chem 138 textbook notes

Chemistry 138 Textbook Notes

Study Plan

Chapter 152/25: The White & Grey Pages2/26: The Suggested Problems2/27: C.15 workshop2/28-3/1: CP 15

Chapter 163/1: The White Pages3/4: The Grey Pages3/5-6: The Suggested Problems3/7: C.16 Workshop3/8-3/11: CP 16

Chapter 173/11: The White Pages3/13: The Grey Pages3/14-15: The Suggested Problems3/16: Workshop3/26-4/1: CP 17

3/16-4/1: Study for Exam 2

Chapter 183/27: The White Pages3/29: The Grey Pages3/30-31: The Suggested Problems4/2: Exam #24/4: Workshop4/9-15: CP 18

Chapter 194/10: The White Pages4/12: The Grey Pages4/13-15: The Suggested Problems4/16: Workshop4/18-22: CP 19

4/14-4/22: Study for Exam 3

Chapter 20-214/19: The White Pages (20)4/20: The Grey Pages (20)4/23: Exam #34/24: The White Pages (21)4/26: The Grey Pages (21)4/28: Suggested Problems: 20-21

Final Exam: 5/2 @ 12:30

Unit 2: Chapters 15-17

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2/25

Chapter 15: Making Accommodations—Solubility and Molecular Recognition

I.) Thermodynamics: Prologue to Reaction

A.) The Rules of Thermodynamics

1.) Energy is conserved: there is only redistribution of energy

2.) Time moves on; Matter moves: Energy is always being more broadly distributed

a.) Global entropy constantly increases

b.) Entropy is how we can tell time has passed

c.) Spontaneity: the universe is more disordered with products

i.) ∆G = GP – GR = ∆H – T∆S < 0

ii.) Spontaneous reactions balance the free energy of the system

B.) Thermodynamics in the Molecular World

1.) Electrical Forces causes exchange in energy, motion, etc.

a.) Thermodynamics dictates what can and cannot happen

2.) How do microscopic attributes determine ∆Gº? We look at basic reactions:

a.) Transfer of the electron

b.) movement of a hydrogen ion

c.) coordination of electrons in noncovalent interactions

C.) The Chemistry of Noncovalent Interactions – “The Chemistry of Accommodation”

1.) One species, the “guest,” is taken up within another, the “host”

a.) The guest is often released later to seek new partners

2.) These relationships are less intimate than a covalent bond: they come and go

3.) Questions to pose:

a.) What aspects of microscopic structure enable this interaction?

b.) Where are/aren’t the electrons?

c.) Where and how is enthalpy changed?

d.) Does local ∆S increase or decrease?

e.) What direction does G flow?

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f.) Where, in the end, will equilibrium lie?

II.) Solutes and Solutions

Example: Sugar water. H2O, with non-covalent bonds to offer, break up the highly polar glucose mole-cule. it is a “thermodynamic battle,” and water is the victor, little clusters of water (host) carry away the sugar (guest) – this is solvation. If something is dissolved in water, it is called hydration. Eventually, it becomes a homogeneous mixture as it reaches equilibrium, at constant temperature. After a certain amount, the solvent cannot hold any more solute, and it is left as a solid at the bottom. The solution is saturated. Heterogeneous equilibrium takes place between the hydrated solute and solid solute.

A.) Order, Disorder, and Equilibrium

1.) Saturation is equilibrium between order and disorder (transition goes both ways)

a.) Disorder: solvated particles

b.) Ordered: Undissolved, crystalized

2.) Similarity: Liquid-solid equilibrium like isothermal compression of a gasa.) Like abrupt condensation, the precipitation is abrupt in solution

b.) Dynamic equilibrium just like before

c.) Concentration holds steady in the presence of undissolved solid (like vaporpressure above pure liquid)

3.) The unwavering concentration is an equilibrium constant

a.) A(s) A(aq)⇌

b.) K´ = [A(aq)] / [A(s)]

i.) We can disregard [A(s)] because it is constant

ii.) K = K´[A(aq)] = [A(aq)]

iii.) The concentration of a saturated solution is the equilibrium constantfor the solid-solution transition

c.) If [A] < K, then the reaction quotient Q is sent below equilibrium

i.) More solute can be added now, until [A(aq)] = K

d.) If [A(aq)] > K, (partial evap of solvent, e.g.) then A(s) ←A(aq) happens untilQ = K. In all of this, K is invariant.

4.) Rules from point 3 do not apply to gaseous solutes

a.) A(aq) A(g)⇌

b.) We can then rewrite K = PA / [A(aq)] into PA = K[A(aq)] <- Henry’s Law

c.) Henry’s Law: The dissolved concentration of gas is proportional to pressureof gas over the solution. (Higher pressure means more gas dissolved into liquid)

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B.) The Solubility Product

1.) The solubility-product constant: Ksp = [Am+]n [Bn-]m

a.) Derived from AnBm(s) nA⇌ m+(aq) + mBn-(aq)

2.) Examples of Solubility products at 25 ºC

a.) AgCl(s) Ag⇌ +(aq) + Cl-(aq)

i.) Ksp = [Ag+][Cl-] = 1.8E-10

3.) Solubility product ≠ Solubility

a.) Solubility is amount of material that will dissolve to saturate something

b.) Solubility product speaks to the ability of hydrated ions to coexist with undissolved solid

4.) Contest for lower free energy (↑∆S) is sometimes won by solid, sometimes solution

5.) Ag+ and Cl- work under altered conditions (e.g., AgNO3 and NaCl)

6.) The common-ion effect

a.) If you add 1.3E-5 AgCl (max solubility) to 1.0L solution 1.0 M AgNO3, the AgCl will not dissolve. Too many Ag ions are already present.

b.) Follows Le Châtelier’s principle: System has relieved itself of stress (too much product)

C.) Selective Precipitation

When 2 reactions are possible with 2 different Ksp, (e.g., AgNO3 to solution of NaCl and NaI, the compound with the lower Ksp will precipitate first.

D.) Space, Time, and Equilibrium

1.) Equilibrium within the solution itself occurs over time

2.) Solute and solvent species continually come together and exchange energyuntil the free energy is uniform throughout the solution

3.) If stress is placed on the system, the equilibrium is destroyed

a.) To restore equilibrium matter most move from high potential to low

4.) Osmosis

a.) dependent only on #solute particles (colligative effect)

b.) Solvent flows toward high concentration of solute, responding to number

c.) Establishes a more comprehensive equilibrium

d.) Internal pressure increases as the incoming water dilutes solution:

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∏ = (n/v)RT

E.) Enthalpy and Entropy

1.) From solid to solution

a.) Take apart the solid (increase the entropy by disorder, put in the lattice energy that the solid originally released)

b.) Prepare the host solvent – also have to pull apart interactions, so entropyand enthalpy increase again

c.) Particles come together, releasing energy and lowering enthalpy: ∆H3 isusually negative; ∆S3 usually increases again, because this new arrangementis still a higher disorder

d.) ∆Hsoln = ∆H1 + ∆H2 + ∆H3

∆Ssoln = ∆S1 + ∆S2 + ∆S3

∆Gsoln = ∆Hsoln – T∆Ssoln

∆Gºsoln = -RT lnK

e.) ∆G can be positive or negative depending on ∆Hsoln

F.) Structure and Solubility

1.) Will it dissolve?

2.) “Like dissolves like:” Polar (water) can dissolve ionic/polar best (usually)

3.) Differences: lattice energy, electronegativity, solvation enthalpy/entropy

a.) Size is important: Smaller attracts water better

b.) Charge matters: higher charge, more attraction between ion and water

G.) Hydrophilic and Hydrophobic Effects

1.) Hydrophilic: water-loving (polar or ionic)

2.) Hydrophobic: water-fearing (nonpolar)

3.) Amphipathic: part one, part other: Forms a protective hydrophilic shell around phobe

a.) Micelle-sac that does that

b.) mono-layer, lipid bi-layer, etc.

III.) Guests and Hosts

The bonds we have learned about are critical to life: hydrogen, dipole-dipole, London, etc. But selective bonding is just as important. The proteins and enzymes in our bodies respond to specific biological struc-tures. These are hosts that can accommodate only one or two guests (substrates). This is how biology works. Ordinary solvents like water are much broader. The goal of solutions is to minimize local enthalpy and maximize local entropy. The solvent will construct a host specific for its guest – efficiency.

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Enzymes are different. They are preorganized, no assembly required. They are already designed for their guests. An enzyme is a lock waiting for its key; it recognizes other species.

A.) Molecular Recognition: One microscopic identity identifies and selects another, to theexclusion of all else.

1.) Molecules Recognize Molecules by Polarity

a.) Attractions and repulsions fall into place, creating a niche that only a certainguest can fit into

2.) Molecules Recognize Molecules by Chirality

a.) Optical isomers

b.) An enzyme is designed for one isomer, but not the other (D, but not L)

3.) Molecules Recognize Molecules by Size

a.) Synthetic hosts use this technique

b.) Example: Crown ether – (OCH2CH2)6

i.) Perfect fit for the potassium ion (K+)

ii.) Used for transporting drugs or other materials in & out of cells

4.) Molecules Recognize Molecules by Shape

a.) Substrates can only fit into enzymes with openings designed for them

b.) Isomerism comes into play again: e.g., ortho/meta/para xylene

c.) Shape-selected hosts can be created to isolate specific isomers

B.) Thermodynamics of Binding

1.) A + B AB; K = [AB] / [A][B]; A is the host, B is the guest, AB the host–guest complex⇌

2.) A good guest offers a large K—it favors products

a.) A preferred guest offers a K much larger than rivals

i.) If B1 and B2 vie for same host, then the winner is determined by K1 and K2

ii.) If [B1] = [B2], then [AB1] / [AB2] = K1[A][B1] / K2[A][B2] = K1 / K2

b.) K1 / K2 expresses selective binding

i.) Understood for precipitation and host-guest association

ii.) The guest offering the largest K occupies the most sites

3.) Which way do enthalpy and entropy go when guests and hosts unite?

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a.) Step 1: guest is removed from one place and put in another

i.) Bonds are broken, so ∆H1 is just like before

ii.) This initial investment is recovered eventually when strongerbonds are formed (equivalent to ∆H3)

b.) Step 2: The host has already been made. ∆H2 is essentially negligiblec.) Step 3: Enthalpy of binding, ∆Hº, comes from ∆H1 and ∆H3

i.) net change in enthalpy

ii.) exothermic binding

d.) Exothermal is the goal, for the union creates order (entropy doesn’t help). In order to compensate it must release energy – lower enthalpy.

e.) Guest-host reactions require pre-organized hosts usually; the entropy cost of restructuring is too high and the enthalpy cannot compensate.

f.) For pre-organized hosts, the entropic cost has been paid already: the hostalready exists

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3/1/12

Chapter 16: Acids and Bases

I.) A Wandering Ion

A.) The transfer of H+ is the most broadly familiar exchange in a reaction

1.) Controls life’s biochemical pathways by its presence or absence

B.) The H+ ion unites with its partner, hydroxide anion OH-

C.) This chapter is about Brønsted-Lowry acids and bases (hydrogen donor/receiver)

D.) H+ is a prominent Lewis acid (electron-pair acceptor)

II.) Conjugate Acids and Bases

HA + B A⇌ - + BH+ (HA is the acid, B is the base)HCl Cl⇌ - ; CH3COOH CH⇌ 3COO- ; H2O OH⇌ - ; NH4

+ NH⇌ 3

Like a redox reaction, the HA loses H and B gains H+

A- is the conjugate base of HA, BH+ is the conjugate acid of B

A.) Strength and Weakness

1.) Strength of acids and conjugate bases are related inversely

a.) Strong: loses proton readily

b.) Weak: Holds onto H+

2.) Weak acids ionize difficultly, A- is a good proton finder (strong conjugate base)

3.) Thermodynamic relationships connect base and acid:

a.) strong acid → weak conjugate base

b.) strong base → weak conjugate acid

c.) weak acid → strong conjugate base

d.) weak base → strong conjugate acid

B.) Stabilization of a Conjugate Base

1.) Many conjugate bases are anions, becoming negative only when H+ leaves

a.) Larger anions can disperse electrons more easily

2.) Conjugate bases can only survive if they can disperse the negative charge

a.) Ex: CCl3COOH (trichloroacetic acid) vs. CH3COOH (acetic acid)

i.) Chlorine withdraws electrons more effectively than hydrogen

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ii.) This makes trichloroacetic acid a more stable base

iii.) Therefore, trichloroacetic acid is stronger than CH3COOH

3.) Conjugate bases can be supported by delocalization

a.) This is especially true for organic acids (aromatic or conjugated π systems)

b.) Delocalization is why R—COOH usually dissociates more easily than R—OH

i.) Carboxylate anions (R—COO-) spreads negative charge over two π bonds (3 p-orbital system)

ii.) This is generally why R—COOH is a stronger acid than R—OH withthe same R group.

4.) Size is important to acidity

a.) Larger structures generally can support an extra electron pair

b.) The acidity of a family of related compounds increase as conjugate basesgrow larger

i.) But NOT always: Sometimes, size leads to poor solvation (the largercompound is less protected from a proton attack than a tightly packedsmaller conjugate base)

-Example: methyl alcohol (CH3OH) vs. tert-butyl alcohol (C4H9OH) methyl alcohol is VERY weak acid, but still stronger than tert-butyl alcohol due to its bulky conjugate base

5.) Bond strength can determine acidity

a.) An easily broken H—A bond is a stronger acid, because its thermodynamicsprevent it from reforming easily

III.) Aqueous Equilibria

Acid-base chemistry is primarily in the solution phase: it is in homogenous equilibrium. Water is almost always the host; it is passive, and it mediates exchange of H+ ions. H2O is both acid and base, being able to become H3O+ (hydronium) and OH- (hydroxide). These are the foundation of Arrehenius acid & base.

A.) Acidic Solutions: pKa and pH

1.) HA (aq) + H2O (l) A⇌ - (aq) + H3O+ (aq) ; Ka = ([H3O+][A-]) / [HA]

2.) Ka is the acid dissociation constant or the acid ionization constant

a.) Large Ka means a large concentration of hydronium and A-, but not much HAb.) Small Ka means a small concentration of hydronium and A- and much undissolved HA (weak acid)

c.) “Large” Ka is > 1 ; 100% ionization means Ka is essentially infinite and we can view the reaction as a 1:1 equations (e.g., HCl, H2SO4)

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d.) Weak acids are shaped by small constants (Ka < 1); most are tabulated

3.) The Model: Acetic acid [Ka(25ºC) = 1.76 × 10-5]

a.) Equilibrium constant increases according to the strength of the acid and weakness of conjugate base (and vice versa): Free energy decreases

b.) ∆Gº = -RT ln Ka → ∆Gº/(2.303 RT) = -log Ka = pKa

i.) 2.303 is the factor to go from base e to base-10

ii.) A lower pKa = higher Ka ; acids with negative pKa are strongest

c.) Acetic acid’s pKa is 4.75. pKa of much stronger HCl (Ka ≈ 107) = -7

i.) pKa increases with every 10-fold decrease in Ka

ii.) pH = –log [H3O+]: pH = 1, [H3O+] = 10-1 M, etc.

B.) Basic Solutions: pKb and pOH

1.) H2O isn’t really acting as a base. It is a proton donor to become OH-

a.) H2O (l) + B (aq) OH⇌ - (aq) + BH+ (aq) ; Kb = ([BH+][OH-]) / [B]

2.) Kb is the base dissociation constant or the base ionization constant

a.) The relationship can be described in a similar way to acidic substances

i.) Basic substances generate hydroxide instead of hydronium

ii.) Hydroxide becomes proton acceptor in solution becoming the newbase

iii.) Strong bases tend to accept H+ from water, producing BH+

iv.) All three are in equilibrium based on Kb

v.) The stronger the base, the higher the Kb and the more OH-

vi.) Strong bases have weak conjugate acids and vice versa.

vii.) 1:1 stoichiometric relationship with OH-

C.) Autoionization of Water

1.) Water can react with its own kind to produce hydronium and hydroxide in an autoionization process.

a.) H2O (l) + H2O (l) H⇌ 3O+ (aq) + OH- (aq) ; Kw = [H3O+][OH-]

2.) Kw is water’s autoionization constant, autodissociation constant, or ion-product constant

a.) Kw(25ºC) = [H3O+][OH-] = 1.0 × 10-14

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i.) No matter what outside source affects it, this constant does not change

ii.) Pure water, to satisfy this equation, starts out with 1.0 × 10-7 M (pH 7)

iii.) Excess hydronium lowers pH, excess hydroxide raises

iv.) Increase in 1 is direct decrease in the other

D.) Kb for a conjugate base

1.) Kw connects concentrations of hydronium and hydroxide, and Ka with Kb

a.) if HA and B are related (B = A-), we have the following:

i.) HA (aq) + H2O (l) A⇌ - (aq) + H3O+ (aq) ; K1 = Ka

ii.) H2O (l) + A- (l) OH⇌ - (aq) + HA (aq) ; K2 = Kb

b.) KaKb = ([H3O+][A-] / [HA]) × ([HA][OH-] / [A-]) = [H3O+][OH-]

i.) KaKb = Kw

IV.) Neutralization

When acids and bases meet in aqueous solution, thermodynamics compels them to form water. The acids and bases in the solution are overpowered by what they have created: hydroxide and hydronium. These two then come together in reaction to form water:

H3O+ (aq) + OH– (aq) H⇌ 2O (l)This reaction is driven heavily toward the product side by an equilibrium constant that is Kw’s reciprocal. Neutralization is the autoionization in reverse. The K constant (1.0 × 1014) is in heavy favor of undissoci-ated water.

A.) Strong Acid and Strong Base

1.) In a strong acid – strong base reaction, everything reacts completely.

a.) Example: HCl and NaOH. HCl in aqueous solution has become hydroniumand the chloride ion. NaOH breaks into sodium ion and hydroxide. Thehydronium and hydroxide meet, one by one, forming water until Kw is reached.

b.) The Cl– and Na+ are “spectator ions,” too weak on their own to be significant

c.) Therefore, the net reaction of a strong acid and strong base is just water.

2.) If the initial concentrations of base and acid are equal, so will the final concentrationsof hydronium and hydroxide

3.) If there is an imbalance, the neutralization will take place until there is nothing leftof the lesser, and the pH will be affected accordingly. Kw will still be the same, though

4.) The value at equilibrium can come a number of ways

a.) Add 0.1 mol hydroxide to 0.2 moles hydronium → 0.1 mole hydronium

b.) Add 0.1 mole hydronium to pure water → 0.1 mole hydronium

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c.) Add 0.2 moles hydronium to 0.1 mol hydroxide →0.1 mole hydrondium

5.) All that matters is the present at equilibrium

B.) Weak Acid and Strong Base

1.) Weak acids – strong base reactions have the same tendency to form water

2.) Example: Acetic acid, which will be mostly undissociated on its own

a.) Add a lot of NaOH, and you have

i.) The dissociation of the weak acid

ii.) the neutralization of hydronium by excess hydroxide

b.) CH3COOH (aq) + H2O (l) CH⇌ 3COO– (aq) + H3O+ (aq) Ka = 1.76E-5 H3O+ (aq) + OH– (aq) 2H⇌ 2O (l) 1/Kw = 1.0E14

Which yields a net neutralization of

CH3COOH (aq) + OH– (aq) CH⇌ 3COO– (aq) + H2O (l) K = Ka / Kw

This reaction is driven to the right by the enormous 1/Kw constant

c.) For every mole of strong base, one mole of weak acid conjugates into base

i.) However, this time the conjugate base is strong, which leads to hydrolysis of the acetate—the solution is now a basic solution with a pH higher than 7

V.) Weak Acids, Conjugate Bases, and Buffers

A.) HA can exist in three ways as a weak acid:

1.) HA, in equilibrium with small concentrations of hydronium and its conjugate base

2.) Its conjugate base A–, with small concentrations of hydroxide and HA

3.) HA and A–, with large concentrations of both

We can apply the mass-action law (equilibrium constant) to all three specific cases

B.) Weak Acid: Mostly HA

1.) Introduced with initial concentration c, weak acid ionizes to an (unknown) equalamount of H3O+ and A–. SO, at equilibrium. . .

a.) [HA] = c – x

b.) [H3O+] = x

c.) [A–] = x

d.) and Ka = [H3O+][A–] / [HA] = x2 / c – x

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i.) In this equation, we can allow ourselves the assumption that c – x ≈ c

ii.) x2 = c • Ka

iii.) the concentrations are 100 times smaller than c

2.) 1 L acetic acid, then, will contain 1.33E-3 M hydronium and 7.52E-12 mole hydroxide, governed by Kw

3.) Water, in this case, ceases to become the primary source of hydronium

C.) Conjugate Base: Mostly A–

1.) This mixture has no undissociated acid initially, rather only the conjugate base

2.) How it got there is unimportant; what matters is the hydrolysis:

A– (aq) + H2O (l) HA (aq) + OH⇌ – (aq)Conjugate base at equilibrium = c – x, concentrations of products = x

3.) The Kb = Kw / Ka, which will be larger than Kw significantly.

a.) This means it will create an excess of hydroxide ions compared to pure water

b.) x is found the same way as above

4.) Existing alone, the acetate gets its share of protons and leaves an excess of hydroxide. The solution is basic. The pH is higher than 7.

D.) Buffers

1.) A buffer contains a sizable concentration of both weak acid and conjugate base

a.) The introduction of a strong acid drastically shifts the delicate balance of hydronium and hydroxide, essentially turning the solution into a strong acid.

b.) The weak acid then has to readjust to the new conditions, changing itsequilibrium – the weak acid which has been dissociated now has to take backits protons:

CH3COOH (aq) + H2O (l) ← CH3COO– (aq) + H3O+

Which adheres to Le Châtelier’s principle.

c.) If the solution was all conjugate base to start, it will react the same if a strongbase comes into play.

2.) This contrasts with a buffer solution where a high amount of both forms are found

a.) The conjugates protect each other from gaining or losing protons

b.) Equilibrium is strained slightly: Q < Ka. The weak acid that is formed must dissociate before it restores balance; it is a weak acid, so this is not a big deal

c.) The buffer solution “fights off” attacks from outside acid, whereas a solutionof only weak acid changes dramatically.

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d.) The reverse process can be applied to the invasion of a strong base

3.) The Henderson-Hasselbalch equation expresses these phenomena

a.) Ka = [H3O+][A–] / [HA]

b.) –log Ka = –log [H3O+] – log [A–] / [HA]

pH = pKa + log [A–] / [HA]

c.) Equilibrium is connected to the concentrations of undissociated acids andconjugate bases (salt), which can be estimated from stoichiometry.

d.) It is a ratio; only the quotient influences pH

i.) This only applies in buffer solutions

4.) Example: What is the pH of a 1.00 L solution with 0.050 mole acetic acid and 0.050 moles sodium acetate? What happens when 0.001 mole of hydroxide is added?

a.) When [HA] = [A–], pKa = pH

i.) pH = 4.75—the pKa of acetic acid

b.) Adding 0.001 mol hydroxide decreases the concentration of [HA] from 0.050 to 0.049. This is balanced by an increase in [A–] from 0.050 to 0.051

i.) pH = pKa + log [A–] / [HA]

ii.) pH = 4.75 + log 0.051 / 0.049 = 4.77 (the buffer keeps the pH from rising up to 11)

5.) pH = pKa ± 1 within the range of mixtures with ratios from 10:1 to 1:10 a.) 10:1 buffer with 10x as much salt

i.) log [A–] / [HA] = log(10) = 1

b.) 1:10 buffer with 10x as much acid

i.) log [A–] / [HA] = log (10–1) = –1

6.) No buffer is perfect, but buffers are essential to life (buffers keep blood pH @ 7.4)

VI.) Titration

A.) Titrations are a carefully controlled reaction: a known amount of B is inserted to discoveran unknown amount of A

B.) From a balanced equation, you can find moles A because it will equal moles B at equilibrium

C.) Primarily done with indicators, which change the color of the species when it goes from acid to base.

1.) This is called the equivalence point

D.) Weak acids go from acid, to buffer, to equilibrium, to a basic solution

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E.) Conjugate bases do the same when a strong acid is added

F.) Strong acid and strong base titrations don’t have the buffer region, it goes straight from oneto the other.

1.) Its equivalence point is always at 7, where moles base = moles acid

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3/13

Chapter 17: Chemistry and Electricity

I.) Putting Reactions to Work

A.) The “universal currencies:” conservation of energy and maximization of entropy; locally, payment is in the form of free energy

1.) Reactants → products + free energy

B.) Gibb’s free energy is the ability to do work (at constant temperature and pressure)

1.) This can be applied to electrical work in a redox reaction

C.) Electrochemistry: the use of chemical energy to push electrons along a wire and into an external circuit.

II.) Pushing Electrons

A.) A skier’s analogy

1.) The Mountain: creates a difference in potential & gravitational energy

2.) The Skier: a mass-endowed body able fall in the gravitational field

3.) The Path: facilitates passage from high to low potential

4.) The Lift: Brings new “skiers” to the top of the “mountain”

B.) Applied to electrochemistry

1.) The Mountain: Electrical potential energy

a.) Measured in voltage (joules per coulomb); electrical work

2.) The Skier: electrons and ions in an electric circuit

3.) The Path: wires, electrolytic solutions

4.) The Lift: Reduction-oxidation reactions

III.) Redox Redux

A.) Terms

1.) Reductant/Reducing agent: gives away/loses electrons; becomes oxidized

2.) Oxidant/Oxidizing agent: causes reductant to give electrons; becomes reduced

B.) The Reactions

1.) Oxidationa.) Reductant → oxidized species + electronsb.) Zn → Zn2+ + 2e–

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2.) Reduction

a.) Oxidant + electrons → reduced speciesb.) Cu2+ + 2e– → Cu

3.) Redox reactions

a.) Reductant + Oxidant → oxidized species + reduced speciesb.) Zn + Cu2+ → Zn2+ + Cu

C.) Thermodynamically equivalent to acids, bases, and conjugates: Strong goes to weak, but hasa hard time going back to strong

D.) The redox equilibrium is determined by the difference in free energy; the largest drop is whenstrong reductant and oxidant go to their respective weak conjugates

IV.) From Chemical Energy to Electrical Work

A.) The “mountain” of chemical energy is the difference in G between products and reactants

1.) ∆G = Gfinal – Ginitial

B.) How many joules?

1.) Work done by the redox electrons outside the system (voltage)

C.) How many coulombs?

1.) 1.6E-19 C/electron x 6.022E23 electrons/mol = 96,485 C • mol-1 = F

D.) ∆G = -nFE (cell potential, measured in J/C: joules = mole • C • moles–1 • joules • C–1)

1.) ∆G grows more negative as V increases with a negatively charged electron

V.) Completing the Circuit

To produce sustainable electricity, you must separate electron donors from the electron receivers

A.) The galvanic cell: a circuit driven by the free energy of a spontaneous redox reaction

1.) Dry cells, wet cells, fuel cells, concentration cells, batteries, etc.

2.) Shorthand notation: Zn(s) | ZnSO4(aq) || CuSO4 (aq) | Cu(s)

B.) What happens in the galvanic cell

1.) Oxidation takes place on the left, reduction on the right

2.) Electrodes are placed in (zinc/copper metal) which conduct the electrons

3.) Oxidation occurs at the anode; reduction occurs at the cathode

a.) Which is which is circumstantial: very rarely is an element always one or other

b.) The identity (reductant/oxidant) is based on the difference in potentials

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4.) The anode loses electrons, putting ions into the aqueous solution

5.) Negative charge builds up in the anode, giving it high electrical energy

6.) Meanwhile, copper ions in solution are being drawn into the metal as electrons enter

7.) There are not enough electrons for nuclei; the electrical energy decreases (more +)

8.) A similar situation happens in solution; too many positive ions in the anode, too manynegative ions in the cathode

9.) The excess electrons on the left travel to the electron-poor cathode on the right

C.) To close the circuit, we put in a salt bridge, providing ions to travel to each side (cations tocathodes, anions to anodes)

VI.) Electrochemical Life and Death

A.) The cell lasts (“lives”) until it reaches equilibrium (“death”)

1.) The difference in free energy between anode and cathode is what the galvanic cell converts into voltage

2.) The reaction quotient changes as the reaction approaches equilibrium

a.) At equilibrium, the cell potential (E) and ∆G are both zero.

B.) The Nernst Equation

1.) Related to the thermodynamic equation of ∆G, ∆Gº, and Q

2.) E = Eº –RT/nF lnQ

a.) Eº is the standard cell potential, derived from Eº = –∆Gº/nF

b.) “standard cell potential” is the value it operates under standard conditions:1 atm, 25º C, and all solutions in 1.0 M concentrations

3.) Eº is intensive, and does not change under standard conditions

a.) It is related to K just like ∆Gº: Eº = RT/nF lnK

b.) E and ∆G only differ by a factor of -nF (charge)

C.) Consequences

1.) Relating voltage and free energy lets us find K through electrical means

2.) Useful for sparingly soluble salts with extremely low concentrations (AgCl, e.g.)

3.) Under constant T, Eº and lnK are directly proportional

4.) We can use electrochemical methods to find Ksp and pH values

VII.) Half-Reactions

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Eº tells us all we need to know about cell thermodynamics, leading to the strength and weakness of elec-tron transfers for various cells. It tells us how highly favored one side is to the other. Negative ∆Gº is a positive Eº: The higher the cell potential, the more spontaneous the reaction. The problem is Eº can only be measured in differences, and so you need a standard, singular reference point for all instead of writing down their various potentials in different combinations.

A.) Standard Electrode Potentials

1.) We proceed by half-reactions, viewing reduction separately from oxidation

a.) Standard Oxidation Potential: Eºox

b.) Standard Reduction Potential: Eºred

c.) Eº = Eºox + Eºred

2.) Give all reductants a specific oxidant partner and measure this: that standard is H+

a.) Given an arbitrary value of 0 for Eºred, and the difference is measured

b.) > 0 = reductant; < 0 = oxidant

c.) Eºox = –Eºred (increase in chemical free energy when reaction is reversed)

i.) If electrons flow to H+ then it is negative, vice versa

ii.) Eºcell = |Eºred|1 + |Eºred|2

iii.) This is the difference between products and reactants

3.) The voltage determines the energy needed to move the electrons: The joules percoulombs rate is constant no matter anywhere.

a.) Eºred is the thermodynamic potential of the redox reaction

B.) Reductants and Oxidants

1.) The species with the more positive Eºred will get the electrons

a.) It is the stronger oxidant

b.) Guarantees largest drop in free energy

2.) Example: Fluorine: Eºred = 2.87 V

a.) Strongest oxidant commonly availableb.) It will take electrons from everything with a potential lower than 2.87, evenif that compound/element would normally be the oxidant

c.) Pecking order of cell potential (V) [examples]

i.) Fluorine F2 2.87ii.) Permanganate MnO4

– 1.51iii.) Gold Au3+ 1.40iv.) Chromate Cr2O7

2– 1.33v.) Oxygen O2 1.23

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vi.) Silver Ag+ 0.80

3.) Elements/compounds with negative reduction potentials will donate

a.) Alkali metals are the most common, with only one electron to give

VIII.) Charge and Mass: Balancing the Redox Equation

Oxidation and reduction is everywhere in the world, a seemingly continuous process, from the tarnishing of silver to the transport of ions across the membrane of a beating heart. Some are simple, like zinc and copper, and others are more complex, like the discharge of a lead battery. But they all have one thing in common: a balanced equation. To write the balanced equation, we have rules to follow.

1.) Identify the reductant and oxidant.2.) Write the half-reactions.3.) Balance the atoms in each half-reaction, leaving O and H for last.4.) Balance the electrons, again separately for half-reactions.5.) Combine the half-reactions into one complete equation.

A.) Acidic Solution

Example: ___ Cr2O72–(aq) + ___ Cl–(aq) → ___ Cr3+(aq) + ___Cl2(aq)

1.) Total charge on left: -3; Total on the right: +3; unequal atoms on the left and right. Is this a redox reaction? Do the oxidation numbers change?

2.) The half reactions:

a.) Cr2O72– → Cr3+ (reduction: Cr is +6 before and +3 after)

b.) Cl– → Cl2 (oxidation: Cl is -1 before and 0 after)

3.) Balance each separately (worry about all except H and O)

a.) Cr2O72– → 2Cr3+

b.) 2Cl– → Cl2

4.) Balance the hydrogen/oxygens with water, 1 for each missing oxygen

a.) Cr2O72– → 2Cr3+ + 7H2O

b.) This creates an imbalance of 14 hydrogens, which are sent to the leftc.) 14H+ Cr2O7

2– → 2Cr3+ + 7H2O ; 2Cl– → Cl2

5.) Count up the charges, and balance them

a.) +12 on the left, +6 on the right ; -2 on the left, 0 on the right

b.) 6e– + 14H+ + Cr2O72– → 2Cr3+ + 7H2O ; 2Cl– → Cl2 + 2e–

c.) The right must match the left, so really its 6Cl– → 3Cl2 + 6e–

6.) Add the half reactions together:

6e– + 14H+ + Cr2O72– → 2Cr3+ + 7H2O

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+6Cl– → 3Cl2 + 6e–

14H+(aq) + Cr2O72–(aq) + 6Cl–(aq) → 2Cr3+(aq) + 3Cl2(g) + 7H2O(l)

a.) 6 moles of electrons are transferred (n = 6)

b.) The electrons disappear in the addition, but they never really go away

B.) Basic Solution

The process is fairly similar, except OH– complicates it a little bit. To avoid this, we pretend firstthat the solution is actually acidic.

Example: Cd(s) → Cd(OH)2 (s) (assumed to be oxidized in the basic solution)

1.) To balance, first add water molecules on the left: 2H2O + Cd(s) → Cd(OH)2 (s)

2.) Then add 2 hydroniums on the right: 2H2O + Cd(s) → Cd(OH)2 (s) + 2H+

3.) Add hydroxide on the left to balance: 2H2O + Cd(s) + 2OH– → Cd(OH)2 (s) + 2H+

4.) Reduce and add electrons: Cd(s) + 2OH– → Cd(OH)2 (s) + 2e–

IX.) Going Up-Hill: Electrolysis

A system at equilibrium, on its own, is entirely dead. It will not change. If you find the energy, however, you can force a non-spontaneous change. This is how all living beings stay alive, and how recharging batteries works as well. In redox reactions, this backward process is electrolysis.

A.) Back to the galvanic cell

1.) To induce electrolysis, you must supply voltage from an external source

2.) It changes from a spontaneous galvanic cell to an electrolytic cell

3.) The anode grows more negative, but from the outside instead of the inside now

a.) This rate of delivery is the current, measured in amperes (A = C/s)

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chem 138 textbook notes

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