REVIEW FOR ALL MATERIAL AFTER EXAM 3
I. Titration Curves:
1.These curves plot the pH change during an acid-base neutralization.
2.Titration is one of the most important ways of accurately finding the quantity of an acid, a base or
some other substance in a mixture or of determining the purity of a substance.
3.Acid-base behavior can be seen with titration curves. Types of reactions we will deal with:
a. Rxn of strong acid w/ strong base
b. Rxn of weak acid w/ strong base
4.Things to remember:
a. Six Strong Acids (all other acids are weak acids):
♦ HCl
♦ HI
♦ HBr
♦ HNO3
♦ H2SO4
♦ HClO4
b. Strong Bases:
♦ Soluble ionic compounds that directly release OH- such as NaOH or Ba(OH)2
♦ Ionic compounds that have either H- or NH2- such as CaH2 or NaNH2
♦ Organics with a negatively charged carbon
c. When dealing with titration problems, treat each situation as “BRAND-NEW”
5.Strong acid/Strong base:
a. Titrate 15mL of 1.0M HBr w/ 0.400 M NaOH. Find pH after the addition of:
♦ No base
♦ 20mL of base
♦ 37.5mL of base
♦ 42mL of base
HBr +
NaOH → NaBr
+
H2O
This titration curve is the same
for all strong acid/strong base
titrations.
♦ No Base—Point A:
Since no base is added yet, [HX]i = [H3O+]f
Therefore, for this problem:
1.00 M HBr = 1.00M H3O+
pH = 0
So, pH = -log(1.00)
HX=strong acid
♦ 20mL of Base—Region B
In this region, the pH increases very slowly—pH is a log scale so to go up by one pH unit,
90% of the remaining acid must be neutralized.
− So if you go from pH of 1 to a pH of 2, you used up 90% of the acid
− If you go from pH of 1 to pH of 3, you used up 99% of the acid, etc.
Also, in this region, moles base added < initial moles of acid. You can recognize
you are in region B b/c these conditions apply. Then:
− Since moles base added < initial moles of acid, all of the strong acid is not
consumed.
− So what is left in the solution?—excess acid and NaBr(this has no effect on pH)
− So, have to do limiting reactant problem to find moles of acid left over
which is equal to the moles of H3O+ left over.
(mol HX left over = moles H3O+ left over)
− Then, divide these moles of H3O+ left over by total volume in liters to get
[H3O+] →pH
♦ 37.5mL Base Added—Point C:
Here, initial moles acid = moles base added—equivalence point.
− Equivalence point—the point where one reactant has been completely consumed
by the addition of another reactant—happens in acid-base titrations ONLY when
strong acid titrated w/ strong base.
− What is in solution here?—only have NaBr and H2O; no strong acid or strong
base left over.
− Therefore, pH = 7—this is only true if and only if rxn is between a strong acid
and a strong base.
♦ 42mL Base Added—Region D:
Here, moles base added > initial moles acid.
− Do a limiting reactant problem to find excess moles of OH− Divide these moles by the total volume in liters
− [OH-] →pOH→pH
6.Titration of weak acid with a strong base:
a. Titrate 15mL of 1.2M HF (Ka = 7.2 × 10-4) with a 2.0M NaOH solution. Find pH:
♦ When no base added
♦ 4.5mL of base added
♦ 6.25mL of base added
♦ 9mL of base added
♦ 13mL of base added
♦ General equation of weak acid w/ strong base:
HA + OH- → H2O + A-
♦ Equation for this problem:
HF + NaOH →
NaF
+
H2O
♦ No Base Added—Point A:
− What is in Solution?—only weak acid (HA), in this case HF
− Since [HA]i ≠ [H3O+], need to do a Ka ICE table equilibrium
calculation to find [H3O+].
♦ 4.5mL of Base Added—Region A':
− Has some OH− OH- reacts immediately to form some H2O and A− HA + H2O ↔ H3O+ + AStress: [HA]↓, [A-]↑
Equilibrium shifts left, suppresses formation of [H3O+]
pH takes an initial “jump”
♦ 6.25mL Base Added—Region B:
− More OH- added causes more A- formation
− Initial mol HA > mol OH- added
− What is in solution here?—some A- and some unreacted HA
− This is a BUFFER! b/c weak acid + conj. base—use H-H eqn. to get pH
− Do a limiting reactant problem to get mol A- and mol HA
♦ 9mL Base Added—Point C:
− Initial mol HA = mol OH- added
− Any HA left?—No
− Any OH- left?—No
− What is in solution here?—NaA—really A-, since it is conj. base of a weak acid,
you know you have a pH>7 (use this as a check to make sure you’re doing the
problem correct)
− Need to do a Kb ICE table calculation
− Before this though, need:
ƒ Value of Kb
ƒ Initial [A-] for your “I” line in ICE table
[A-] = mol A- formed = mol HA initial
total volume (L) total volume (L)
− Kb →[OH-] →pOH →pH
♦ 13mL Base Added—Region D:
Here, moles base added > initial moles acid.
− Do a limiting reactant problem to find excess moles of OH− Divide these moles by the total volume in liters
− [OH-] →pOH→pH
REMEMBER!!—Treat every titration problem as a brand new problem!
II. Solubility:
1.Why care?
a. Environmental aspects—can recover valuable metals; remove toxic heavy metals from waste
streams
b. Medical—imaging; teeth/bones
c. Structural—cement/limestone
2.Solubility Rules:
a. Compounds with alkali metal cations (first column) or NH4+ are SOLUBLE
b. Compounds with NO3-, ClO4-, ClO3-, CH3COO- are SOLUBLE
♦ These two rules are always true
♦ These two rules apply regardless of the counterion
c. Compound containing halide anions (2nd to last column—F-, Br-, Cl-, I-) are SOLUBLE
♦ Except when F- with Pb2+ or an alkaline earth cation (2nd column)—then INSOLUBLE
♦ Except when Cl-, Br-, or I- with Ag+ or Pb2+, then INSOLUBLE
d. Compounds containing sulfate (SO42-) are SOLUBLE
♦ Except when with Ca2+, Sr2+, Ba2+, Pb2+, or Hg2+, then INSOLUBLE
♦ Example: BaSO4 is safe for GI (gastrointestinal) imaging regardless of the fact that barium
is very toxic. A patient drinks a “cocktail” containing BaSO4 and the process of the
compound through the digestive organs can be followed by x-ray analysis since the
compound is opaque to x-rays. The Ba2+ makes the SO42- insoluble and so it just passes
through the system. However, if Ba2+ was paired with anything from RULE b, then it would
be very toxic b/c it would be soluble.
e. Compounds containing CO32-, PO43-, S2-, C2O42- (oxalate) are INSOLUBLE
♦ Except if RULE a applies—then SOLUBLE
f. Compounds containing hydroxide are INSOLUBLE
♦ Except if RULE a applies OR if Ba2+ present—then SOLUBLE
3.Know how to tell if a compound will precipitate or not using theses rules, b/c this will most likely
be on the exam (don’t need to memorize)
4.Example: Toxico Chemical Company has a waste stream with Sr2+, Ag+, Fe2+ and K+. Add anions
in a particular order to cause 1 metal cation to precipitate from solution at a time.
a. Adding carbonate (CO3-) would cause the iron, strontium and silver to precipitate at the same
time, so that is a bad idea.
b. First, add Cl- so the Ag+ only will precipitate
c. Then add SO42- so the Sr2+ only will precipitate
d. Then add CO3- so the iron will precipitate. Will be left with K+ which is okay to have in water.
5.“Soluble/insoluble” is an over-simplification b/c even “insoluble” ionic compounds dissolve to a
small extent.
a. Example: If Hydroxyapatite (tooth enamel and bone—“insoluble”) is put in some water, a tiny
amount of this compound will dissolve and an equilibrium will be established:
Ca5(PO4)3OH(s) ↔ 5Ca2+(aq) + 3PO43-(aq) + OH-(aq)
b. If you eat drink anything sugary and don’t brush your teeth before you go to bed:
♦ Unbrushed teeth + sugar → lactic acid—this increases [H3O+]
♦ [H3O+] reacts w/ small amount of OH- from solubility equilibrium
♦ Stress on equilibrium: [OH-]↓
Response: system shifts R to raise [OH-]—therefore, more hydroxyapatite dissolves and this
leads to tooth decay.
c. Certain metabolic disorders prevent absorption of Ca2+ from food
♦ Then, body will use tiny amount of Ca2+ from hydroxyapatite equation.
♦ Stress: [Ca2+] ↓
♦ Response: Equilibrium shifts R to make more Ca2+→does this by dissolving bones—leads
to osteoporosis
6.Ksp—Solubility Product Constant: the value of the equilibrium constant that reflects the solubility
of the compound.
a. sp = solubility product
b. Ksp for Hydroxyapatite equilibrium expression = [Ca2+]5[PO43-]3[OH-] ≈ 10-36
c. F- can be substituted for OH- in hydroxyapatite:
♦ Ca5(PO4)3F(s) ↔ 5Ca2+(aq) + 3PO43-(aq) + F-(aq)
♦ Ksp = [Ca2+]5[PO43-]3[F-] ≈ 10-60
This is 1024 less soluble than normal
d. Using Ksp, you can calculate
♦ Molar solubility—s (mol/L)
♦ Solubility—g/L
♦ Also, can predict if a precipitation rxn. will occur by comparing Ksp to Qsp
7.Find the molar solubility and the solubility of BaSO4 in pure water. Ksp=1.1×10-10
8.To predict whether or not a precipitation will occur, compare Ksp with Qsp(uses initial
concentrations)
MX(s) ↔ M+(aq) + X-(aq)
a. If Ksp = Qsp, then the solution is saturated and at equilibrium
b. If Ksp > Qsp, then the solution is not saturated—no precipitate will form—can still make more
soluble species.
c. If Ksp > Qsp, then the solution is supersaturated—rxn. goes to L, solid formation occurs and
precipitation occurs.
9.Know that CaCO3 (limestone) used in buildings is said to be “insoluble” also, but still slightly
soluble.
10.If 50mL of a 3.6×10-4M AgI solution is mixed with 75mL of a 1.25×10-3M NaCl solution, will
AgCl precipitate?
11.Common Ion Effect:
a. Look at effect on solubility if one of the component ions is already present in solution.
b. AgCH3CO2(s) ↔ Ag+(aq) + CH3CO2-(aq)
♦ If [Ag+] ↑, rxn goes L to compensate and use some of the Ag+ making more solid
(precipitating) and therefore, becoming less soluble
c. Solids are less soluble in the presence of a common ion.
12.Example: Zn(OH)2: Ksp= 1.8×10-14; solubility (s) = 1.6×10-5 mol/L in pure H2O
Find solubility of Zn(OH)2 in a pH = 12 solution
pOH=2
[OH-] = 10-2 = 0.01
(use this in ICE table)
Thermodynamics
I. Entropy:
1.Enthalpy = ∆H
a. Both of these types of reactions can occur:
♦ Exothermic—heat out of the system—∆H<0
♦ Endothermic—heat into system—∆H>0
2.Entropy = ∆S: measure of inherent disorder or randomness
a. If ∆S>0, that means that there is increased entropy—increased disorder
b. If ∆S<0, means entropy is negative—increased order
c. Usually, can predict sign of ∆S by inspection. Ex:
♦
NaCl(s) → Na+(aq) + Cl-(aq) ∆S>0
more ordered
more disordered
♦
d.
e.
f.
g.
h.
2H2(g) + O2(g) →
2H2O(g)
∆S<0
3 moles of gas
2 moles of gas
Both of the examples above can happen, therefore, you can’t predict whether or not a reaction
will occur just based on entropy.
When comparing similar substances, the entropies of gases are much higher than those for
liquids, and entropies of liquids are large than those for solids.
Larger molecules have a larger entropy than smaller molecules
Molecules with more complex structure have larger entropies than simpler molecules.
∆So = ∑So (products) - ∑So (reactants)
So = entropy at standard conditions
o
= naut—standard conditions—25oC, 1atm, 1.0M
3. Second Law of Thermodynamics: The entropy of the universe must always increase
a. ∆Suniverse = ∆Ssystem + ∆Ssurroundings
b. For a spontaneous process, or a process that will occur, ∆Suniverse>0
c. It is possible for a process to occur that causes a more ordered system (like hydrogen gas and
oxygen gas combining to form water vapor), but the entropy of the surroundings must be
increased to compensate (for water vapor to form, heat must be given off). Ex: Penny Demo.
♦ 10 pennies in a stack
♦ Throw them up in the air—going from ordered to disordered, so this is spontaneous.
♦ However, to put them back into a stack, the system is becoming more ordered—therefore,
surroundings have to become more disordered
♦ Students have to expend energy looking for pennies and putting them back into stack—
surroundings becoming less disordered
4.Compute ∆Sorxn for this reaction:
2NO(g) + O2(g) → 2NO2(g)
210.8J/K. mol 205.1 J/K. mol 240.0 J/K. mol
Predict ∆Sorxn < 0 because becoming more ordered
∆Sorxn = [(2 mol NO2)( 240.0 J/K. mol)] – [(2 mol NO)( 210.8J/K. mol) + (1mol O2)( 205.1 J/K. mol)]
∆Sorxn = -146.7 J/K (see if it matches with what you predicted)
5.Gibbs Free Energy:
a. ∆G = ∆Ho – T∆So
♦ ∆G = Gibbs Free Energy—measure of spontaneity
♦ The term “free energy” represents the maximum energy available to do useful work.
Therefore, in this context, the word “free” means “available.”
♦ If ∆Grxn < 0, the reaction is spontaneous
♦ If ∆Grxn = 0, the reaction is at equilibrium with K = 1
♦ If ∆Grxn > 0, the reaction is not spontaneous
b. If reaction is spontaneous, ∆Grxn < 0
♦ ∆Suniverse > 0 always
♦ If ∆Ssystem ↑, therefore, ∆Ssystem > 0 (this is good for spontaneity), then, -T ∆S < 0 which
helps ∆G < 0.
♦ Also, as ∆Ssurroundings ↑ by sending heat out of system into the surroundings, it is an
exothermic process—∆H < 0 which causes ∆G < 0
c. If at standard conditions, ∆G = ∆Go → easy to compute if you know ∆Gof values of all
components of reaction—just uses same method as ∆Ho or ∆So calculations.
∆Go = ∑ ∆Gof(products) − ∑ ∆Gof(reactants)
d. What if Temp. ≠ 25oC?
Use ∆G = ∆Ho - T ∆So
♦ ∆G = kJ/mol
♦ ∆Ho = kJ/mol
♦ T=K
Pay attention to units!
♦ ∆So = J/K. mol → must convert to kJ/K. mol
e. If ∆G = 0, then reaction is at equilibrium—this means process “crosses-over” from
spontaneous to non-spontaneous and vice versa.
♦ Finding cross-over temp.
∆G = ∆Ho - T ∆So
0 = ∆Ho - Tcross ∆So
Tcross = ∆Ho
← kJ
o
∆S
← J/K →MUST CONVERT to kJ/K
♦ What if ∆Ho < 0 (exothermic) and ∆So > 0 (more disorder)?—then you would get a negative
temp.—negative Kelvins is not possible!
− Means there is no cross-over T
− If you get –K, means the reaction is either always or never spontaneous
− Here, ∆G = ∆Ho - T ∆So
(−)
(+)
− So, ∆G must be negative—therefore, always spontaneous.
♦ What if ∆Ho > 0 (endothermic) and ∆So < 0 (less disorder)?— Tcross would be negative K—
IMPOSSIBLE! So no Tcross
− Here, ∆G = ∆Ho - T ∆So
(+)
(−)
− So, ∆G must be positive. The reaction is always non-spontaneous.
− However, the opposite reaction would be spontaneous.
f. What if ∆G not at 1atm P or 1.0 M solution?
♦ ∆G = ∆Go + RTlnQ
− ∆G = kJ/mol
− ∆Go = kJ/mol
− R = 8.314 J/K×mol = 0.008314 kJ/K×mol
−T = K
− Q = Reaction quotient—written just like K but uses initial molarities
♦ If ∆G = 0—then at equilibrium
Then Q = K
Therefore, 0 = ∆Go + RTlnK
∆Go = -RTlnK
6.Reduction/Oxidation (redox) reactions:
a. Acid-base reactions were H+ (proton) transfer.
b. Redox reactions are e- (electron) transfer
c. Why are redox reactions so important?
♦ ATP ↔ ADP
♦ Photosynthesis
♦ Batteries
♦ Fuel Cells
d. Reduction—gain in e-, therefore, REDUCTION in positive charge.
e. Oxidation—loss of e♦ Lion “LEO” goes “GER”
LEO = Loss electrons oxidation
GER = Gain electrons reduction
f. Rusting: Fe → Fe2O3
♦ Fe0 → Fe3+ + 3eg. Balancing redox reactions (simple) → balance atoms and charge
Simple: Cu + Ag+(aq) → Ag(s) + Cu2+(aq)
Cu + 2Ag+(aq) → 2Ag(s) + Cu2+(aq)
♦ Ag+ is being reduced—called oxidizing agent
♦ Cu is being oxidized—called reducing agent because it causes reduction while being
oxidized
♦ Reduction and oxidation must occur together.
h. Balancing redox reactions (harder)—use blue sheet (know for exam)
♦ “Pretend” that all compounds are ionic
− Not really true, only do this for bookkeeping
♦ Assign oxidation numbers to all compounds to find reduction and oxidation
− What is oxidation number of O in CH3CH2OH? −2
− What is oxidation number of Cr in Cr2O72-? 6
Because:
i. Balance the following redox equation that occurs in basic solution.
Fe(OH)3(s) + Cr(aq) → Fe(OH)2(s) + Cr(OH)3(aq)
♦ Step 1: Assign Oxidation Numbers
♦ Step 2: Separate the reaction into an oxidation half-reaction and a reduction half-reaction
♦ Step 3: Balance all non-oxygen and non-hydrogen atoms first
♦ Step 4: Add ONLY H2O to balance oxygen atoms. Then add ONLY H+ to balance
hydrogen atoms.
♦ Step 5: Add enough electrons to balance the charge in each half-reaction
♦ Step 6: Multiply each half-reaction by a whole number to equalize the number of electrons
♦ Step 7: Add the two half-reactions together. Cancel items that appear on both sides of the
final reaction either partially or fully. ALL ELECTRONS MUST CANCEL. Check to see if
charges and atoms balance. (If Acidic solution, STOP HERE. If Basic, go on to step 8)
♦ Step 8: Add enough OH- to both sides to equal the quantity of H+. Combine the H+ ions and
the OH- ions to form H2O (acid-base neutralization). Now cancel items that occur on both
sides either partially or fully. Check to see if charges and atoms balance.
7. If the reduction and oxidation half-reactions are physically separated, the e- transfer can be made
to conduct electricity and power a device.
a. First Battery made by Galvani—called solution cells or galvanic cells
b. Both reduction and oxidation happen in a voltaic cell:
♦ Red: Cu2+ + 2e- → Cu
♦ Ox: Zn → Zn2+ + 2ec. In all electrochemical cells, oxidation occurs at the anode
d. Reduction occurs at the cathode.—“RED CAT”
e. The salt bridge allows cations and anions to move between the two half-cells. If there was no
salt bridge, the battery would short circuit. The salt bridge causes the Cl- to go towards the
anode and the K+ to go toward the cathode (since anode is losing e-, it attracts the Cl-)
f. The negative electrons and negatively charged anions make a “circle.” The electrons move
from anode to cathode through the wire and the negative anions move from the cathode to the
anode through the salt bridge.
g. Now, we use alkaline batteries—mostly just solid with some liquid for moisture.
8.For Cu/Zn cell, Eo = 1.10 V
a. Eo = voltage difference between separated half reactions
b. If Eo > 0 then reaction is spontaneous and product favored
c. If Eo < 0 then reaction is reactant favored
d. Using a table of standard reduction potential, you can predict Eo for a redox reaction.
♦ Usually give Eo only of reduction half-reaction
o
♦ How to get Eo for oxidation?—flip reaction to make it oxidation and flip sign of E
− You can flip sign of Eo but you CANNOT multiply it with coefficients
e. Figure out Eo for this redox reaction:
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag
♦ Step 1: split into half reactions:
red:
2e- + 2Ag+(aq) → 2Ag
ox:
Zn(s) → Zn2+(aq) + 2e-
♦ Step 2: find Eo and do necessary calculations:
Eo = 0.7999
red:
2e- + 2Ag+(aq) → 2Ag
2+
ox:
Zn(s) → Zn (aq) + 2e
Eo = 0.736
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag
Eo = 1.5359
So this reaction is spontaneous and product favored
f. For a reaction to be spontaneous, Eo > 0 and ∆Go < 0. Therefore, they are proportionally
related by the following expression:
♦ If you put in a constant, then you get the following equation:
∆Go = −nF Eo
− ∆Go = J → convert to kJ; Also, using ∆Go, you can find K
− n = moles of electrons transferred—can find this from a redox equation before you
cancel the electrons.
− F = Faraday’s Constant: 96500 J/(volt×mol e-)
− Eo = volts
♦ Example: The standard cell potential, Eocell, for the reduction of silver ions with copper
metal is +0.46 at 25oC. Calculate the ∆Go for this reaction.
In this cell, copper is the anode and silver is the cathode
Overall cell reaction: Cu(s) + 2Ag+(aq) → Cu2+ (s) + 2Ag(s)
If you break this into two half reaction, you get:
red:
2e- + 2Ag+ → 2Ag
ox:
Cu → Cu2+ + 2eFrom this, we can see that two moles of electrons are transferred. So, n = 2
∆Go = −nF Eo
∆Go = −(2 mol e-) (96500 J/(V×mol e-)) (0.462 V)
∆Go = -89,200 J OR -89.2 kJ
9.Nuclear Chemistry:
a. Nuclear chemistry is about the transformation of the atomic nuclei.
b. We care about this because this type of chemistry is used to make energy, weapons and in
medicine.
c.
♦ A = atomic mass = #protons + #neutrons
♦ Z = atomic number = number of protons in element, defines the element—if you change the
number of protons, then you have a different element.
♦ X = element
d. Types of Nuclear Reactions:
♦ α emission : alpha particle is a He (helium) nucleus ejected from a larger nucleus.
− Example: must balance both A and Z
♦ β emission: gives off a beta particle—an electron ejected from the nucleus
− How is this possible? There are no electrons in the nucleus!—actually, neutrons are
just pairs of protons and electrons, so one of these electrons is separated from the
proton it is paired with and emitted from the nucleus.
− Example:
♦ With α emissions, Z ↓ by 2.
With β emission, Z ↑ by 1.
− Alpha particles are 8000 times heavier than beta particles. Therefore, they move much
more slowly and are less harmful. They can be stopped by a piece of paper. However,
they are very harmful if ingested.
ƒ For example, when you smoke, you ingest polonium. The polonium goes through the
cell walls of the alveoli and disturbs the nucleus and the DNA in the cell which leads
to lung cancer.
− Beta particles are lighter and faster. They can be stopped with a 1 inch thick sheet of
lead.
ƒ Example: Technetium is given to patients getting a renal scan. It can be a little
radioactive when inside the body.
♦ If Thorium-231 decays to Protactinium-231, what particle is emitted?
− Since the atomic mass stays the same, we know it is not an alpha particle b/c alpha
particles cause the atomic mass to decrease by 4. Also, since the element is changing,
two know the number of protons is changing, so it cannot be a gamma particle, because
it does not cause a change in number of protons. Therefore, it has to be a beta particle.
♦ Nuclear decay causes an unstable configuration in the new nucleus—new nucleus “snaps”
into a more stable configuration and gamma rays are released.
− Gamma Rays (γ) are very high energy photons that are extremely toxic.
− Gamma knife: Can obliterate a tumor with a gamma knife. Take 2 gamma rays and
place the tumor at the point at which they intersect.
− Example:
♦ Fission: larger nucleus fragments. Example:
− Can get chain reactions b/c you get more neutrons out than you put in, and these
neutrons then react with more uranium.
ƒ Controlled Chain Reactions—can harness energy
ƒ Uncontrolled—atomic bomb
♦ Fusion: smaller nuclei forming larger nuclei—SUN:
− In this reaction, products have 0.7% less mass than reactants—but how is this possible
b/c it is going against law of conservation of mass!
ƒ Einstein made great theory: E = mc2
This tells us that mass and energy are the same thing. A tiny mass loss gives off huge
amounts of energy.