Download Chemistry Section 2

Chemistry Section 2
Heat Exchange

Three Kinds of Heat Exchange:
1) Convection: Fluid movement caused by the hotter portions of a fluid rising and the cooler portions of
a fluid sinking. Air currents and convection currents in water are examples.
2) Radiation: Electromagnetic waves emitted from a hot body into the surrounding environment.

Light colors radiate and absorb less

Dark colors radiate and absorb more

Q1. What is a black body radiator?
3) Conduction: Molecular collisions carry heat along a conduit. Recall that temperature is a reflection of
the average kinetic energy of the molecules. High energy molecules collide with their neighbors, which
in turn collide with their neighbors until eventually the energy is spread equally throughout.

Heat conduction is roughly analogous to current flow through a wire.

Q2. A pan of water is placed upon an electric heating element on a stove. Describe all types of
heat exchange expected to occur in this scenario.
Heat Capacity & Specific Heat Capacity

Heat Capacity: The amount of energy (Joules or Calories) a system must absorb to give a unit change in
temperature (J/K or cal/˚C).
o
Formula, where C is the heat capacity, q is heat (or other energy) and T is temperature:

o

C = q/∆T
Constant Volume vs. Constant Pressure Heat Capacities: Systems can be confined by rigid walls
(constant volume) or can be open to the atmosphere like water in a beaker (constant pressure).

Q3. What happens when heat enters a system? Does the temperature always increase? Is any
increase in temperature always exactly proportional to the heat absorbed by that system? (Hint:
Think of adding energy to a sealed steel container vs. adding energy to a balloon; remember that
temperature is the average kinetic energy of the molecules.)

Q4. For the same system, which heat capacity will be greater, the constant volume heat capacity
or the constant pressure heat capacity?
Specific Heat Capacity: Above we defined Heat Capacity as the energy absorbed/unit change in
temperature for a system. This “system” could include a solution, the container holding the solution, and
even a thermometer or stirring rod. Specific Heat Capacity, however, describes energy absorption for one
individual substance only and is defined per unit mass. Little “c” is used instead of big “C”.
o
Formula:

c = q/m∆T ; often re-written as: q = mc∆T
o
Specific Heat of Water = 1.0 cal/g˚C or 4.18 J/g˚C
o
Q5. Beakers 1 and 2 contain 0.25 L of water and 0.5 L of water, respectively. How does the heat
capacity of the water in Beaker 1 compare to that of the water in Beaker 2? How do the specific heat
capacities compare?
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Sample MCAT Question
1) Beaker A contains 2.55 g of FeCl3 dissolved in 100 mL of water. Beaker B contains 10.20 g of FeCl3
dissolved in 400 mL of water. If the specific heat capacity of the solution in Beaker A is known to be
1.33 cal/g˚C, what is the expected specific heat capacity of the solution in Beaker B?
A)
B)
C)
D)
1.33 cal/g˚C
2.66 cal/g˚C
5.32 cal/g˚C
Solutions do not have specific heat capacities.
Solution: The two solutions have the same molar concentration of FeCl3, Beaker B simply contains four times as much of the
solution. Specific Heat Capacity is an intensive property and thus should not vary with amount. This makes Answer A the best
choice. Note, however, that had the question asked for heat capacity—which refers to the heat absorption of the system as a
whole instead of per gram or per mole—then Beaker B would indeed have approximately four times the heat capacity. The
way the data are presented here is a common MCAT trend. You are given two sets of mass and volume that appear different
and unrelated, but when multiplied actually give identical molarities. Be on the lookout for these!
Calorimeters

Purpose: Device used to calculate enthalpy change (∆H). We are assuming that q (which is what the
calorimeter actually measures) is equal to ∆H. This is true at constant pressure.
o
Know how each type of calorimeter works.

Coffee Cup Calorimeter: Solve using: q=mc∆T

Bomb Calorimeter: Solve using: q = C∆T. This does NOT give enthalpy, but change in internal
energy, ∆U or ∆E. Use heat capacity (big C) instead of specific heat capacity (little c).


Q6. Which type of calorimeter provides constant pressure and which type provides constant
volume?
Pressure-Volume (PV) Work:
o
Work is energy transfer via a force (physics), or via a change in volume at constant pressure
(chemistry).
o
PV Work = P∆V (requires constant pressure, any change in volume tells you there is pv work)

On a pressure vs. volume graph, the area under the curve is pv work.

Q7. Which of the calorimeters described above allows for pv work?
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Thermodynamics

The First Law of Thermodynamics: Energy neither be created nor destroyed
o
IMPORTANT NOTE: By convention, work done on the system is positive, work done by the system is
negative (chemistry definition); Q8. How is the sign of work defined in physics?
o
Two Common Ways to Define the First Law of Thermodynamics:
1) The total energy of an isolated system is always constant.
2) The energy change in a closed system is equal to the heat absorbed by that system plus any work
done on that system by its surroundings.


∆E = q + w

Q9. Explain verbally and/or mathematically how definitions 1 and 2 say the same thing.
The Second Law of Thermodynamics:
o
Heat cannot be changed completely into work in a cyclical process.
o
Entropy in an isolated system can never decrease.
The Third Law of Thermodynamics:
o


Pure crystalline substances at absolute zero have an entropy of zero.
The Zeroth Law of Thermodynamics:
o
If object A is in thermal equilibrium with object B, and object C is also in thermal equilibrium with
object B, then objects A and C must be in thermal equilibrium with each other.
o
What the heck does that mean? This is the honest reaction of most students upon reading the Zeroth
Law for the first time. This definition is indeed a rather obscure construct. Some texts state the Zeroth
Law simply as “temperature exists.” Others explain that the transitive nature of thermal equilibrium
(A = B and B = C, therefore A = C) had to be defined to allow for the definition of a measurable
physical parameter, temperature. (Did that help? Probably not.)
o
For the MCAT, Just Remember This: Everything tends to move toward thermal equilibrium with
everything else. Objects with higher temperatures will always equilibrate over time with their
surroundings, including other objects with which they are in contact. Finally, if two objects are in
thermal equilibrium, by definition they have the same temperature.

Q10. Provide a conceptual definition for temperature.

KE = 3/2kBT (where kB is Boltzmann’s constant; this shows the direct relationship between
temperature and kinetic energy)

Know how to convert Celsius to Kelvin; always use Kelvin in formulas, unless Celsius is specified.
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Enthalpy (∆H)


Definition: THINK OF ENTHALPY AS: Enthalpy = the energy contained within chemical bonds. This
is a simplified definition, but one that works perfectly for the purposes of the MCAT. It is illustrated by the
fact that we can calculate ∆H for a reaction (i.e., the change in enthalpy) by finding the difference in total
bond energy between the products and the reactants.
o
Units: Joules
o
Standard State: A set of specific conditions chosen as the reference point for measuring and reporting
enthalpy, entropy, and Gibbs free energy. You may remember seeing these conditions specified at the
top of a table of ∆H values. When values are given for standard state conditions, a superscript, called
“naught,” is added. The naught symbol resembles a degree symbol (∆H°).

Do NOT confuse standard state with STP. Do NOT memorize any values for standard state
conditions because they can be different for different tables. Later on, you will need to memorize
STP conditions. If you confuse these two you can easily get into trouble. For example, a common
standard state temperature is 25°C or 298K, but the standard temperature for STP conditions is
0°C or 273K. To make this worse, the MCAT authors often use the answer obtained when you
confuse the two concepts as one of the wrong answer choices.

Elements in their standard state have Hformation° = zero. This is because elements in their
standard state are used to define the enthalpy scale and thus there is no change in energy to
create them from themselves. The enthalpy change for graphite (solid carbon in its standard state)
for example, is zero. However, the enthalpy change for diamond is 2.0 kJ/mol because energy is
required to form diamond from graphite. There is a similar definition for Gibbs free energy, which
you will encounter later in this section, i.e., Gformation° = zero for elements, but the standard
state entropy of elements is not zero.
Common Forms of ΔH:
o
ΔHRXN the enthalpy change for a reaction: This is usually calculated by adding reactions (and their
associated enthalpy changes) from a table. You must select the reactions from the table that—when
added together—will produce the net reaction for which you are calculating ∆HRXN. To calculate ∆HRXN
you will add all of the values given for each of the reactions you use, paying careful attention to signs
and stoichiometry. If the reaction proceeds in the same direction as it would in the net reaction, use
the value given directly. If it proceeds in the opposite direction, change the sign of the value given.
You MUST multiply the number given in the table by the coefficient in the balanced net
reaction. For example, suppose one of the reactions given was the formation of liquid water from the
elements hydrogen and oxygen: 1/2O2 + H2  H2O ∆H°f = -285 kJ/mol, but in your net reaction water
is a reactant rather than a product. You would need to change the sign of ∆H° to be positive and
combine it with the enthalpy changes from the other reactions used. If your balanced net reaction
contained two moles of water, you would need to double the value given, and so forth.

Calculating ΔHRXN Using Bond Energies: You can also calculate the enthalpy change of a reaction
using bond energies. This is not quite as common on the MCAT as the method of adding reactions
(but it is a bit more intuitive in our opinion). To do so, simply add up the bond energies of all of
the products and reactants. If a bond is broken during the reaction, energy is required, so the
bond energy should be given a positive sign. If a bond is formed, energy is released, so the bond
energy should be given a negative sign. Once again, multiply all bond energy values by their
coefficients in the balanced equation.

Looking Ahead: In a future chemistry lesson, you will be asked to calculate the standard reduction
potential, E for electrochemical reactions. To do so, you will be adding together half-reactions in
a very similar fashion to what we’ve just explained here. When calculating E, remember that
you do NOT multiply by the coefficients in the balanced equation. However, you DO
change the sign if the half-reaction is reversed.
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o
ΔHformation: The enthalpy value for the formation of a compound from its elements in their standard
states. If the number is negative, formation is an exothermic process, if it is positive, the process is
endothermic.
o
ΔHcombustion: The enthalpy value for the combustion of a compound with O2 to form CO2 and water. A
high heat of combustion is associated with an unstable molecule and a low heat of
combustion with a stable molecule.
o
ΔHsolution: The enthalpy value associated with the dissolution of a species into solution. We’ll discuss
this in more detail when we cover solution chemistry.
o
ΔHvaporization: The enthalpy value associated with the phase change from liquid to gas. The reverse
process (condensation) simply interchanges products and reactants and thus the sign is just changed.
o
ΔHfusion: The enthalpy value associated with the phase change from liquid to solid. The sign changes
for the reverse process (melting).

Each of the above items has an analogous definition for changes in entropy (∆S) and Gibbs free
energy (∆G) (i.e., ∆Gformation, ∆Gvaporization, etc. Because the basic science principles are identical,
but these other quantities aren’t seen as frequently, they could very likely appear on the MCAT.)
Entropy (∆S)

Definition: THINK OF ENTROPY AS: Entropy = a measure of the randomness or disorder in a system.
Reactions and other processes will be more likely to be spontaneous if they increase entropy.
o
o
Units: Joules/K

As a reaction proceeds, if randomness increases, energy will be released and thus be available to
do work. If randomness decreases, energy is required to “create” this increased order, and that
amount of energy will thus be unavailable to do work.

positive ∆S = increased randomness, and thus more energy available to do work.

negative ∆S = decreased randomness, and thus less energy available to do work.

Reactions at equilibrium are at maximum entropy.
Entropy increases with increasing:
1) number of items/particles/etc.

Caveat: The number of moles of gas trumps the number of moles of species in any other
phase. Thus, even if a reaction turns two moles of reactants into one mole of product, if that
one mole of product is a gas, and the reactants are not, entropy has increased; and S will
therefore be positive.
2) volume
3) temperature
4) disorder (e.g., S is greater for an amorphous material than for a crystal)
5) complexity (e.g., S is greater for C2H6 than for CH4)
Gibbs Free Energy (∆G)

Definition: THINK OF GIBBS FREE ENERGY AS: ∆G = the amount of “free” or “useful” energy
available to do work (excluding pv work; as a result of running an isothermal, isobaric reaction)
o
If energy is available and the system can do work, Gibbs Free Energy is negative. If energy must be
added to the reaction (e.g., heat must be added to the system) to make it proceed, Gibbs Free Energy
is positive.
o
Units: Joules
o

negative ∆G = Spontaneous process; free energy available to do work.

positive ∆G = Non-spontaneous process; no free energy available; energy is required.
Q11. Provide a conceptual definition for the terms isobaric and isothermal.
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
The Fundamental Thermodynamic Relation:
o
∆G = ∆H - T∆S

o
o
Recall that enthalpy is basically the change in bond energy from reactants to products. If there
were no change in randomness during the reaction, the amount of energy available to do work
(ΔG) would be exactly equal to enthalpy (∆H). As described above, if randomness increases
(positive ΔS), energy will be released and that energy (in addition to ΔH) will also be available to
do work (creating a larger, more negative ΔG). By contrast, if randomness decreases, energy will
be “used” to create this order, decreasing the amount of energy available to do work. The “T” term
in the equation converts entropy into joules (J/K*K = J). You may recall that energy can also be
used to increase temperature or to expand the volume (PV work), but neither occurs here because
the system is both isothermal and isobaric.
Use the Fundamental Thermodynamic Relation to predict the answer to MCAT questions such as:

Q12. If ∆H is positive and entropy change is negative, what will the sign of ∆G be?

Q13. If the change in entropy is positive, and enthalpy is negative, the reaction is: a) spontaneous,
b) non-spontaneous, c) can be either spontaneous or non-spontaneous depending on temperature.

Q14. If a reactant is dissolved in solution, causing the temperature of the reaction vessel to
increase, the G for this reaction must be a) positive, b) negative, or c) cannot be determined.
∆G° = - RTlnKeq ; relates the equilibrium constant to the Gibbs free energy (Note: Remember that the
ln of any positive number less than 1 is negative).

A useful rearrangement of the above equation: Keq = e-ΔG/RT

Q15. If the value of Keq is known, what can we infer about ΔG°? If a reaction is spontaneous, what
can we infer about the rate of that reaction?
Sample MCAT Question
2) The hydration of ammonium nitrate is a highly exothermic dissolution reaction. Which of the following
statements is NOT true of this process?
A) The reaction must be spontaneous because it is both exothermic and has a favorable entropy change.
B) The reaction could be spontaneous or non-spontaneous depending on the temperature at which the
reaction is run.
C) The products of the reaction have greater entropy than do the reactants.
D) The total bond energy of all the products exceeds the total bond energy of all the reactants.
Solution: Answer A states in words what is demonstrated by the fundamental thermodynamic relationship: ∆G = ∆H - T∆S.
Namely, if the enthalpy term is negative and the entropy change is positive (as it must be for a dissolution reaction), and thus
the -T∆S term is negative, the reaction must be spontaneous. Answer C states the conditions for having a positive entropy
change, which we know this reaction does have because it is a dissolution reaction. Answer D states the conditions for being
exothermic, which this reaction is, as given in the stem. Answer B would be true if the signs of enthalpy change and entropy
change were both negative, or both positive, but they are not in this case. We are looking for a statement that is NOT true, so
B is the correct answer.
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Acid-Base Chemistry

Acid-Base Definitions:
o
Arrhenius: Acids produce H+ ions in solution; bases
produce OH- ions in solution.
o
Bronsted-Lowry: Acids donate protons (H+); bases
accept protons (H+).
o
Lewis: Acids accept a pair of electrons; bases donate
a pair of electrons.


When the MCAT tests Lewis acids or
Lewis bases, they always specifically label
them as such. Otherwise, when you see
“acid” on the MCAT, think of the
Bronsted-Lowry definition. In other words,
think of species that donate an acidic
hydrogen. The Arrhenius definition is oversimplified and rarely seen on the MCAT.
o
Acid/Conjugate Acid and Base/Conjugate Base: The “conjugate base” of an acid is the acid minus
its hydrogen (e.g., HCl = acid; Cl- = conjugate base). The “conjugate acid” of a base is the base plus
a hydrogen (e.g., NH3 = base; NH4+ = conjugate acid). Which species you call the acid/base or the
conjugate acid/base is arbitrary (e.g., you could call HCl the acid and Cl- the conjugate base or call Clthe base and HCl the conjugate acid).
o
Amphoteric substances can act as either an acid or a base (e.g., H2O).
pH Scale: The pH scale is a mathematical ranking system for the acidity or basicity of aqueous solutions.
The solution being ranked could be water only, water + acid, water + base, or water + base + acid. Notice,
however, that water is always there. In fact, as you’ll see in the next section, the pH scale ranks solutions
based not so much on the acids or bases themselves, but on how those acids or bases influence the
equilibrium for the ionization of water.
o
pH = -log[H+]
o
pH of pure H2O at 25°C = 7.0, and the [H+] = [OH-]; pH > 7 = basic, and the [H+] < [OH-]; pH < 7 =
acidic, and the [H+] > [OH-]; pH = 7 is defined as “neutral” and the [H+] = [OH-]


AlCl3 and BF3 are two common examples of Lewis
acids. The electrophiles in all organic chemistry
reactions are acting as Lewis Acids. NH3, OH- and
anything else with an electron pair to donate will
act as a Lewis base.
IMPORTANT NOTE
Q16. How does the hydrogen ion concentration differ for two solutions, one with pH = 2 and the
other with pH = 4?
o
pOH = -log[OH-]
o
pH + pOH = 14
Acid-Base Equilibria:
o
All equilibrium constants (Keq, Ka, Kb. Kw or Ksp) are written via the law of mass action, with
pure liquids (l) and solids (s) omitted. You should think of Ka or Kb just as you do Keq. A large Ka
(or a small pKa) indicates that at equilibrium there are far more products than reactants. For an acid
dissociation, this would mean a lot of dissociation (i.e., a lot of H+ formed) and thus a very strong acid.
Similarly, a large Kb (or a small pKb) indicates a very strong base (i.e., a lot of OH- formed—either
from dissociation of a hydroxide base such as NaOH, or from deprotonating water).
o
Ionization of Water: H2O + H2O  H3O+ + OH- (H3O+ is the same as H+)

Kw = [H3O+][OH-] = 10-14 (at 25° C; This caveat should make sense because you learned earlier
that temperature is the one thing that does change Keq.)

pKw = pH + pOH = 14


Q17. Show how the equation above for pKw is derived from the equation for Kw.
pKa + pKb = 14
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o
o
o
Acid Dissociation: HA + H2O  H3O+ + A- (H3O+ is the same as H+)

Ka = [H+][A-]/[HA]

Q18. Acid HX dissociates 80% in water. Would you expect its Ka to be greater than, less than, or
equal to one?
Base Dissociation: A- + H2O  OH- + HA

Kb = [OH-][HA]/[A-]

Ka*Kb = Kw = 10-14 (at 25°C); because ([H+][A-]/[HA])*([OH-][HA]/[A-]) = [H+][OH-] = Kw

Q19. T/F An aqueous solution with a pH of 8 is basic and therefore by definition it does not contain
any unreacted H+ ions.

Q20. The conjugate base of Acid A has a Kb of 1.0 x 10-9 and the conjugate base of Acid B has a
Kb of 1.0 x 10-10. Which acid will create the largest decrease in pH when added in equimolar
amounts to pure water?
Summary and Clarification: Many students do not understand how the three equilibriums described
above (Ionization of Water, Acid Dissociation, and Base Dissociation) relate to one another. The
ionization of water is often called the “autoionization of water” precisely because it happens
automatically in all water. So, when an acid or a base is added to water, the autoionization equilibrium
is already on-going in that solution—a process we can describe with an equilibrium constant, Kw. If we
are at 25°C, we know that Kw will be exactly 10-14 and the concentrations of H+ ions and OH- ions will
both be 10-7M.
After adding an acid or a base, we suddenly have two equilibriums present in the same solution: 1)
the ionization of water, and 2) the equilibrium for the acid or the base we added. Just as we described
the ionization of water with an equilibrium constant, Kw, we can also describe the dissociation of the
acid or base we added with Ka or Kb. The important key many students cannot visualize is this:
When we add an acid or base to water, the equilibrium of that acid or base will directly
impact the equilibrium for the ionization of water according to Le Chatelier’s Principle.
Looking at the formula, H2O + H2O  H3O+ + OH-, we can see that adding an acid will shift the
equilibrium to the left. This will use up hydroxide ions. Each hydroxide ion that reacts will also use up
one hydronium ion, but remember that we just added extra hydronium ions in the form of the acid.
The net result will be more hydronium ions relative to hydroxide ions, and therefore a lower pH
(Remember that the [OH-] equaled the [H+] before we added the acid). Similarly, if we add a base to
neutral water, it will also shift the reaction to the left, but this time we will be using up H3O+ ions. The
net result will be more hydroxide ions relative to hydronium ions and therefore a higher pH.
Notice that the addition of either an acid or a base shifts the equilibrium for the ionization
of water to the left! Of course, we could add both an acid and a base to the same solution of water—
creating three equilibriums in the same solution. In that case, the acid and base equilibriums would
have competing influences on the equilibrium for the ionization of water. If the acid and base were of
equal strength, there would be no net effect and the pH would remain neutral.

The equilibrium for the ionization of water is always present in aqueous solutions.

Adding an acid shifts the reaction to the left, and increases the relative [H+]

Adding a base shifts the reaction to the left, and increases the relative [OH-]

Kw, Ka and Kb are used to describe these three equilibriums.

At 25°C, Kw = Ka*Kb; this should make sense because 1) we demonstrated above that this is
mathematically true, and 2) if we always remain at 25°C the Kw for the ionization of water should
never change—per our rule that temperature is the only thing that changes Keq.
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o
Calculating pH:


For Strong Acids/Bases: (e.g., pH of a 1 x 10-3M HCl solution)

The pH or pOH = –log[strong acid or base]

Q21. What must we assume to make this calculation?
For Weak Acids: (e.g., pH of a 1 x 10-4M CH3COOH solution)
1)
2)
3)
4)
5)
6)

Write out the equilibrium equation (HA  H+ + A-).
Use x to represent the concentration of each of the two products (or 2x, 3x etc. depending on
the coefficients in the balanced equation).
Use “[HA] – x” for the concentration of the original acid.
Solve for x from the resulting equation: Ka = (x)(x)/[HA - x].
If this results in a quadratic equation, assume that x is much smaller than [HA] (in step #3
above) and omit it.
Use –log[H+] to solve for the pH.
Acid/Base Strength & Dissociation:
o
Strong Acids/Strong Bases: All strong acids and strong bases dissociate 100% in water (making
them good electrolytes). There is obviously a continuum of strength, not a hard, fast line, but for the
MCAT we will consider a species “strong” only if it is included in the following lists:

Strong Acids List:

HI, HBr, HCl, HNO3, HClO4, HClO3, H2SO4, H3O+

Note: H3O+ is a borderline strong acid. It does have a pKa < 0, but just barely (pKa = -1.7).
Many texts use it as a line of demarcation: acids stronger than hydronium ion are “strong”
and acids weaker than hydronium ion are “weak.”

Note: HF is NOT a strong acid, but is often mistakenly labeled as such.


Strong Bases List:

o
Q22. Explain why HF is NOT a strong acid and yet the closely related HCl is a strong acid?
Group IA hydroxides (NaOH, KOH, etc.), NH2-, H-, Ca(OH)2, Sr(OH)2, Ba(OH)2, Na2O, CaO.
Weak Acids/Weak Bases: Weak acids and bases do not dissociate readily in solution. As a general
rule, an acid with a pKa greater than zero, or a Ka less than one, can be considered a weak acid.
Similarly, a base with a pKb greater than zero, or a Kb less than one, can be considered a weak base.

Examples of Weak Acids: Anything NOT on the strong acid list.


Examples of Weak Bases: Anything NOT on the strong base list.


H2O, H2S, NH4+, HF, HCN, H2CO3, H3PO4, acetic acid, benzoic acid, etc.
H2O, NH3, R3N, pyridine, Mg(OH)2, etc.
Impact of Salts on the Dissolution of Weak Acids and Weak Bases:

The percent dissociation of benzoic acid (weak acid) decreases in a sodium benzoate solution.

The percent dissociation of ammonium hydroxide (weak base) decreases in an ammonium
chloride solution.

Q23. Provide a potential explanation for the two observations stated above.
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Hydrolysis of Salts:


This terminology can be a bit tricky: The “salt of a weak acid” refers to the conjugate base of
that weak acid combined with a cation to form a salt (see the example below).


Similarly, the “salt of a weak base” refers to the conjugate acid of that weak base combined
with an anion to form a salt (see the example below).


HCO3- = “weak acid”; CO32- = “conjugate base”; Na2CO3 = “salt of a weak acid”
NH3 = “weak base”; NH4+ = “conjugate acid”; NH4NO3 = “salt of a weak base”
When the salts of weak acids or weak bases dissolve in water one of the ions will
undergo hydrolysis to re-form the weak acid or the weak base:

Reforming of a weak acid: 1) Na2CO3  2Na+ + CO32- 2) CO32- + H2O  HCO3- + OH-

Reforming of a weak base: 1) NH4NO3  NH4+ + NO32- 2) NH4+ + H2O  NH3 + H3O+

Q24. Hydrolysis of which of the following salts in solution will increase the pH of the
solution? a) NaNO2, b) NH4Cl, c) NaF, d) NaClO2, e) CH3COONa, f) NaCl.
Sample MCAT Question
3) The conjugate base of a strong acid is usually a weak base. Therefore, it can also be said that the Kb of
the conjugate base of a strong acid compares in what way to the Kb of a strong base?
A)
B)
C)
D)
The Kb of the conjugate will exceed the Kb of the strong base.
The Kb of the conjugate will be approximately equal to the Kb of the strong base.
The Kb of the conjugate will be less than the Kb of the strong base.
The Kb of the conjugate cannot be determined without also knowing the Ka.
Solution: This is an excellent illustration of one of the ways in which the MCAT will require you to think conceptually. There
is no memorized rule that will allow you to quickly answer. You must stay calm, re-read with careful scrutiny and produce an
answer based on your conceptual mastery of each term. The entire stem can basically be rephrased to ask, “How do the Kb
values of a strong base and a weak base (the conjugate of the strong acid) compare?” Because Kb*Ka is equal to a constant, if
one is high, the other must be low, and vice versa. A strong base must have a high Kb and thus has a low Ka. The weaker base
would have a lower Kb, also allowing it to have a relatively higher Ka. Answer C is thus correct.
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
Titration: Drop by drop mixing of an acid and a base with an indicator.
o
Q25. If a question states: “A strong base is titrated with a strong acid,” which one is being added dropwise and which one is in the beaker? Which solution is referred to as the titrant? Which solution is
referred to as the anylate?
o
One “equivalent” = the amount of acid or base
necessary to produce or consume one mole of
[H+] ions.

o
o
Titration Curves: Q27. Draw each of the following
titration curves and label the axes.

Strong Acid (SA) titrated with a Strong Base
(SB)

Strong Base (SB) titrated with a Strong Acid
(SA)

Weak Acid (WA) titrated with a Strong Base
(SB)

Weak Base (WB) titrated with a Strong Acid
(SA)
Pay special attention to where the titration
curve starts. Strong acids titrated with
anything (strong or weak base) should start
low on the y-axis. Strong bases titrated with
anything should start high on the y-axis. The
end of the titration curve should correlate
with the pH of the titrant. For example, if
you’re titrating with a strong base, near the
end of the titration the curve should flatten
out at a point high on the y-axis. If titrating
with a weak base, however, the curve should
end much lower.
Titration of a SA w/ SB or SB w/ SA:

o
Q26. How many equivalents of base can be
neutralized by one equivalent of H2SO4?
IMPORTANT NOTE
Equivalence Point/Stoichiometric Point = midpoint of the nearly vertical section of the graph.
At this point [titrant] = [analyte]. For example, for a solution of NaOH being titrated with HCl,
at the equivalence point [HCl] = [NaOH] in the flask. Put another way, the moles of HCl in the
beaker = the moles of NaOH in the beaker. Because HCl and NaOH are both considered “strong”
(i.e., dissociate 100% in water), they will both produce the same amount of ions per mole. Thus,
for titrations involving a SA and a SB, [H+] = [OH-] at the equivalence point (this is NOT true
if a WA or WB is involved, as we’ll discuss below). By definition, if [H+] and [OH-] are exactly
equal, pH = 7 at the equivalence point.
Titration of a WA w/ SB or WB w/ SA:

Equivalence Point/Stoichiometric Point = midpoint of the nearly vertical section of the
graph. Again, at this point [titrant] = [analyte]. However, because the acid and base are not
both “strong” (i.e., dissociate 100% in water) the [H+] does NOT equal [OH-]. This also tells us
that the pH does NOT equal 7. Notice that this is different than for SA/SB titrations. This is why
it is very important that you clearly master these two unique types of titrations and keep their
behavior clearly separated in your mind. An MCAT question will simply ask: “What is the pH of the
solution at the equivalence point?” or an answer choice will say “The concentration of hydroxide
ions equals the concentration of hydrogen ions at point B.” You will need to remember that the
first thing you must do on such a question is decide which type of titration is being performed.




For WB w/ SA:
For WA w/ SB:
For SA w/ SB:
For WA w/ WB:
pH < 7
pH > 7
pH  7
pH  7 (assuming equal strength; these titrations are rarely attempted)

Half-Equivalence Point = midpoint of the nearly horizontal section of the graph. Here the
pH = pKa. We can also say that [HA] = [A-] at the half-equivalence point. HA will continue to
be deprotonated until at the equivalence point the solution contains 100% A- and 0% HA.
This is not true of a titration involving a strong acid and strong base because they both dissociate
100%. Therefore, by definition, almost immediately after adding HA 100% has been changed to
A-. Therefore, SA/SB titrations do NOT have half-equivalence points.

Q28. For which of the following titrations will the [OH-] = [H+] at the equivalence point? For which
titrations will [titrant] = [analyte] at the equivalence point? a) SA with SB, b) WB with SA, c) WA
with SB, d) WA with WB.
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o
Indicators: Indicators are weak acids that change color as they dissociate from HA into H+ and A-. To
set up a titration, you must know beforehand the approximate pH of your equivalence point; you then
select an indicator that will change color at that approximate pH. The dissociation of the indicator and
the acid/base reaction we are analyzing run simultaneously in the same beaker but are otherwise
unrelated. The amount of indicator is so small compared to titrant and analyte that we can assume it
has no impact on pH.
o
Equivalence Point vs. End Point: There is always some confusion between these terms. To review, the
equivalence point is where [titrant] = [analyte]. The end point is simply the point when the
indicator causes the color change. There is no causal relationship whatsoever between the solution
reaching the equivalence point and the indicator changing color (i.e., the endpoint). The extent to
which they do or do not correlate depends entirely on the indicator chosen.


Q29. Three common indicators used in the lab are methyl orange, litmus, and phenolphthalein.
These indicators change color at a pH of 3.7, 6.5, and 9.3, respectively. Which indicator should be
chosen to identify the equivalence point for the titration of a weak acid with a strong base?
Buffers:
o
A buffer solution contains a weak acid and weak base, often the conjugates of each other. In a buffer
there is an equilibrium between a weak acid and its conjugate base, or between a weak base and its
conjugate acid.
o
The nearly horizontal area surrounding the half-equivalence point is called the “buffer
region.” Adding a relatively large amount of titrant at this point in the titration will have little effect
on pH. In fact, that is the very reason the graph is flat in this region—the volume of titrant added
increases (x-axis) with little or no increase in pH (y-axis).
o
pH = pKa + log[A-]/[HA] (Henderson-Hasselbalch equation) or pH = pKa – log[HA]/[A-]

o
Q30. Use the Henderson-Hasselbalch equation to demonstrate that pH = pKa at the halfequivalence point.
Recognizing Buffer Problems: Buffer problems have a way of “hiding” on the MCAT. Usually
students simply fail to recognize that what was described to them in the passage or question stem was
a buffer. Here are some clues that you are dealing with a buffer:
1)
Watch for equimolar amounts. The maximum buffering strength occurs when the [HA] is equal
to the [A-]. This is the ratio one would start with if making a buffer in the lab. Therefore, be on
the lookout for equimolar amounts of a weak acid and its conjugate base, or a weak base and its
conjugate acid.
2)
Watch for conjugates. To be a buffer, the two equimolar substances are often conjugates of
each other, such as: NH3 and NH4, CH3COO- and CH3COOH, or HCO3- and CO32-.
3)
Watch for WEAK acids or bases. The equimolar pair must be a weak base or a weak acid and
its conjugate. Strong acid or strong base conjugate pairs do NOT form buffers.

Q31. What would be the result if you tried to make a buffer out of a strong acid and a strong
base?
4)
Watch for resistance to pH change. Any time an acid or base is added to a solution and the pH
does not change “very much,” or “changes slightly,” this should be a dead giveaway that the
solution is a buffer.
5)
Watch for the half-equivalence point. Remember that only solutions of weak conjugate
acid/base pairs have a buffer region, and therefore they are the only solutions that have a halfequivalence point.
6)
Watch for pH = pKa. Many students memorize this principle, but fail to recognize that what it is
really saying is that pH = pKa at the midpoint of the buffer region. This is another unmistakable
clue that you are dealing with a buffer.
7)
Watch for the ratio of [HA]/[A-] or [A-]/[HA]. One particular problem on the CBT practice
exams seems to stump the majority of students. The question stem asks about the ratio of an acid
to its conjugate base. Those who recognize it as a buffer problem usually answer it correctly. Those
who do not guess.
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Sample MCAT Question
4) A student mixes 2.5 mL of 1.0 M C6H5OH with 5.0 mL of 0.5 M C6H5O-. The pH is measured and
found to be 9.9. Based on this information, phenol is best described as a:
A)
B)
C)
D)
weak base with a pKa of approximately 10.
strong base with a pKa of approximately 7.
weak acid with a pKa of approximately 10.
the conjugate base of a weak acid, whose pKa cannot be determined from the information given.
Solution: The ratios of volume and molarity given result in adding equimolar amounts of phenol (C6H5OH) and its conjugate
base. This is a dead giveaway that this is a buffer problem. By adding exactly equimolar amounts the student has recreated the
half-equivalence point where [HA] = [A-]. Therefore, pH = pKa, which in this case means the pH is known to be exactly 9.9.
Phenol is a weak acid, so combining these two known facts makes answer C correct.
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