles of the substance.
Don’t be fooled into thinking that boiling points are “high” or freezing points are “low”
based on our familiarity with water, which is really one of a few substances that undergo
both those phase changes under “normal” conditions. Elemental nitrogen boils
vigorously at –196°C, and elemental iron freezes at nearly 1500°C.
It is assumed that you already know how solids, liquids and gases compare with regard to
definite shape and volume. Together, liquids and gases are known as fluids.
There are several important intermolecular forces. Some are stronger, others weaker, and
with a little knowledge and common sense, one can usually determine what forces are at
work between particles by making some simple observations about what phase a
substance is in at a particular set of conditions and what its freezing and boiling
temperatures are.
Categories of intermolecular forces:
1. Ion-ion attractions: Electrostatic attractions are very strong and virtually all ionic
compounds are solids at STP. Such compounds are collectively called salts.
2. Ion-dipole: Some molecules (of which water is the classic and most important
example) contain a permanent dipole, as discussed in Chapter 9. Such molecules
are strongly attracted to ionic charges, which is why water dissolves so many
ionic compounds well.
3. Dipole-dipole: For molecules with a permanent dipole, the center of positive
charge of one molecule is attracted to the center of negative charge of
another. Such attractions are not particularly strong unless the molecules are
fairly close together.
4. Hydrogen bonding: In molecules where H is bonded to O, N or F, there is a high
degree of polarization, resulting in particularly highly charged positive and
negative poles. In such molecules, the attraction between molecules is stronger
than simple dipole-dipole attractions. Hydrogen bonding explains many unusual
phenomena that will be discussed shortly.
5. London dispersion forces: Sometimes called van der Waal’s forces (although
technically, van der Waal’s forces also include dipole-dipole and hydrogen
bonding), these are the weakest of the intermolecular forces and exist between all
molecules, even non-polar ones. They are the result of induced dipolar
charges. Draw a diagram below to illustrate this.
As the molecular masses of the particles increase, London dispersion forces become
stronger. The more electrons there are, the greater the induced dipole. So we see that for
example, CH4 (molar mass 16g) is a gas at STP, but CCl4 (molar mass 152g) is a liquid.
The more attraction there is between particles, the higher the melting and boiling points
will be.
The different bonds range in strength as follows:
Ion-ion > ion-dipole > hydrogen bonding > dipole-dipole > London dispersion
Together, dipole-dipole, hydrogen bonding and London dispersion forces are known as
van der Waal’s forces.
Hydrogen Bonding
During hydrogen bonding, the more electronegative element (O, N, or F) pulls the shared
electrons strongly to itself, leaving a high partial positive charge on the hydrogen and a
high partial negative charge on the other atom.
Even in large molecules where there are –OH or –NH2 groups found, there will be
polarity, even if the main molecule itself is mainly non-polar.
Example: cyclohexane vs. glucose
This unusually high degree of dipole-dipole attraction explains some interesting things
about water in particular.
1. Capillary action:
2. Surface tension:
3. Unusually high boiling and freezing temperatures:
Because of its polarity and bond angle (104.5º), water also has the unusual characteristic
of expanding when it freezes, leaving the solid less dense than the liquid phase. This is
extremely important to life on earth for several reasons.
Other compounds that also demonstrate H-bonding between molecules dissolve VERY
well in water. Examples include NH3 and HF or HCl.
Vapor Pressure and Changes of State
Liquids can evaporate or change to gas at temperatures below the boiling point. The
greater the intermolecular forces, the lower the vapor pressure will be. Vapor pressure
(VP) also depends on temperature. As temperature increases, more molecules have the
kinetic energy to overcome the intermolecular forces and escape to the vapor phase. The
relationship between vapor pressure and temperature can be represented graphically and
mathematically by this simple formula:
Ln(Pvap) = -Hvap/R (1/T) + C where y = ln(Pvap), x = 1/T (in Kelvins), m = slope = Hvap/R. and b = y intercept = C. C is a constant particular to the liquid in question.
Probably a more useful version of this equation involves 2 pressures and 2 temperatures.
It allows a scientist to calculate the new vapor pressure at a new temperature, or the
temperature needed to achieve a certain vapor pressure (such as calculating a boiling
point at a different pressure or elevation.) The equation looks like this:
Ln PvapT1 = Hvap 1 - 1
PvapT2
T2 T1
Phase Changes
You should know the names for all possible phase changes between solid, liquid and gas
phases.
Heat of Fusion: Hfus is the heat required to melt one mole of a solid. Melting is always
endothermic, leaving Hfus positive. Conversely, freezing is exothermic and Hfus is
negative.
Heat of Vaporization: Hvap is the heat required to vaporize one mole of liquid to
gas. Vaporization is endothermic, and condensing is exothermic. This is one of the
reasons that steam burns can be so severe.
Specific Heat Capacity: c is the heat required to change the temperature of 1 gram of a
substance by 1ºC. It is different for different substances, and is even different for
different phases of the same substance. For example, c for water is 1.0 cal/gºC but for
steam and ice, c is only about 0.5cal/gºC.
To calculate heat energy change within a given phase for a substance H = mcT.
A Heating Curve (or Cooling Curve) has this basic profile. Fill in the missing
information.
Triple Point Diagrams tell us a lot about a substance’s phase change conditions.
Triple point diagrams look like this. Fill in the missing information.
Two important features of the diagram are the triple point and the critical point.
Critical Temperature is the highest temperature at which a substance can exist as a
liquid. Beyond that temperature, the gas can no longer be compressed to liquid phase
regardless of pressure.
Critical Pressure is the pressure required to liquefy the gas at the critical temperature.
Compare triple point diagrams for water and carbon dioxide.
Vapor Pressure: All substances possess a vapor pressure, that is, pressure caused by
evaporating (or subliming) molecules. Vapor pressure depends on intermolecular
attractions (viscosity) and temperature. Non-polar liquids like ether or gasoline have
higher vapor pressures than polar liquids like water. The vapor pressure of solids is
lower still.
Tungsten metal is said to have a vapor pressure of
Vapor pressure is directly proportional to temperature.

When a liquid is heated until its vapor pressure equals the pressure of the
atmosphere above the liquid, the boiling point is reached.
The molecules have so much kinetic energy and are moving so fast that they cannot
remain in the liquid phase, even if they are not at the liquid surface. The result is the
formation of vapor bubbles in the liquid that rise to the surface and burst.


A liquid’s normal boiling point is the boiling temperature at 1 atmosphere of
pressure.
Change the pressure, change the boiling point.
Solids
When molecules slow down enough so that the intermolecular forces of attraction are
strong enough to keep the molecules from moving past each other, the substance freezes
to solid form.
Non-polar substances (like the components of air) have very low freezing (and boiling)
points.
Polar and ionic substances have higher freezing/melting points.
If allowed to cool slowly enough, most substances will freeze into some sort of
crystals. Crystals are regular geometric arrangements of particles that have flat cleavable
surfaces and straight edges. There are many different crystal shapes.
If substances are cooled too rapidly for crystals to form, the material becomes an
“amorphous” solid. Amorphous literally means “without form,” but here is simply means
non-crystalline.
Glass is an example of an amorphous solid.
Some solids are classified as metallic or covalent network solids. Here, there are no
simple intermolecular forces holding particles together.
1. Metallic bonding: This type of bonding occurs in metals and metal alloys
(solutions). Think of this as a group of positive nuclei all sharing a collective
group of electrons. Scientists call this the “sea of electrons” or “electron gas”
bonding model. Since electrons do not adhere to a particular nucleus, but are free
to move around from one nucleus to another, most metals have the properties of
electrical conductivity and malleability. Metallic bonding is typically quite
strong, making metals hard solids at room temperature.
A second theory of metallic bonding is the MO (molecular orbital) theory,
which suggests that the s or s and p orbitals of neighboring metallic atoms form
overlapping molecular orbitals, allowing the collective sharing of outer electrons
by groups of metallic atoms. It may be thought of as being similar to the sharing
of delocalized electrons in molecules through the formation of resonance orbitals.
2. Covalent network solids: Within molecules there are covalent bonds holding the
atoms of the molecule together. Covalent bonds are extremely strong; much
stronger than intermolecular forces. In covalent network solids, atoms are bonded
covalently in a 3-dimensional array, making for a very strong crystalline
solid. The hardest substances known to man are covalent network solids. They
include diamond, silica (silicon dioxide or quartz) and silicon carbide. A diamond
may be thought of as one huge molecule of carbon atoms.
Solutions and Solution Properties
11.1 Solution Composition
Concentrations:
So far our focus in terms of solution concentration has been on molarity, which is the most
commonly used by chemists.
Concentration
There are however, several other useful ways for measuring concentration. Here are some of
them:
1. % by mass
2. % by volume
3. Molarity (M)
4. molality (m)
5. normality (N)
6. mole fraction (X)
7. ppm
8. ppb
Be sure you know the definitions for each and when they are appropriately used.
The terms “dilute” and “concentrated” are relative to the solute in question.
Types of Solutions
A solution can be formed in many ways other than solids in liquids. Put in an example of each
of the solutions in the table below:
Solute phase
Solvent phase
Example
Gas
Gas
Gas
Liquid
Gas
Solid
Liquid
Gas
None
Liquid
Liquid
Liquid
Solid
Solid
Gas
None
Solid
Liquid
Solid
Solid
11.2 The Energies of Solution Formation
Attractive Forces and Solubility
The surrounding of a solute particle by solvent molecules is called _________________, or in
the case of water as the solvent, _____________________.
There are three sets of attractive forces involved in the dissolving process:
1. Attractions between solvent molecules. The solvent must expand to make room for the
solute particles.
2. Attractions between solute particles. The solute expands.
3. Attractions between solvent molecules and solute particles form the solution.
The 3 energy changes may be represented as H1, H2, and H3. Heat of solution is then
calculated:
Hsln = H1 + H2 + H3.
If the attractions between solute particles (or between solvent molecules) are greater than the
attractions of the solvent molecules to the solute particles, the solute will not dissolve.
If however, the solvent-solute attractions are strong enough, the solute will be pulled apart by the
solvent molecules particle by particle and the solute will dissolve.

Because truly dissolved solute particles are on the order of 1 nm or less in diameter,
they are too small to reflect or refract light. Therefore, except for metal alloys, true
solutions are transparent. (Transparent and colorless do NOT mean the same
thing.)
Eventually, when all the solvent molecules are involved surrounding solute particles, so that
overall no more solute can dissolve, the solution is __________________.
Actually, although it appears that dissolving has stopped, the system has really reached a
dynamic equilibrium in which the rate of crystallization has caught up to the rate of dissolving.
Solute + Solvent
 is dissolving,  is crystallizing (precipitating)
Solution
We can easily see also that when 2 solutions are poured together in which oppositely charged
ions (a cation from one solution and an anion from the other) have very strong attraction for each
other (greater than their attractions to the solvent molecules) a ___________________ will form.
It is possible for a solution to have more dissolved solute than theoretically possible. This
special circumstance is referred to as a _____________________ solution.
Heat of Solution
When dissolving occurs an energy change will accompany the process, as witnessed by a change
in solution temperature. Sometimes the change is slight, and at others it is quite dramatic.
Hsln is measured in kJ/mol of solute dissolved.
If there is a -Hsln value, it means dissolving is ______________ and solution temperature will
____________.
If the value is +Hsln , then dissolving is ______________ and solution temperature will
_______________.
Please think about this carefully, as it often seems backwards in terms of the temperature. It may
help to think about the fact that it is the solvent that is giving up or absorbing the energy
involved in the dissolving process.



An example of a solute with a –Hsln is _______________________.
An example of a solute with a +Hsln is _______________________.
A salt in which there is very little change in temperature with dissolving is ___________.
When might knowledge of a substance’s heat of solution be useful?
The solubility of a solute at various temps is called a solubility curve. A solute with a curve of
increasing slope indicates a ____Hsln.
A decreasing slope would indicate a ____Hsln.
A standard, simple equation for calculating a H value based on mass of substance and
temperature change is H = mcT where “m” is the mass of the substance that is changing
temperature (T) and “c” is the specific heat capacity of that substance. You may remember this
from our calorimetric problems from an earlier unit.
Please remember that there are many solvents besides water. But the same principles
apply whether for calculating concentrations, heats of solution or colligative properties.
11.3 Factors Affecting Solubility
Factors Affecting Rate of Dissolving
Please be very careful not to confuse these ideas.
The RATE at which a solid solute dissolves can be increased by
1. _________________ the solute.
2. _________________ the mixture.
Solubility Factors
Besides the actual nature of the solvent and solute, the only things that affect SOLUBILITY
(how MUCH solute will dissolve are
1. Temperature and
2. Pressure (mostly for gases in liquids.)
Like Dissolves Like
The more alike solvent and solute molecules (particles) are in terms of their polarity, the easier
dissolving will be.
Water is our most important polar solvent. Water molecules are polar and capable of
hydrogen bonding. It dissolves polar solutes and many ionic solutes well. Most any solute with
–OH or –NH2 groups can hydrogen bond to water and will likely dissolve very well.
Except for small alcohols, many organic solvents are non-polar and will not dissolve things that
water dissolves easily.
Examples of organic solvents: Polar: Methanol, ethanol, ethylene glycol, glycerol.
Non-polar: ethers, acetone, benzene, methylchloride, naphthalene,
gasoline
Another nice example can be found in vitamins. Some vitamins are water soluble (B and C) and
others are fat soluble (A, D, E and K). The body can store fat soluble vitamins, and taking too
much can actually be harmful. The body does not store water soluble vitamins and they must be
consumed daily.
A look at the molecules makes it easy to see why their solubilities differ.
Vitamin A
Vitamin C
It’s pretty easy to guess the polarity of substances (that are not ionic) by seeing whether they will
dissolve in water or not. For example, the ink in a ballpoint pen is non-polar while table sugar is
a polar molecule.
Another way to make a pretty intelligent guess about a substance’s polarity is to see what
physical state it is in under normal conditions, or how easily it is converted to a gas or liquid.
The reason all of the main gases of air (except water vapor) are gases is that they are
_______________.
If a substance has a relatively low freezing and boiling point, it is _________________.
Pressure Effects
The solubility of gases in liquids is affected by pressure.
Henry’s Law: Cg = kPg
C is the gas solubility, P is the pressure of the gas and k is Henry’s law proportionality constant.
Simply put, gases dissolved best at high pressure. Gas solubility is directly proportional to the
pressure of the gas above the liquid.
That’s why soft drinks and other carbonated beverages are bottled at 4 to 5 times normal
atmospheric pressure. Carbon dioxide is non-polar and doesn’t dissolve well in water. High
pressure forces the gas to become more soluble.
Temperature Effects
Gas solubility also depends heavily on temperature. Gases become ____________soluble as
temperature _________________.
“Thermal pollution” in environmental waters is a serious problem, because as waters get warmer,
CO2 that is needed by _________________ and O2, needed by ________________ both become
less soluble.
Even in the very coldest water, the maximum dissolved oxygen concentration is about
__________.
Why do all marine mammals breathe air (rather than having gills)?
Most substances have a + heat of solution, becoming more soluble as solution temperature
increases.
11.4 Vapor Pressure
Colligative Properties
These are the physical properties of solutions. They include
1. Vapor pressure (VP)
2. Boiling Point (BP)
3. Freezing Point (FP)
4. Osmotic Pressure (OP)
Colligative properties change with the concentration of dissolved particles in liquid solvents.
Vapor pressure is the gaseous pressure caused by evaporating solvent molecules. For a pure
solvent, VP depends on 2 things: the type of solvent and temperature. As temp. increases, VP
_________________.
When a nonvolatile solute is dissolved in a solvent, fewer molecules of solvent are at the surface
of the liquid where they have the chance to escape to the vapor phase. The result is a lower VP
for the solution.
Raoult’s Law calculates the new vapor pressure of a solution based on solute concentration.
PA = XAPA where PA is the VP of the solution, XA is the mole fraction of the solvent in the
solution and PA is the VP of the pure solvent at a particular temperature. Your text has example
problems of Raoult’s Law.
Raoult’s Law predicts ideal solution behavior. Most real solutions do not follow predictions
exactly, but some have higher pressures that expected while others have lower depending on the
intermolecular forces existing between solvent and solute particles.
11.5 Boiling Point Elevation and Freezing Point Depression
Boiling Point Elevation
Boiling point is defined as the temperature at which the vapor pressure of a liquid equals the
pressure of the gas above the liquid. (Remember how boiling points are affected by pressure:
triple point diagrams.)
If the presence of a solute reduces the solvent’s VP, the liquid must be hotter to boil. It has
farther to go to get the VP up to the pressure of the gas above the liquid resulting in boiling
point elevation.
Calculating changes in BP is easy: Tb = Kbmi
 Tb is the change or increase in the BP (normally 100C for water).
 Kb is the molal boiling point constant for the solvent. For water, the value is 0.51C/m,
but it is different for other solvents. The AP loves to give this kind of problem using a
solvent other than water.
 m is the molality of the solution (moles solute/1000g of solvent).
 i is a value called the van’t Hoff factor that will be discussed shortly.
Freezing Point Depression
Freezing occurs when the solvent molecules slow down enough that intermolecular forces of
attractions begin to lock them into solid form.
The freezing of solutions is a difficult thing to define. Solutions, especially concentrated ones
don’t seem to freeze cleanly or solidly. Freezing point changes are calculated the same way as
BP changes.
Tf = Kfmi Kf is the molal freezing point constant. (The value for water is 1.86C/m and the
normal FP for water is 0C.
The presence of solute particles “interferes” with solvent molecules getting next to and locking
onto each other in the freezing process. As the solvent particles lock together, crystals of pure
solvent begin to freeze out of the solution. As the liquid solvent is removed to the solid form, the
remaining solution becomes more and more concentrated, its freezing point continuing to
decline. Solutions therefore often turn to “slush” and may not ever freeze completely solid.
Ever suck on a Popsicle?
11.7 Osmotic Pressure: Don’t worry too much about osmotic pressure. That is more of
biological interest and will probably not be on the AP exam. Osmotic pressure is measured in
atmospheres and is calculated
 = MRT where is osmotic pressure in atmospheres, M is molarity, R is 0.0821 L atm/mol K
and T is temperature in K.
But be sure you can solve problems that find molar mass from freezing point depression or
boiling point elevation.
11.7 Colligative Properties of Electrolyte Solutions
The van’t Hoff Factor (i)
When strong electrolytes dissolve, they produce 2 or more ions from each dissolved
molecule or formula unit.
The van’t Hoff factor tells us how many particles there will be and therefore how may
times the expected effect the solute will have on the BP or FP.




NaCl  __________________
i = _____
HBr  ___________________
i = _____
Ca(NO3)2  ______________________
i = _____
Al2(SO4)3  ______________________
i = _____
Technically, the van’t Hoff factor is calculated as
i = T actual/T calculated for the solute as a non-electrolyte
What we determined above was the ideal van’t Hoff factor. The actual changes in
temperature are never ideal because when oppositely charged ions come together, they
temporarily “stick” to each other acting like one particle instead of two. So the factor is
always somewhat less than predicted ideally. For weak electrolytes (like acetic acid), it
will be much less than predicted, because these electrolytes do not ionize completely.
The factor for acetic acid may only be 1.1.
11.8 Colloids
Colloids
When you have a strange substance and you can’t seem to come up with a good answer
when you ask yourself if it’s a solid, liquid or gas, it is probably a colloid.
Colloids include things that seem to defy simple description, like Jello, clouds and Silly
Putty.
A colloid is a mixture, but it is not a solution. Where the dispersed particles of a true are
less than one nanometer in diameter and too small to settle out by gravity or reflect light,
the dispersed particles of a colloid are ______________ nanometers in diameter. Such
particles may include macromolecules (proteins, starch or DNA), cells (red blood cells)
or small liquid drops, gas bubbles or solid particles (like smoke).
Even though not truly dissolved, colloidal particles resist settling because of
1. Brownian movement- the constant bombardment of the particles by surrounding
molecules, and
2. like electrostatic charges, that cause the colloidal particles to repel each other.
These particles are too small to be separated by gravity or normal filtration, but are large
enough to reflect/refract light.
This quality results in the Tyndall Effect. Whereas a beam of light is undetectable as it
passes through a true solution, a beam is visible in a colloidal suspension. A car’s
headlights are invisible in dry, clear air, but are visible in air that is foggy or dusty. The
small droplets of water or particles of solid dust make the air a colloid.
Dispersing Phase
(Solvent like
phase)
Gas
Gas
Gas
Liquid
Liquid
Liquid
Solid
Solid
Solid
Dispersed
Phase
Colloid
Type
Example
Gas
Liquid
Solid
Gas
Liquid
Solid
Gas
Liquid
Solid
None
Aerosol
Dry aerosol
Foam
Emulsion
Sol
Solid foam
Solid emulsion
Solid sol
All are true solutions.
Fog
Smoke
Whipped cream
Milk, mayonnaise
Paint
Marshmallow/Styrofoam
Butter
Ruby glass
Liquid in liquid emulsions are common and important colloids. Two liquids that will
dissolve in each other in any proportion (e.g. water and ethanol) are called miscible.
Liquids that won’t mix and dissolve (e. g. water and oil) are immiscible. Substances
which are added to immiscible liquids that allow them to mix and stay mixed are called
emulsifying agents. Mayonnaise is made of vegetable oil and water, but requires the
proteins in egg albumin (white) to stay mixed. Detergents are emulsifying agents that
allow the grease and oil on clothing and dishes to mix with water be carried away.
Then, of course, the picture becomes even more interesting with substances like milk or
blood, which are both true solutions and colloids at the same time.
Removal of Colloidal Particles
If the electrostatic charges on the colloidal particles can be masked or removed, the
colloid breaks down. Lightening discharges colloidal water droplets. The droplets
coalesce into raindrops. Salt ruins Jello because the ions mask the static charges of the
colloidal protein molecules in the Jello. Salts are also used to remove precipitated
“flocculent” from wastewater. Heat also destroys colloids nicely. Again, the Jello
collapses when heated.
ACIDS & BASES
14.1 The Nature of Acids and Bases
Arrhenius definition:
 Acids produce
 Bases produce
Traditional acids only exhibit what are considered characteristic acidic properties
in aqueous solution.
Ex. HCl, not acidic in pure form or if dissolved in benzene.
But is highly soluble in water and is electrolytic.
Con. HCl is up to 37% HCl by weight in a saturated solution.
That is 450L of HCl(g) at 1 atm dissolved in a liter of water.
The characteristic species for acids is the hydrated proton.
For simplicity, this is usually represented as: ______
Polar water molecules cause the ionization of HCl molecules: HCl  H+ + Clor HCl(aq) + H2O(l)  H3O+(aq) + Cl-(aq)
H3O+ is the hydronium ion: seen frequently in chemical reaction
representations.
This however is probably not very accurate. Realistically, the species is probably
more like H5O2+ or H9O4+
Whatever representation is used, in acidic solutions, acid molecules transfer
protons to water.
Brønsted-Lowery Acid/Base Definition
In this definition
 Acids are

Bases are
*Water need not necessarily be involved (but it usually is.)
OH- is an excellent proton acceptor. **Important note: the hydroxide ion is the
strongest basic speciess that can exist in water solution. Other ions may be
stronger bases, but they form hydroxides immediately upon contact with
water. Ex. N-3.
N-3 + 3 H2O 
H+(aq) +OH- (aq)  H2O(l) is the equation for a traditional acid base
neutralization. It is the common net ionic equation for all such reactions.
Here is another important reaction that you should know well:
NH3(aq) + H2O(l) 
Base
Acid
The reaction is an equilibrium and so is reversible. In this case, NH4+ is the acid
and OH- the base.
Conjugate Acid-Base Pairs
When a Brønsted acid donates a proton, it becomes a potential proton acceptor,
making it a Brønsted base. This is called a
____________________________________.
Similarly, a base accepts a proton becoming its __________________________.
Strong acids have ______________ conjugate bases.
Weak acids have _______________ conjugate bases.
Similar for bases.
Ex: HCl is a very strong acid. Its conjugate base the ______ion is such a weak
proton acceptor that it doesn’t even act like a base. It is most often a spectator
ion in solution.
 See p. 574 for an excellent table of conjugate acid/base pairs and their
relative strengths.
**Important note: In aqueous solution, the H+ ion is the strongest acid
possible.
The Autoionization of Water
Write the equation for the autoionization of water:
___________________________________________________
This process occurs spontaneously but in low proportions.
At room temperature, only one molecule in every 108 transfers a proton. This
equates to about 1 g of protons in an Olympic sized swimming pool full of water.
*Note: the forward reaction is endothermic, and so the degree of ionization goes
up with rising temperature.
A simpler way to write the equation is: H2O(l)  H+(aq) +OH- (aq)
14.2 Acid Strength
The difference between strong and weak acids involves equilibrium position.
 For strong acids, the equilibrium lies far to the right, and K is a very
large value. So, we think of strong acid dissociations as being
complete. We typically represent strong acid equations as: HA(aq)
+ H2O(l)  H3O+(aq) + A-(aq)
 The conjugate bases of strong acids are very weak.
 For weak acids, the equilibrium typically lies to the left, and K is
much smaller. Weak acid equilibria are usually represented:
HA(aq) + H2O(l)  H3O+(aq) + A-(aq)
 The weaker the acid, the stronger its conjugate base.
Strong Acid
Weak Acid
Water is amphoteric, that is, it can be both an acid and a base.
 It undergoes autoionzation.


It can be a hydrogen ion donor (acid). H2O(l) + NH3(aq)  NH4+(aq) + OH(aq)
It can be a hydrogen ion acceptor (base). H2O(l) + HCl(aq)  H3O+(aq) +
Cl-(aq)
14.3 The pH Scale.
We can write Kc for this equilibrium (remember, as a pure liquid, water is left
out since its concentration is essentially constant.)
This is a “special K” (Part of a balanced breakfast, along with 2 eggs, bacon,
buttered toast with jelly, coffee, milk, orange juice, a One-A-Day® vitamin and a
cheese Danish.)
K values that apply to special, common equilibrium equations are given a
subscript that designates them as such. For the ionization of water, we use
_______.
Kw is called the ‘ionization product constant for water.’
 Both H+ and OH- have equal concentrations.
 At 25ºC, [H+] = [OH-] = 1.0 x 10-7M
 Therefore, Kw = _________________________
This is an important value, and you should remember it.
Since hydrogen and hydroxide ions appear in equal concentration, this ‘solution’
is neutral.
**Note: Contrary to popular belief, rumor and opinion, water is hardly ever
neutral, unless it has been distilled and heated to about 80ºC to drive all gases
from the liquid.
In acidic solutions [H+] > [OH-] and in basic solutions [H+] < [OH-].
The concentrations of the 2 ions are inversely proportional.
Write the mathematical formula for calculating pH. (The symbol ‘pH’ comes
from the French pouvoire of hydrogen. Pouvoire means ‘strength’ or ‘power’.)

pH = __________________

Note: pH is a logarithmic scale. So a change in pH of 2 units means
a change in [H+] of 100 fold, not 2 fold.

pH 7 is neutral.

The normal range of the pH scale is from _________________.

It is possible to have pH values outside this range.

Solutions with pH < 7 are _________________.

Solutions with pH > 7 are _________________.
When calculating pH for any problem, one must focus on the major species
present in the solution.
 Write a balanced equation (or equilibrium equation) for the chemistry
occurring in the system, and determine which species are the major ones
contributing to the solution pH.

This skill cannot be stressed strongly
enough.
pOH can be calculated in the same way as pH: pOH =
_________________________.

pH + pOH = _______.

[H+] x [OH-] = 1.0 x 10-14
While it is not difficult to calculate pOH, convention usually calls for pH, even
with a basic solution. If given a basic solution (like 0.4M NaOH) go ahead and
calculate pOH and simply take 14 - pOH to get pH. Simple.
*Note: all pH and pOH problems use base10 logs, and not natural logs. Be
sure you can do these problems comfortably with your calculator. Example
problems are in the chapter.
Measuring pH.
PH can be measured in a couple of different ways.
One way to measure pH or pH changes is with color indicators.
Most indicators are molecular dyes which are weak acids. When the molecule is
in tact (with hydrogen(s) attached, it is one color. When the molecule ionizes,
releasing an H+, it becomes a different color.
Most indicators have ______ possible colors, although some have _____ with 2
transition points.
There is a good table showing several indicators, their colors and transitional
pH’s on p. 715.
____________________________ is a mixture of several dyes that gives a rainbow
of colors with changing pH.
Sometimes the changes are not very sharp making the indicator of questionable
value, however, they are still widely used especially in acid/base titrations.
Another, more accurate way to measure pH is with an electronic pH meter.
These use a special probe sensor to measure electrical potential in a solution.
There is still some debate as to exactly how these things work, and they are
notoriously fickle instruments.
14.4 Calculating pH of Strong Acids
Strong acids ionize completely in water. These are the strong acids you should
know. Memorize their names and formulas.
HCl
_____________________
HClO3____________________
HBr
_____________________
HClO4____________________
HI
_____________________
H2SO4 ____________________
HNO3
_____________________
There are others, such as chromic or tungstic acids that are also very strong, but
they are also very uncommon, so the list above should serve. Consider that all
other acids not listed above are weak.
If you wish to review the system for naming acids, you must be extra nice to the
instructor and perhaps offer some bribe or small sacrifice (peanut M&Ms® work
well.)

Generally, calculating pH for strong acids is pretty easy since they
dissociate completely. The original [HA] = [H+]. Just take –log and you’re
done.
14.5 Calculating pH of Weak Acids
Weak acids form aqueous equilibria due to incomplete ___________________.
HX(aq)  H+(aq) + X-(aq) HX, HA or HB are common
representations for a
generic weak acid.

Ka = acid dissociation constant. Write the equilibrium expression for the rxn
above:
Ka =
The weaker the acid the smaller the Ka value.
If Ka is known [H+] and pH can be calculated.
Notice that ICE tables are used extensively for weak acid and base
problems.
There is an example problem on p. 638-39 in the textbook.
Steps for solving weak acid equilibrium problems: (commonly referred to as
‘x2’ problems)
1. List the major species in solution.
2. Choose the species that can produce H+ and write balanced equations for
these equilibria.
3. Using the equilibrium constants (Ka) decide which reaction will produce
the most H+ ions.
4. Write the Ka expression for this dominant equilibrium.
5. Set up an ICE table and list the initial concentrations of the species
participating in the dominant equilibrium.
6. Define the change in concentrations in terms of x.
7. Write the equilibrium concentrations into the ICE table in terms of x.
8. Write the equilibrium concentrations (in terms of x) into the K expression.
9. Solve for x the ‘easy way’ by ignoring x in the denominator of the
expression. We assume that [HA]0 – x ≈ [HA]0.
10. Use the ‘5% rule’ to see if the assumption in (8) is valid. (If the value of x
is more than 5% of the [HA]0, then you will need to use the quadratic
equation or successive approximations to solve the problem.)
11. Calculate [H+] and pH.
Weak acid solutions show much lower electrical conductivity than strong ones.
They are classified as weak ___________________.
Percent Dissociation
 Sometime pH problems ask you to calculate the % dissociation of
the weak acid. This is easily done once the [H+] is known. Simply
divide [H+]/[HA]0 x 100%.
 These problems can also be worked backwards. Given %
dissociation, we can determine pH and Ka.
14.6 Bases
Strong bases include the column________and _________ metal hydroxides and
oxides.
These are strong mostly because they dissolve in water (Col. IA better than IIA),
whereas the oxides and hydroxides of other metals dissolve poorly and so do not
put enough hydroxide ions in solution to be strong.
Even some of the Column IIA hydroxides do not dissolve very well (e. g.
Mg(OH)2 and Be(OH)2. Calcium hydroxide is the most common Column II base.
It is used in __________.
(Special tip: If you mix Philips® Milk of Magnesia (Mg(OH)2) with orange juice,
you get a ‘Philips screw driver.’) Hold the applause, it will only get worse.
Active metal oxides (ex. Na2O or CaO) create OH- ions when they dissolve:
Write the equation here:
Hydrides and nitrides are also strongly basic anions, forming hydroxide ions on
contact with water.
Ex. NaH(s) + H2O(l) 
Mg3N2(s) + 6H2O(l) 
 Again, calculating pH of strong bases is easy. [Base] = [OH-]. Calculate –
log [OH-] and subtract the answer from 14.0.
16.6 Weak Bases
Weak base + H2O  conjugate acid + OHB(aq) + H2O(l)  BH+(aq)
+ OH-(aq)
The H of water is attracted to an unbonded electron pair (often on a nitrogen
atom).
Ex. NH3(aq) + H2O(l)  NH4+(aq) +OH- (aq)
Kc =
Once again, because water is a pure liquid, it is left out of the Kc expression. This
is another ‘special K.’
Kb =
constant.
Kb is called a base dissociation
Many amines (--NH2), HS-, CO3-2 and ClO- ions cause this effect of ionizing
water to form OH- ions resulting in a basic solution.
..
H
CH3
HO N
H
hydroxylamine
CH3--NH2(aq) + H2O(l)  CH3--NH3+ (aq) + OH-(aq)
..
H
N
H
methylamine
Amines react readily with acids to form salts, much as ammonia does. If ‘A’ =
amine,
A + HCl  AHCl: a ‘hydrochloride’ salt. Even without water, this is a type of
acid/base reaction.
Many drugs are amines that are volatile, unstable and have a short shelf life.
Stability and shelf life improve dramatically if the drug is converted to the
hydrochloride salt. Effectiveness is largely unhampered.
There are several amine-based drugs: examples include quinine, codeine,
caffeine and amphetamines.
Many amines have an unpleasant, fishy smell. They are what make dead fish
and rotting meat smell rotten. One such compound is called ‘cadaverine.’
Others include indigestine, trash-cannine, and locker-roomine (just kiddine.)
CH2--CH--NH2 + HCl --->
|
CH3
Amphetamine
CH2--CH--NH3+Cl
|
CH3
Amphetamine hydrochloride
The salt is more stable than the straight amine.
A second class of weak bases are the anions (conjugate bases) of weak acids.
Since weak acids have fairly strong conjugate bases, these anions act readily as
hydrogen ion acceptors.
C2H3O2-(aq) + H2O(l) 
weaker side
stronger side
When molecules or ions cause the ionization of water molecules, the process is
called
___________________.
The weaker side of such equilibria (the molecular side vs. the ionic side) is
always favored. That is, there will be more molecules in the system than ions.
Due to hydrolysis, salts with the anions of weak acids tend to form basic
solutions with pH values _____________________.
Ex. of such salts include: ______________________________________.
Example problem: Find the pH of a 0.010M solution of NaClO, a common
ingredient in most household bleaches and bleaching products (like Tilex®).
ClO-(aq) + H2O(l)  HClO(aq) + OH-(aq)
I
0.010M
0
0
C
-x
+x
+x
E
0.010-x
x
x
Table in book  Kb for ClO- = 3.3
Kb =
x 10-7
P. A 22 (appendix)
Since Kb is very small, the value of x will also be very small and 0.010M - x 
0.010M.
So, Kb = x2/0.010 = 3.3 x 10-7 and x2 = 3.3 x 10-9
x = [OH-] = 5.7 x 10-5 M
pOH = -log [OH-] = 4.2
pH = 14 - pOH = 14 - 4.2 = 9.8
**Important note: if K < 10-5 you may avoid the quadratic equation by
making the assumption that compared to original concentrations, x is going to
be very small and so the change in the original concentration will be small
enough to be ignored.
The solution of sodium hypochlorite in water is distinctly basic (although not
strongly so.) If you have ever spilled bleach on your fingers, you know that it
feels slippery. This is a characteristic of basic solutions. You may also have
noticed that it does interesting things to colored clothing. Some years this is
considered hip fashion, but usually not.
14.7 Polyprotic Acids
Ka for first, second and even third proton ionization is different.
Ex. H2SO3  H+ + HSO3Ka1 = 1.7 x 10-2
HSO3-  H+ + SO3-2
Ka2 = 6.4 x 10-8
P. A-23 (appendix) has Ka values for polyprotic acids.
With such acids, nearly all the hydrogen ion in solution comes from the first
ionization.
For most polyprotic acids, Ka1 is 103 or more times larger than Ka2.
Because Ka2 is usually so small, pH can be calculated quite accurately from Ka1.
Relationship Between Ka and Kb.
NH4+(aq)
 NH3(aq) + H+(aq)
NH3(aq) + H2O(l)  NH4+(aq) +OH- (aq)
Net Eq:
H2O(l)  H+(aq) +OH- (aq)
Ka = 5.6 x 10-10
Kb = 1.8 x 10-5
Kw = 1.0 x 10-14
Auto ionization of water.
*When 2 reactions are added together to give a third, the equilibrium constants
of the third equation is the product of the equilibrium constants of the first 2
equations.
If Reaction 1 + Reaction 2 = Reaction 3, then K1 x K2 = K3.

From the above equations, K1 x K2 =
________=______
=

So, **Ka x Kb always = _____ = __________
If either Ka or Kb is known the ionization constant for the conjugate acid or base
can be calculated easily.
14.8 Acid-Base Properties of Salt Solutions
Many ions that are the conjugate partners of weak acids or bases cause the
hydrolysis of water.
Salts can be thought of as being formed by the neutralization of a parent
_____________
and ____________.

If both parents are strong, no hydrolysis will occur: Solution pH _______.
Examples include _____________________.

Salts of strong acid and weak base parents: pH ____________.
Examples: NH4Cl and AlCl3
Hydrolysis equations:

Salts of weak acid and strong base parents: pH ____________.
Examples: NaC2H3O2, KF, Ca(CN)2
Hydrolysis equations:

Weak acid and weak base parents: pH varies depending on which ion causes
the
greatest degree of hydrolysis.
Examples: NH4CN, FeCO3
To find pH we must compare Ka and Kb.
NH4+; Ka = 5.6 x 10-10
CN-; Kb = 2.0 x 10-5

The cyanide ion causes more hydrolysis than ammonium, so the solution
will be basic.
14.9 Acid-Base Character and Chemical Structure
A couple of simple rules will help you predict whether a substance will be acidic,
basic or neutral in a solvent.
 Ionization of H depends largely on _____________ and _______________.
 Ex. NH4+ and CH4 are isoelectric, but N has the greater nuclear charge. The
bonds
are more polar and so the ammonium has a greater tendency to release
H. Methane
shows no acidic properties.
 NH4+  H+ + NH3
Ka = 5.6 x 10-10



CH4  CH3- + H+
No reaction
The more polar the bond character, the greater the value of Ka.
Or, the greater the electronegativity of the central atom, the greater
the acidity.
This explains why HCl is such a strong acid. However, even though the
electronegativity of HF is also very high, the short, strong bond present because
of F's small radius, prevents the acid from being strong. HF does not ionize 100%
and so is a weak acid.
We can make general predictions about any hydride. Metallic hydrides tend to
be ______. Non-metallic hydrides tend to be _______. The closer the element is
to the end of its row, the stronger acid or base it is likely to be.
 Acidic character increases left to right across a row and as we go down a
column.
Hydroxides and Oxyacids
These compounds usually contain oxygen, hydrogen and one other element. If
this other element is a metal that gives up electrons more easily than hydrogen,
the hydroxide ion forms and the compound is a base.
If the ‘other atom’ is a nonmetal with higher electronegativity, the release of
hydroxide becomes less likely. If the release of H is more likely, the compound is
an acid.
Note that Oxygen gets an extra electron in either case. It just depends on
whether the O lets go of the H or not.
For oxyacids with the same O and H structure but __________ central atoms, acid
strength goes _______ as electronegativity goes up.
For oxyacid series with the ________ central atom, acid strength goes up as the
number of _____________ goes up.
Examples include the sulfate and chlorate acid series.
14.10 Acid-Base Properties of Oxides.
Most active metal and nonmetal oxides act as acids or bases. They are often
referred to as anhydrous acids or anhydrous bases. Anhydrous means “no
water.”
The rules are simple.
 In order to be a good acid or base, the oxide must be water soluble, so
some oxides are not very acidic or basic. For example, Al2O3 has very
low solubility and so is a poor base.
 Metal oxides are bases. They react with water to form aqueous
hydroxides.
o Ex. Na2O(s) + H2O(l)  2 Na+(aq) + 2 OH-(aq)
 Nonmetal oxides are acids. They combine with water to form acid
molecules.
o Ex. SO3(g) + H2O  H2SO4(aq) sulfuric acid.
o NOx and SOx gases are the primary culprits in acid precipitation
that does so much environmental damage. CO2 also makes water
including rain naturally acidic.
14.11 Lewis Theory of Acids and Bases
Lewis acids and bases are defined in terms of electron pairs rather than protons.
To be a proton acceptor, a species must have an unshared electron pair.
Therefore, a Lewis base is referred to as an ________________________________________.
A Lewis acid is then described as an ____________________________.
The Lewis concept of acids and bases is broader than the Brønsted concept and
includes many reactions that do not even involve H transfer.
Draw here the classic Lewis A/B example of NH3 and BF3 reacting.
Where aquatic systems and/or H+ are involved, the B-L definition is most useful.
The Lewis system is not often used unless called for specifically in context.
BF3 is not referred to as an acid normally, but it may be called an acid in the
Lewis sense.
‘Ragsdale definition’- any positive ion is an acid, and any negative ion is a base.
Fe+3 + 6:CN:-  [Fe(CN)6-3] ferricyanide ion
L. acid
L. base
This reaction may also be classified as a complexation.
There are many complexing agents that will be attracted to these Lewis acid
metal ions. Many are anions, but all must have at least one unbonded electron
pair.
The number of complexing agents or ligands that surround the metal ion is
called the coordination number.
Acid strength of metal ions depends on charge/size ratio. The larger the ratio,
the stronger the hydrolytic effect.
 A large +1 ion has essentially no hydrolytic effect: Ex.
 A small +3 ion has a large hydrolytic effect: Ex.
The metal ion polarizes the water molecule, which then releases a H+ ion while
the OH- ion remains attracted to the metal ion.
General form: M(H2O)n+z  M(H2O)n-1(OH)+(z-1) + H+(aq)
M = metal; n = #water molecules hydrating the ion; z = ionic charge
For most +3 ions, n = 6; for most +2 or +1 ions, n is probably 4.
Record the formation of aluminum hydroxide from aluminum hexahydrate ion:
Kh = 1 x 10-5 (pH = 5) Kh is an acid hydrolysis constant.
Aqueous Equilibria: Further Considerations
15.1 Solutions of Acids or Bases Containing a Common Ion
It’s possible to drive an equilibrium system in a desired direction by adding a
common ion to the system.
It’s also possible to effectively add only one of the ions of the equilibrium by
making sure that the other ion of the pair is a ___________________.
Ex. HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)
By adding HCl(aq), we add H+(aq) and Cl-(aq) ions, but chloride is not
part of this system and so has no effect.
NaC2H3O2 will do the same thing. What is another substance that might
be added to accomplish the common ion effect?_________________
Adding any of these substances causes the equilibrium to shift _________.
Adding a common ion does not affect K.
Acid Solutions Containing Common Ions
A 1.0 M solution of HF has a [H+] = 0.027 M and a % dissociation of 2.7%.
Calculate the [H+] and % dissociation of a solution that is 1.0 M HF and 1.0 M
NaF.
 Important species in the solution: HF, F-, Na+ and H2O.
 We may ignore the water since it’s a much weaker acid than HF.
 We may also ignore the Na+ since it is not a hydrolytic ion (strong base
parent).
 So the really important species are the HF and the F-.
HF  H+ + FI 1.0M
0
1.0 M
C -x
+x
+x
E 1.0-x
x
1.0+x
Ka = [H+][ F-]
x(1.0+x)
[HF]
=
1.0-x
= 7.2 x 10-4
Ignoring x in 1.0+x and 1.0-x and solving for x: x = 7.2 x 10-4 and % ionization =
0.072%
We see that the ionization is much reduced in the acid/salt solution mixture.
This makes sense with respect to Le Chatelier’s principle: if the [F-] on the
product side is raised, the equilibrium shifts left, and the [H+] goes down.
**Note: % ionization of weak acids increases as [HA] decreases, even though
pH goes up.
The AP folks really like these kinds of problems, which involve 2 sets of
calculations. For problems like this one that involves the partial neutralization
of a weak acid by a strong base or of a weak base by a strong acid, do the
problems in this order:
Do the stoichiometry of the acid-base reaction, THEN
Do the equilibrium calculations on the remaining solution.
Assume that acid-base reactions involving H+ and OH- are complete.
15.2 Buffered Solutions
Composition and Action of Buffered Solutions
Buffers are solutions designed to maintain a (relatively) constant pH with
the addition of _______________________________.
That constant pH is not necessarily pH 7.
Important examples of buffer systems include
__________________________.
Buffers neutralize or “absorb” both H+ and OH-.
This is accomplished by creating weak acid/base or base/acid systems.
Ex. HC2H3O2/C2H3O2- or NH4+/NH3
Such a system is easily established by using a weak acid and a salt with
the same anion.
Ex.__________________________
Or using a weak base and a salt with the same cation.
Ex_____________________
General acid dissociation equilibrium: HX  H+ + XKa = [H+] [X-]/[HX]
[H+] = Ka ([HX]/[X-])
[H+] and therefore pH depend on
the acid Ka value
the ratio of HX/XAdding Acid or Base to a Buffer
Adding base: OH- + HX  H2O + X-: with respect to the equilibrium; HX
 H+ + XOH- + H+  H2O; as [H+] drops due to neutralization,
the equilibrium shifts _______ to replace the disappearing hydrogen ions.
There are plenty of HX molecules to accomplish this. The pH remains
nearly unchanged.
Adding acid: H+ + X-  HX: again with respect to the equilibrium; HX 
H+ + X-; as the [H+] rises, the equilibrium shifts _________ to use up the
extra hydrogen ions. The excess X- from the salt allows this shift to take
place easily. Again, the pH remains mostly constant. This is not the case
with a solution of just the weak acid.
In both cases, the [HX]/[X-] ratio changes, but if the change is small, the
change in pH is also small.
** Buffers are most effective if [HX]/[X-] is about 1. If [HX] = [X-], then [H+] =
Ka.
** It is best to choose a buffer whose acid form (HX) has a pH close to the pKa of
the
desired pH.
pKa = _____________ If [HX]/[X-] = 1, then pH = pKa.
Calculating Buffer pH
Use the same basic procedure as a common ion effect problem.
Ka = [H+] [X-]/[HX] so
[H+] = Ka [HX]/[X-]
-log[H+] = -log(Ka [HX]/[X-])
= -logKa - log[HX]/[X-]
-log = p therefore pH = pKa - log[HX]/[X-] or more commonly
pH = pKa + log[X-]/[HX] where X- is the base and HX is the acid.
The general form of this equation is known as the Henderson-Hasselbach
equation.
Write the general form here:_________________________
There is also a corresponding base form of the
equation:_________________________
Ex. Find the pH of a 0.12M lactic acid (HC3H5O3) solution with 0.10M
sodium lactate.
Method 1:
HC3H5O3  H+ + C3H5O3I
0.12M
0
0.10
C
-x
+x
+x
E
0.12 - x
x
0.10 + x
Ka = 1.4 x 10-4 = [H+] [C3H5O3 ]/[HC3H5O3] = x(0.10+x)/(0.12-x)
x will be small compared to 0.12 and 0.10 and so may be ignored.
Therefore, 1.4 x 10-4 = 0.10x/0.12 and x = 1.7 x 10-4
and pH = 3.77
OR
Method 2 using the Henderson-Hasselbach equation:
pH = pKa + log [base]/[acid]
= -log(1.4 x 10-4) + log (0.10/0.12)
=
3.85
+
(-0.08) = 3.77
Which do you think is easier??
**Special note: The Henderson-Hasselbalch equation always has a conjugate
weak acid/base pair as the A and B in the equation. Never will H+ or OH- be
the acid or base.
Addition of Acids or Bases to Buffers and pH.
**Any reaction with a strong acid or strong base goes ____________________.
These include:
Strong acid with strong base.
Strong acid with weak base.
Strong base with weak acid.
Weak acid/weak base reactions are more difficult.
As mentioned before, deal with the stoichiometry of the neutralization first,
then do the equilibrium pH problem with the remaining ions in the solution.
0.300 mol HC2H3O2 + 0.300 mol NaC2H3O2 has a solution pH = 4.74.
Find the pH after the addition of 0.020 mol NaOH.
Find the pH after the addition of 0.020 mol HCl.
Solve the 2 problems here:
15.3 Buffer Capacity and pH
Along with pH, the other important characteristic of a buffer is its buffering
capacity.
How much acid or base can the solution absorb without a significant change in
pH?
Buffer capacity () depends on the amount of acid and conjugate base from
which the buffer is made.
Ex. Consider 2 systems:
1M HC2H3O2 + 1M NaC2H3O2
0.1M HC2H3O2 + 0.1M NaC2H3O2
pH is the same for both: pH = pKa since the [HX]/[X-] = 1
However, the 1M system can buffer far more acid or base than the 0.1M
system.
To help illustrate how dramatic the buffering effect can be, consider the
following examples:
Adding 2.0 mL of 10M HCl (0.020 mol HCl) to
a plain acid solution of pH = 4.74 (1.8 x 10-5 M HCl) causes the pH
to drop to 1.70: a change of 3.04 units. This solution has become
about ______times more acidic.
a solution of 0.100 mol HC2H3O2 buffered with 0.10 mol NaC2H3O2
also with a pH = 4.74. Here the pH drops to 4.56, a change of only
0.17 units.
Adding 2.0 mL of 10M NaOH (0.020 mol NaOH) to the same 2 solutions.
pH goes from 4.74 to 12.3, a change of 7.6 units, over 40,000,000 x
more basic. Why so much greater than the example above?
pH goes from 4.74 to 4.92, a change of only 0.18 units, nearly
identical to that in example 2 above.
A Quick and Dirty Summary of Acid/Base/Buffer Problems
There are essentially only about 5 types of these problems:
Strong acid: pH = -log [H+] or [H+] = 10-pH
Weak acid (HA): Ka = x2/[HA]i where x in [HA]i -x is ignored. x = [H+]
Buffer: Acid pH = pKa + log [A-]/[HA] (OR Base  pOH = pKb + log
[HA]/[A-])
Weak base: Kb = x2/[Base]i where x in [Base]i -x is ignored and x = [OH]. Kb = Kw/Ka
Strong base: pOH = -log [OH-] or [OH-] = 10-pOH and pH = 14.00 –
pOH
**Titrations that are part way to the equivalence point are buffer
problems.
**Titrations at the equivalence point are salt hydrolysis problems
(usually weak base, Kb problems: A- + H2O  HA + OH-).
15.4 Titrations and pH Curves
The _________________ point of an acid-base reaction is the point at which
stoichiometrically equivalent quantities of acid and base have been added.
HX + MOH  H2O + MX
Color indicators show when this point is reached.
There is a problem here: if either or both of the ions of the salt MX cause
the hydrolysis of water, the equivalence point pH may not be at pH 7.
Different indicators change colors at different pHs.
Choosing the right indicator for the titration is important.
Modern industry uses auto-titrators which employ constant monitoring
by a pH meter and stops automatically when the programmed end point
pH is reached. This is called a potentiometric titration.
Draw a titration curve for a strong acid-base titration here:
14
13
12
11
10
9
8
pH 7
6
5
4
3
2
1
0
0
10
20
30
40
50
60
70
mL 0.100M NaOH added to 50 mL 0.100M HC l
Write in the indicators phenolphthalein, bromthymol blue and methyl red on the
diagram.
Any indicator that falls into the steep portion of the curve will work for this
titration.
Strong Acid-Base Titrations
pH increases very slowly at first.
pH can be calculated at any stage of the titration based on the stoichiometry of
the neutralization and how much acid or base subsequently remaining.
There is a rapid rise in pH near the equivalence point.
**pH at the equivalence point is the pH of the resulting salt solution (based on
Ka or Kb.)
Since neither Na+ nor Cl- cause hydrolysis, pH = 7 at the equivalence point for
these titrations.
Phenolphthalein is used most often for such titrations.
Remember that pH is a logarithmic scale.
Titration of a Weak Acid with a Strong Base
Draw the titration curve for the titration of a weak acid with a strong base here.
14
13
12
11
10
9
8
pH 7
6
5
4
3
2
1
0
0
10
20
30
40
50
60
70
mL 0.100M NaOH added to 50 mL 0.100M HC 2H3O2
Note the important differences in this curve compared to the strong acid curve:
The starting pH is ~3 rather than 1.
pH rises more rapidly early, but more slowly at the equivalence point.
There is a less dramatic change at the equivalence point with weaker
acids.
Equivalence point is NOT pH 7. It is the pH of a 0.05M acetate solution.
(It started as 0.1M acetate, but the volume doubled.)
Equivalence Points of
Strong bases/weak acids will be at pH ____________.
Strong acids/weak bases will be at pH ____________.
Would methyl red (color change ~pH 5) be a good indicator for the titration
above?___________
An indicator must be chosen whose color change occurs within the range of the
steep portion of the graph that includes the equivalence point.
Would phenolphthalein (color change ~pH 9) be a suitable indicator for the
titration above?_____________
When titrating one solution against another, use NaVa = NbVb.
N = normality = equivalents/L sln., a = acid, b = base.
Calculating pH with titration systems: (adding base to an acid)
 Up to the equivalence point, the solution has a mixture of acid and salt.
Calculate the pH like you would for a buffer.
 At the equivalence point, the solution contains essentially only the salt.
Calculate the pH as you would based on the hydrolysis caused by the ions
present using Ka for a hydrolytic cation (like NH4+ or Al3+) or Kb for a
hydrolytic anion (like CN- or C2H3O2-).

Beyond the equivalence point, the solution contains salt and strong base. pH
is almost entirely controlled by the presence of the hydroxide from the strong
base. (Hydrolysis is usually insignificant compared to the presence of the
hydroxide from the strong base.
We could easily exchange the terms acid and base in the above
description.
Do an example problem here: What is the pH of a solution of 50.0mL of 0.100M
acetic acid to which 30.0 mL of 0.100 M NaOH has been added.
**Remember: Do the stoichiometry of the neutralization first, then calculate pH
based on the equilibrium concentrations of the remaining species.
**Another important note: The stoichiometry must be done with moles rather
than concentration (molarity). Why?
_________________________________________
Mol OH-:
Mol HC2H3O2:
Balanced equation for the neutralization:
OH- + HC2H3O2  C2H3O2- + H2O
I
C
E
M of HC2H3O2 =
M of C2H3O2- =
Ka = 1.8 x 10-5 = [H+][C2H3O2-] / HC2H3O2
) = 1.2 x 10-5M
pH = -log(1.2 x 10-5M) = 4.92
[H+] = Ka([HC2H3O2] /[C2H3O2-]
Titrating Polyprotic Acids
With more than 1 ionizable H, titration of polyprotic acids takes place in more
than 1 step.
Ex. carbonic acid: H2CO3 + OH-  H2O + HCO3- ; then HCO3- + OH-  H2O+
CO3-2
A graph of such a titration will have 2 equivalence points.
15.5 Acid/Base Indicators
The best indicator for a titration is one that changes colors at or very near the
equivalence point of the titration (in the steep portion of the titration curve.)
 Most indicators are complex weak acid molecules, represented as HIn.
The molecular form (HIn) has one color, and the ionized form (In-) has a
different color. The indicators are usually such weak acids that they do
not appreciably affect the solution pH or the titration.
 Indicators will change color at a pH equal to the pKa ± 1of the weak acid,
HIn. For example, bromthymol blue has Ka = 1.0 x 10-7 and has a
detectable color change range of pH = 6-8.
 Below is a chart with some common acid/base indicators and the pHs at
which they change color. Note that thymol blue has 2 color changes
indicating that it is probably a diprotic weak acid rather than monoprotic.
15.6 Solubility Equilibria
These equilibria deal with systems where not all components are in the same
phase, specifically with the precipitation and dissolution of ionic precipitates.
Ex. include tooth decay, gall and kidney stones and limestone caves.
The problems are more specific than the general solubility rules.
Solubility equilibria deal with Ksp, the solubility product equilibrium
constant.
There must be a saturated solution in contact with solid solute.
Ex. BaSO4(s)  Ba+2(aq) + SO4-2(aq) Photo from Zumdahl of lower g. i. tract
using BaSO4.
Ksp =
Ca3(PO4)2(s)  3Ca+2(aq) + 2PO4-3(aq)
Ksp =
= 2.0 x 10-29
How soluble is calcium phosphate? Do you know anything made of
calcium phosphate derivatives?_________________________
Solubility and Ksp
Solubility is usually measured in g/L or M. Ksp’s units vary.
The solubility of many substances change as the [ ]’s of other species change.
Ex. Mg(OH)2 gets more soluble as pH drops. All bases become more
soluble in acids.
Let’s examine this with respect to the solubility equilibrium:
Mg(OH)2(s)  Mg+2(aq) + 2OH-(aq)
As [H+] goes up, [OH-] goes __________ and the equilibrium. shifts
____________.
As the equilibrium. shifts, more of the solid dissolves.
Ksp remains the same.
Ex. CaF2(s)  Ca+2(aq) + 2F-(aq) Ksp = 3.9 x 10-11 (@25°C)
Calculate the solubility of calcium fluoride in g/L.
Ksp = 3.9 x 10-11 = [Ca+2] [F-]2 =
The Common Ion Effect and Solubility Equilibria
Adding CaCl2(aq) or NaF(aq) to the solution above causes further
precipitation of CaF2(s).
Calculate the solubility of calcium fluoride in 0.010M Ca(NO3)2.
Ksp does not change.
[Ca+2] = 0.010 + x; [F-] = 2x
Ksp = 3.9 x 10-11 = (0.010 + x)(2x)2 = (0.010)(2x)2 (since x is very small
compared to 0.010)
x2 = 3.9 x 10-11 /4(0.010) = 9.8 x 10-10
x = 3.1 x 10-5M (as compared with 2.1 x 10-4M without the calcium nitrate)
Repeat the problem only with 0.010M NaF
x = 3.9 x 10-7M The solubility has decreased by about 2000x from the
original solution.
15.7 Precipitation and Qualitative Analysis
Equilibrium can be achieved from either direction.
Ions in solution can come together to form a solid precipitate, or
Solid and solvent can produce an aqueous solution.
You may recall from the unit on equilibrium systems that we could calculate “Q”
for a system not at equilibrium to determine which way the system was going to
shift in order to reach equilibrium. We can do the same thing here.
Ex. Q = [Ba+2] [SO4-2]
If Q = Ksp  the solution is at equilibrium (saturated).
If Q > Ksp  precipitation of the solid will occur until Q = Ksp.
If Q < Ksp  dissolving will occur until Q = Ksp.
Ex. 0.100L of 3.0 x 10-3 M Pb(NO3)2 is added to 0.400L of 5.0 x 10-3 M
Na2SO4. Will a precipitate form?
Could both new compounds precipitate?________________
If there is to be a precipitate, it will be______________
Ksp for PbSO4 = 1.6 x 10-8
Write the equation for the equilibrium______________________
Ksp and Qsp = ____________________
[Pb+2] =
[SO4-2] =
Qsp =
Qsp ___Ksp
Precipitation____________occur.
Solubility of any basic anion is affected by pH (both up and down).
Ex. Mg(OH)2(s)  Mg+2(aq) + 2OH-(aq)
1. If [H+] goes up, equilibrium shifts right to replace neutralized
hydroxides.
As the equilibrium shifts right, more solid dissolves, solubility
goes up.
This helps explain why Milk of Magnesia works effectively on
stomach acid.
2. If [OH-] goes up, the common ion effect shifts the equilibrium left,
precipitating more
solid.
Ex. Ksp Mg(OH)2 = 1.8 x 10-11 What is the solubility in a solution buffered
at pH = 9?
pOH = ________ and [OH-] = _______________
Ksp = _____________________ = [Mg+2] (1.0 x 10-5)2 = 1.8 x 10-11
[Mg+2] = 0.18M, so solubility of Mg(OH)2 = 0.18M.
It is actually quite soluble in a buffered, slightly basic solution.
In an unbuffered solution, [Mg+2] = 2.62 x 10-4M and pH = 10.7.
For compounds with weakly basic anions:
CaF2(s)  Ca+2(aq) + 2F-(aq)
H+(aq) + F-(aq)  HF(aq)
Calcium fluoride is more soluble in acid than in neutral or basic solutions.
The solubility of compounds with anions of strong acid anions are largely
unaffected by pH.
Ex. AgCl and BaSO4
Here’s a lovely little example problem from the text: The following solutions are
mixed.
1. 150.0 mL of 0.0100 M Mg(NO3)2 and
2. 250.0 mL of 0.100 M NaF.
Calculate the concentrations of Mg2+ and F- ions at equilibrium.
1. Calculate the new concentrations of Mg2+ and F- ions after mixing. C1V1 =
C2V2 (Nitrate and sodium will be spectators, right?)
2. Calculate Qsp.
3. Compare Qsp to Ksp. (Ksp = 6.4 x 10-9)
4. Run the reaction essentially to completion. Which ion will precipitate out
completely?
5. How much F- will be left after the precipitation? Calculate [F-].
6. Using x to represent [Mg2+] at equilibrium, calculate the ion
concentrations at equilibrium using Ksp.
Selective Precipitation
This allows you to choose whether (weather?) you get rain or snow. Woops,
that’s meteorology.
Metal ions can be separated based on solubility.
Ex. A mixture of Ag+ and Cu+2 ions in the same solution.
Add HCl: AgCl precipitates (Ksp = 1.8 x 10-10)
The copper remains dissolved as CuCl2 is quite soluble.
Sulfide ions are often used to selectively separate metal ions.
[S-2] can be easily controlled by pH.
Saturated H2S solution (0.10M) is used:
H2S(aq)  H+(aq) + HS-(aq)
Ka1 = 5.7 x 10-8
Net
HS-(aq)  H+(aq) + S-2(aq) Ka2 = 1.3 x 10-13
__________________________________
H2S(aq)  2H+(aq) + S-2(aq)
Ka = Ka1x Ka2 = 7.4 x 10-21
Ka = ____________________: at saturation 7.4 x 10-21 = [H+]2[S-2]/0.10
So, [H+]2[S-2] = 7.4 x 10-22
This equation can be easily used to calculate the [S-2] at any pH, since [H+]
is known.
If the pH goes down, [H+] goes ______ and [S-2] goes _______.
Here is another, tougher example problem:
Consider a mixture of 0.10M Zn+2 and 0.10M Cu+2
To what should the pH be adjusted to precipitate copper sulfide, but NOT
zinc sulfide?
First, write the solubility equilibrium equations for both compounds:
ZnS(s) _______________________ Ksp = 1.1 x 10-21
CuS(s) _______________________ Ksp = 6.3 x 10-36
The pH should be adjusted to bring the sulfide ion concentration
just to the point where the ZnS will almost begin to precipitate.
Note that since the CuS is much less soluble than ZnS, it will begin
precipitating long before the ZnS.
Using saturated H2S(aq), [S-2] = Ksp/[Zn+2]: Ksp = 1.1 x 10-21 and
[Zn+2] = 0.10M
Therefore [S-2] = 1.1 x 10-21/0.10, and [S-2] = 1.1 x 10-20 is needed to
begin the precipitation of ZnS.
So, next, what will be the [H+] that gives the necessary sulfide value?
[H+]2(1.1 x 10-20) = 7.4 x 10-22 and [H+]2 = 7.4 x 10-22/1.1 x 10-20
[H+] = (7.4 x 10-22/1.1 x 10-20)1/2 = 0.24M
pH = -log 0.24 = 0.60
Thus, if the pH = 0.6 or lower the [S 2] is too low for ZnS to precipitate.
(Q < Ksp)
If the pH goes down, [H+] goes up and [S-2] must drop since K is constant.
Let’s check and make sure that Q for the CuS is greater than K and that
the compound will indeed precipitate as predicted.
Q = [Cu+2][S-2] = 0.10(1.1 x 10-20) = 1.1 x 10-21
1.1 x 10-21 > 6.3 x 10-36 and Q >> Ksp, so the CuS will precipitate, but ZnS
will not.
Qualitative Analysis
Chemists are frequently called upon to identify “mystery substances” or to
determine what is dissolved in an unknown solution.
Such identification involves several methods of qualitative analysis.
 Some are simple, like flame tests.
 Others involve more sophisticated flame tests measurements involving
atomic absorption (AA) spectrophotometers.
 Still others involve precipitating out groups of cations and anions by a q.
a. scheme.
15.8 Complexations and Complex Ions
Formation of complex ions can also influence solubility.
Ex. AgCl becomes more soluble in NH3(aq) solution than in plain water.
Lewis bases, known as complexing agents or ligands share unbonded
electron pairs with metal ions forming a complex ion.
The number of ligands surrounding the Lewis acid (metal ion) is called
the coordination number.
**The most common ligands include: NH3, CN-, Cl- or Br-, S2O3-2, SCNand OH-.
Ex. AgCl(s)  Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq)  Ag(NH3)2 +(aq)
Net: AgCl(s) + 2NH3(aq)  Ag(NH3)2 +(aq) + Cl-(aq) Coordination
number = 2
A solution of ammonia will dissolve a silver chloride precipitate.
General rule: positive charge x 2 = # complexing ligands
Notable Exception, Al(OH)4Coordination number = 4
However, the rule still holds if you consider the water molecules that are also
attached as ligands.
Al(OH)4(H2O)2Aqueous Equilibria: Further Considerations
17.1 The Common Ion Effect
It’s possible to drive an equilibrium system in a desired direction by adding a
common ion to the system.
It’s also possible to effectively add only one of the ions of the equilibrium by
making sure that the other ion of the pair is a ___________________.
Ex. HC2H3O2(aq) ⇌ H+(aq) + C2H3O2-(aq)
By adding HCl(aq), we add H+(aq) and Cl-(aq) ions, but chloride is not
part of this system and so has no effect.
NaC2H3O2 will do the same thing. What is another substance that might
be added to accomplish the common ion effect?_________________
Adding any of these substances causes the equilibrium to shift _________.
There are example problems in the text on pp. 616-17.
Adding a common ion does not affect K.
Common Ions Generated by Acid-Base Reactions
HC2H3O2(aq) + OH-(aq) ⇌ C2H3O2-(aq) + H2O(l)
*Note that this is just the opposite of water hydrolysis by the acetate ion.
K = 1/Kb for acetate = Ka for acetic acid/Kw = 1.8 x 10-5/1.0 x 10-14 = 1.8 x
109
The extremely large value of K suggests that the reaction goes
virtually__________.
Ex. Adding 0.10 mol NaOH to 0.20 mol acetic acid to total 1.00L. The
neutralization of 0.10 mol of HC2H3O2 is complete leaving 0.10 mol
HC2H3O2 of to come to equilibrium with H+ and C2H3O2-.
**However, the pH of the resulting solution will NOT be the same as a 0.1M
acetic acid solution (pH = 2.87). Why not?
The value of Ka stays the same.
The presence of additional acetate ions pushes the equilibrium to the left,
reducing the [H+] and raising the pH.
Calculations: Ka = [H+] [C2H3O2-]/[HC2H3O2]
for 0.1M HC2H3O2: x2/0.1 = 1.8 x 10-5 x = ____________ pH =
__________
for the partially neutralized acid solution above: 0.1(x)/0.1 = 1.8 x
10-5
x = _____________ pH = ____________
The AP folks really like these kinds of problems, which involve 2 sets of
calculations. For problems like this one that involves the partial neutralization
of a weak acid by a strong base or of a weak base by a strong acid, do the
problems in this order:
Do the stoichiometry of the acid-base reaction, THEN
Do the equilibrium calculations on the remaining solution.
Assume that acid-base reactions involving H+ and OH- are complete.
See the sample exercise on p. 623.
17.2 Acid/Base Titrations and Titration Curves
The _________________ point of an acid-base reaction is the point at which
stoichiometrically equivalent quantities of acid and base have been added.
HX + MOH ---> H2O + MX
Color indicators show when this point is reached.
There is a problem here: if either or both of the ions of the salt MX cause
the hydrolysis of water, the equivalence point pH may not be at pH 7.
Different indicators change colors at different pHs.
Choosing the right indicator for the titration is important.
Modern industry uses auto-titrators which employ constant monitoring
by a pH meter and stops automatically when the programmed end point
pH is reached. This is called a potentiometric titration.
Draw a titration curve for a strong acid-base titration here:
14
13
12
11
10
9
8
pH 7
6
5
4
3
2
1
0
0
10
20
30
40
50
60
70
mL 0.100M NaOH added to 50 mL 0.100M HCl
Write in the indicators phenolphthalein, bromthymol blue and methyl red on the
diagram.
Any indicator that falls into the steep portion of the curve will work for this
titration.
Strong Acid-Base Titrations
pH increases very slowly at first.
pH can be calculated at any stage of the titration based on the stoichiometry of
the neutralization and how much acid or base subsequently remaining.
There is a rapid rise in pH near the equivalence point.
**pH at the equivalence point is the pH of the resulting salt solution (based on
Ka or Kb.)
Since neither Na+ nor Cl- cause hydrolysis, pH = 7 at the equivalence point for
these titrations.
Phenolphthalein is used most often for such titrations.
**See Table 17.1 on p. 620 to understand the big change in pH near the
equivalence point.
Remember that pH is a logarithmic scale.
Titration of a Weak Acid with a Strong Base
Draw the titration curve for the titration of a weak acid with a strong base here.
14
13
12
11
10
9
8
pH 7
6
5
4
3
2
1
0
0
10
20
30
40
50
60
70
mL 0.100M NaOH added to 50 mL 0.100M HC 2 H3 O2
Note the important differences in this curve compared to the strong acid curve:
The starting pH is ~3 rather than 1.
pH rises more rapidly early, but more slowly at the equivalence point.
There is a less dramatic change at the equivalence point with weaker
acids. (See figure 17.8 p.625)
Equivalence point is NOT pH 7. It is the pH of a 0.05M acetate solution.
(It started as 0.1M acetate, but the volume doubled.)
Equivalence Points of
Strong bases/weak acids will be at pH ____________.
Strong acids/weak bases will be at pH ____________.
Would methyl red be a good indicator for the titration above?___________
An indicator must be chosen whose color change occurs within the range of the
steep portion of the graph that includes the equivalence point.
Would phenolphthalein be a suitable indicator for the titration
above?_____________
Calculating pH with titration systems: (adding base to an acid)
Up to the equivalence point, the solution has a mixture of acid and salt.
Calculate the pH like you would for a buffer (Section 17.3).
At the equivalence point, the solution contains essentially only the salt.
Calculate the pH as you would based on the hydrolysis caused by the ions
present (Ka or Kb).
Beyond the equivalence point, the solution contains salt and strong base.
pH is almost entirely controlled by the presence of the hydroxide from the
strong base. (Hydrolysis is usually insignificant compared to the presence
of the hydroxide from the strong base.
We could easily exchange the terms acid and base in the above
description.
Do an example problem here: What is the pH of a solution of 50.0mL of 0.100M
acetic acid to which 30.0 mL of 0.100 M NaOH has been added.
**Remember: Do the stoichiometry of the neutralization first, then calculate pH
based on the concentrations of the remaining species.
**Another important note: The stoichiometry must be done with moles rather
than concentration (molarity). Why?
_________________________________________
Mol OH-:
Mol HC2H3O2:
Balanced equation for the neutralization:
OH- + HC2H3O2  C2H3O2- + H2O
I
C
E
M of HC2H3O2 =
M of C2H3O2- =
Ka = 1.8 x 10-5 = [H+][C2H3O2-] / HC2H3O2
) = 1.2 x 10-5M
pH = -log(1.2 x 10-5M) = 4.92
[H+] = Ka([HC2H3O2] /[C2H3O2-]
Titrating Polyprotic Acids
With more than 1 ionizable H, titration of polyprotic acids takes place in more
than 1 step.
Ex. carbonic acid: H2CO3 ⇌ H+ + HCO3- ⇌ H+ + CO3-2
Titrating with NaOH:
H2CO3 + OH-  HCO3- + H2O
HCO3- + OH-  CO3-2 + H2O
Note: there are 2 distinct equivalence points.
Two different indicators are needed.
There are 2 color changes that occur.
14
12
10
pH
8
6
4
2
0
mL HCl added to Na2CO3(aq)
17.3 Buffered Solutions
Composition and Action of Buffered Solutions
Buffers are solutions designed to maintain a (relatively) constant pH with
the addition of _______________________________.
That constant pH is not necessarily pH 7.
Important examples of buffer systems include
__________________________.
Buffers neutralize or “absorb” both H+ and OH-.
This is accomplished by creating weak acid/base or base/acid systems.
Ex. HC2H3O2/C2H3O2- or NH4+/NH3
Such a system is easily established by using a weak acid and a salt with
the same anion.
Ex.__________________________
Or using a weak base and a salt with the same cation.
Ex_____________________
General acid dissociation equilibrium: HX ⇌ H+ + XKa = [H+] [X-]/[HX]
[H+] = Ka ([HX]/[X-])
[H+] and therefore pH depend on
the acid Ka value
the ratio of HX/XAdding Acid or Base to a Buffer
Adding base: OH- + HX  H2O + X-: with respect to the equilibrium; HX
⇌ H+ + XOH- + H+ ---> H2O; as [H+] drops due to neutralization,
the equilibrium. shifts _______ to replace the disappearing hydrogen ions.
There are plenty of HX molecules to accomplish this. The pH remains
nearly unchanged.
Adding acid: H+ + X-  HX: again with respect to the equilibrium; HX ⇌
H+ + X-; as the [H+] rises, the equilibrium shifts _________ to use up the
extra hydrogen ions. The excess X- from the salt allows this shift to take
place easily. Again, the pH remains mostly constant. This is not the case
with a solution of just the weak acid.
In both cases, the [HX]/[X-] ratio changes, but if the change is small, the
change in pH is also small.
** Buffers are most effective if [HX]/[X-] is about 1. If [HX] = [X-], then [H+] =
Ka.
** It is best to choose a buffer whose acid form (HX) has a pH close to the pKa of
the
desired pH.
pKa = _____________ If [HX]/[X-] = 1, then pH = pKa.
Buffer Capacity and pH
Along with pH, the other important characteristic of a buffer is its buffering
capacity.
How much acid or base can the solution absorb without a significant change in
pH?
Buffer capacity () depends on the amount of acid and conjugate base from
which the buffer is made.
Ex. Consider 2 systems:
1M HC2H3O2 + 1M NaC2H3O2
0.1M HC2H3O2 + 0.1M NaC2H3O2
pH is the same for both: pH = pKa since the [HX]/[X-] = 1
However, the 1M system can buffer far more acid or base than the 0.1M
system.
Calculating Buffer pH
Use the same basic procedure as a common ion effect problem.
Ka = [H+] [X-]/[HX] so
[H+] = Ka [HX]/[X-]
-log[H+] = -log(Ka [HX]/[X-])
= -logKa - log[HX]/[X-]
-log = p therefore pH = pKa - log[HX]/[X-] or more commonly
pH = pKa + log[X-]/[HX] where X- is the base and HX is the acid.
The general form of this equation is known as the Henderson-Hasselbalch
equation.
Write the general form here:_________________________
There is also a corresponding base form of the
equation:_________________________
Ex. Find the pH of a 0.12M lactic acid (HC3H5O3) solution with 0.10M
sodium lactate.
HC3H5O3 ⇌ H+ + C3H5O3I
0.12M
0
0.10
C
-x
+x
+x
E
0.12 - x
x
0.10 + x
Ka = 1.4 x 10-4 = [H+] [C3H5O3-]/[HC3H5O3] = x(0.10+x)/(0.12-x)
x will be small compared to 0.12 and 0.10 and so may be ignored.**
** Some books give 10-5 as a flat cutoff for when it is acceptable to ignore x in
these calculations, but others use what is called the “5% Rule.” That is, if K is
less than 5% of the smallest concentration value in the calculation, x may be
ignored.
In this problem, 0.10 is the smallest value. Five percent of that is 0.05 x 0.10 = 5 x
10-3. Ka is still smaller than this value, so x may be ignored.
Therefore, 1.4 x 10-4 = 0.10x/0.12 and x = 1.7 x 10-4
and pH = 3.77
Method 1:
OR
Method 2 using the Henderson-Hasselbalch equation:
pH = pKa + log [base]/[acid]
= -log(1.4 x 10-4) + log (0.10/0.12)
=
3.85
+
(-0.08) = 3.77
Which do you think is easier??
**Special note: The Henderson-Hasselbalch equation always has a conjugate
weak acid/base pair as the A and B in the equation. Never will H+ or OH- be
the acid or base.
Addition of Acids or Bases to Buffers and pH.
**Any reaction with a strong acid or strong base goes ____________________.
These include:
Strong acid with strong base.
Strong acid with weak base.
Strong base with weak acid.
Weak acid/weak base reactions are more difficult.
As mentioned before, deal with the stoichiometry of the neutralization first,
then do the pH problem with the remaining ions in the solution.
Ex. problem p. 633. 0.300 mol HC2H3O2 + 0.300 mol NaC2H3O2 has a
solution pH = 4.74.
Find the pH after the addition of 0.020 mol NaOH.
Find the pH after the addition of 0.020 mol HCl.
Solve the 2 problems here:
To help illustrate how dramatic the buffering effect can be, consider the
following examples:
Adding 2.0 mL of 10M HCl (0.020 mol HCl) to
a plain acid solution of pH = 4.74 (1.8 x 10-5 M HCl) causes the pH
to drop to 1.70: a change of 3.04 units. This solution has become
about ______times more acidic.
a solution of 0.100 mol HC2H3O2 buffered with 0.10 mol NaC2H3O2
also with a pH = 4.74. Here the pH drops to 4.56, a change of only
0.17 units.
Adding 2.0 mL of 10M NaOH (0.020 mol NaOH) to the same 2 solutions.
pH goes from 4.74 to 12.3, a change of 7.6 units, over 40,000,000 x
more basic. Why so much greater than the example above?
pH goes from 4.74 to 4.92, a change of only 0.18 units, nearly
identical to that in example 2 above.
A Quick and Dirty Summary of Acid/Base/Buffer Problems
There are essentially only about 5 types of these problems:
Strong acid: pH = -log [H+] or [H+] = 10-pH
Weak acid (HA): Ka = x2/I where x in I-x is ignored. x = [H+] and I =
initial [HA]
Buffer: pH = pKa + log [A-]/[HA]
Weak base: Kb = x2/I where x = [OH-] and I = initial [base] . Kb =
Kw/Ka
Strong base: pOH = -log [OH-] or [OH-] = 10-pOH and pH + pOH = 14
17.4 Solubility Equilibria
These equilibria deal with systems where not all components are in the same
phase, specifically with the precipitation and dissolution of ionic precipitates.
Ex. include tooth decay, gall and kidney stones and limestone caves.
The problems are more specific than the general solubility rules.
Solubility equilibria deal with Ksp, the solubility product equilibrium
constant.
There must be a saturated solution in contact with solid solute.
Ex. BaSO4(s) ⇌ Ba+2(aq) + SO4-2(aq)
Ksp =
Ca3(PO4)2(s) ⇌ 3Ca+2(aq) + 2PO4-3(aq)
Ksp =
= 2.0 x 10-29
How soluble is calcium phosphate? Do you know anything made of
calcium phosphate derivatives?_________________________
Solubility and Ksp
Solubility is usually measured in g/L or M. Ksp’s units vary.
The solubility of many substances change as the [ ]’s of other species change.
Ex. Mg(OH)2 gets more soluble as pH drops. All bases become more
soluble in acids.
Let’s examine this with respect to the solubility equilibrium:
Mg(OH)2(s) ⇌ Mg+2(aq) + 2OH-(aq)
As [H+] goes up, [OH-] goes __________ and the equilibrium. shifts
____________.
As the equilibrium. shifts, more of the solid dissolves.
Ksp remains the same.
Ex. CaF2(s) ⇌ Ca+2(aq) + 2F-(aq) Ksp = 3.9 x 10-11 (@25¡C)
Calculate the solubility of calcium fluoride in g/L.
Ksp = 3.9 x 10-11 = [Ca+2] [F-]2 =
The Common Ion Effect and Solubility Equilibria
Adding CaCl2(aq) or NaF(aq) to the solution above causes further
precipitation of CaF2(s).
Calculate the solubility of calcium fluoride in 0.010M Ca(NO3)2.
Ksp does not change.
[Ca+2] = 0.010 + x; [F-] = 2x
Ksp = 3.9 x 10-11 = (0.010 + x)(2x)2 = (0.010)(2x)2 (since x is very small
compared to 0.010)
x2 = 3.9 x 10-11 /4(0.010) = 9.8 x 10-10
x = 3.1 x 10-5M (as compared with 2.1 x 10-4M without the calcium nitrate)
Repeat the problem only with 0.010M NaF
x = 3.9 x 10-7M The solubility has decreased by about 2000x from the
original solution.
17.5 Criteria for Precipitation and Dissolution
Equilibrium can be achieved from either direction.
Ions in solution can come together to form a solid precipitate, or
Solid and solvent can produce an aqueous solution.
You may recall from the unit on equilibrium systems, that we could calculate
“Q” for a system not at equilibrium to determine which way the system was
going to shift in order to reach equilibrium. We can do the same thing here.
Ex. Q = [Ba+2] [SO4-2]
If Q = Ksp  the solution is at equilibrium (saturated).
If Q > Ksp  precipitation of the solid will occur until Q = Ksp.
If Q < Ksp  dissolving will occur until Q = Ksp.
Ex. 0.100L of 3.0 x 10-3M Pb(NO3)2 is added to 0.400L of 5.0 x 10-3M
Na2SO4. Will a precipitate form?
Could both new compounds precipitate?________________
If there is to be a precipitate, it will be______________
Ksp for PbSO4 = 1.6 x 10-8
Write the equation for the equilibrium______________________
Ksp and Qsp = ____________________
[Pb+2] =
[SO4-2] =
Qsp =
Qsp ___Ksp
Precipitation____________occur.
Solubility of any basic anion is affected by pH (both up and down).
Ex. Mg(OH)2(s) ⇌ Mg+2(aq) + 2OH-(aq)
1. If [H+] goes up, equilibrium. shifts right to replace neutralized
hydroxides.
As the equilibrium. shifts right, more solid dissolves, solubility
goes up.
This helps explain why Milk of Magnesia works effectively on
stomach acid.
2. If [OH-] goes up, the common ion effect shifts the equilibrium. left,
precipitating more
solid.
Ex. Ksp Mg(OH)2 = 1.8 x 10-11 What is the solubility in a solution buffered
at pH = 9?
pOH = ________ and [OH-] = _______________
Ksp = _____________________ = [Mg+2] (1.0 x 10-5)2 = 1.8 x 10-11
[Mg+2] = 0.18M, so solubility of Mg(OH)2 = 0.18M.
It is actually quite soluble in a buffers, slightly basic solution.
In an unbuffered solution, [Mg+2] = 2.62 x 10-4M and pH = 10.7.
For compounds with weakly basic anions:
CaF2(s) ⇌ Ca+2(aq) + 2F-(aq)
H+(aq) + F-(aq) ⇌ HF(aq)
Calcium fluoride is more soluble in acid than in neutral or basic solutions.
The solubility of compounds with anions of strong acid anions are largely
unaffected by pH.
Ex. AgCl and BaSO4
Selective Precipitation
This allows you to choose whether (weather?) you get rain or snow. Woops,
that’s meteorology.
Metal ions can be separated based on solubility.
Ex. A mixture of Ag+ and Cu+2 ions in the same solution.
Add HCl: AgCl precipitates (Ksp = 1.8 x 10-10)
The copper remains dissolved as CuCl2 is quite soluble.
Sulfide ions are often used to selectively separate metal ions.
[S-2] can be easily controlled by pH.
Saturated H2S solution (0.10M) is used:
H2S(aq) ⇌ H+(aq) + HS-(aq)
Ka1 = 5.7 x 10-8
Net
HS-(aq) ⇌ H+(aq) + S-2(aq) Ka2 = 1.3 x 10-13
__________________________________
H2S(aq) ⇌ 2H+(aq) + S-2(aq)
Ka = Ka1x Ka2 = 7.4 x 10-21
Ka = ____________________: at saturation 7.4 x 10-21 = [H+]2[S-2]/0.10
So, [H+]2[S-2] = 7.4 x 10-22
This equation can be easily used to calculate the [S-2] at any pH, since [H+]
is known.
If the pH goes down, [H+] goes ______ and [S-2] goes _______.
Here is another, tougher example problem:
Consider a mixture of 0.10M Zn+2 and 0.10M Cu+2
To what should the pH be adjusted to precipitate copper sulfide, but NOT
zinc sulfide?
First, write the solubility equilibrium equations for both compounds:
ZnS(s) ⇌_______________________ Ksp = 1.1 x 10-21
CuS(s) ⇌_______________________ Ksp = 6.3 x 10-36
The pH should be adjusted to bring the sulfide ion concentration
just to the point where the ZnS will almost begin to precipitate.
Note that since the CuS is much less soluble than ZnS, it will begin
precipitating long before the ZnS.
Using saturated H2S(aq), [S-2] = Ksp/[Zn+2]: Ksp = 1.1 x 10-21 and
[Zn+2] = 0.10M
Therefore [S-2] = 1.1 x 10-21/0.10, and [S-2] = 1.1 x 10-20 is needed to
begin the precipitation of ZnS.
So, next, what will be the [H+] that gives the necessary sulfide value?
[H+]2(1.1 x 10-20) = 7.4 x 10-22 and [H+]2 = 7.4 x 10-22/1.1 x 10-20
[H+] = (7.4 x 10-22/1.1 x 10-20)1/2 = 0.24M
pH = -log 0.24 = 0.60
Thus, if the pH = 0.6 or lower the [S-2] is too low for ZnS to precipitate.
(Q < Ksp)
If the pH goes down, [H+] goes up and [S-2] must drop since K is constant.
Let’s check and make sure that Q for the CuS is greater than K and that
the compound will indeed precipitate as predicted.
Q = [Cu+2][S-2] = 0.10(1.1 x 10-20) = 1.1 x 10-21
1.1 x 10-21 > 6.3 x 10-36 and Q >> Ksp, so the CuS will precipitate, but ZnS
will not.
There will be a question similar to this one on the next exam. (Was that subtle
enough?)
Complexations
Formation of complex ions can also influence solubility.
Ex. AgCl becomes more soluble in NH3(aq) solution than in plain water.
Lewis bases, known as complexing agents or ligands share unbonded
electron pairs with metal ions forming a complex ion.
The number of ligands surrounding the Lewis acid (metal ion) is called
the coordination number.
**The most common ligands include: NH3, CN-, Cl- or Br-, S2O3-2, SCNand OH-.
Ex. AgCl(s) ⇌ Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2 +(aq)
Net: AgCl(s) + 2NH3(aq) ⇌ Ag(NH3)2 +(aq) + Cl-(aq) Coordination
number = 2
A solution of ammonia will dissolve a silver chloride precipitate.
Other examples may be found on p. 626 in the text.
General rule: positive charge x 2 = # complexing ligands
Notable Exception, Al(OH)4Coordination number = 4
Please read the sections at the end of Chpt. 17 on Amphoterism and
Qualitative Analysis.