I. Base TheoryA. Concepts

1) Arrhenius Concept: base produces OH- in water2) Bronsted-Lowery Model: base is a H+ acceptor

3) Strong Hydroxide Basesa) Alkali Metal Hydroxides: NaOH, KOH, etc…

i. Completely Dissociated in Waterii. NaOH Na+ + OH- K = very large

b) Alkaline Earth Hydroxides: Ca(OH)2, Mg(OH)2, etc…

i. Not very soluble in waterii. What does dissolve is completely dissociatediii. Ca(OH)2 Ca2+ + 2 OH- K = very large

c) Example: pH of 0.05 M NaOH12.70 1.30-14 pH1.30 log(0.05)- pOH

4) Non-Hydroxide Bases (Weak Bases)a) Any atom with a lone pair of electrons can accept a proton = baseb) Ammonia in water:

c) Other amine molecules are also bases

5) The general base equationB + H2O BH+ + OH-

6) Strong Base: equilibrium lies far to the right ([OH-] ≈ [B]0)

7) Weak Base: equilibrium lies far to the left ([OH-] << [B]0)

a) Calculations are similar to weak acids

H N

H

H

H O H H N

H

H

H

+ + OH

base acid conjugate acidconjugate base

Kb = 1.8 x10-5

CH3CH2 N

CH2CH3

CH2CH3N

triethylamine pyridine

[B]]][OH[BHKb

b) Example: pH of 15.0 M NH3 Kb = 1.8 x 10-5

i. We can find [H+] from KW = [H+][OH-] = 1 x 10-14

or pH + pOH = 14ii. Percent dissociation still means the same thingiii. The 5% rule for approximations: x/[B] x 100% < 5%

c) Example: pH of 1.0 M methylamine Kb = 4.38 x 10-4

II. Polyprotic AcidsA. Carbonic Acid is a diprotic acid

1) Polyprotic means there are more than one ionizable proton2) Carbonic acid is the acid that helps maintain body pH H2CO3

H2CO3 H+ + HCO3-

HCO3- H+ + CO3

2-

3) Ka1 and Ka2 stand for the loss of the first and second protons, respectively

7-

32

31 10 x 3.4

]CO[H]][HCO[HK

a

11--

3

23

2 10 x 6.5][HCO

]][CO[HK

a

4) Usually Ka1 >> Ka2

a) As (-) charge builds up, it is harder to remove the next protonb) H+ from the first ionization forces the second ionization to the leftc) We can usually ignore all but the first ionization in calculations

B. Phosphoric Acid is a triprotic acid1) Ionizations

H3PO4 H+ + H2PO4- Ka = 7.5 x 10-3

H2PO4- H+ + HPO4

2- Ka = 6.2 x 10-8

HPO42- H+ + PO4

3- Ka = 4.8 x 10-13

2) Ka1 >> Ka2 >> Ka3

3) Example: pH of 5.0 M H3PO4, and the concentrations of all of the phosphoric acid derived speciesa) Use Ka1 only to find [H+] and the pH

b) Then, [H+] = [H2PO4-]

c) [H3PO4] = [H3PO4]0 - [H+]

d) Find [HPO42-] and [PO4

3-] from what is already calculated and Ka2, Ka3

58-

-3

a2

a1 10 x 2.110 x 2.610 x 7.5

KK

513-

-8

a3

a2 10 x 3.110 x .8410 x 6.2

KK

C. Sulfuric Acid is a unique diprotic acid1) The first ionization of sulfuric acid is a strong acid (Ka1 = large)

H2SO4 H+ + HSO4-

2) The second ionization of sulfuric acid is a weak acidHSO4

- H+ + SO42- Ka2 = 1.2 x 10-2

3) Example: pH of 1.0 M H2SO4

a) Assume complete dissociation of the first protonb) Use the equilibrium calculation on Ka2

i. [H+]0 does not = 0 because of Ka1

ii. [H+]0 = 1.0 M

c) Approximate 1 + x ≈ 1 in this case (5% rule says this is ok)d) When [H2SO4] > 1.0 M, you can ignore Ka2

4) Example: pH of 0.01 M H2SO4

a) The 5% rule tells us we can’t ignore Ka2 in this case

b) We must use the quadratic equation to solve for xc) When H2SO4 < 1.0 M, we can’t ignore Ka2

III. Acid-Base Properties of SaltsA) Simple Salts1) Salt = ionic compound = one that completely ionizes in water2) Some salts have no effect on pHa) Cations of strong bases have no effect on pH

i. Na+, K+, etc…ii. These cations have no affinity for OH- in water

b) Anions of strong acids have no effect on on pHi. Cl-, NO3

-, etc…

ii. These anions have no affinity for H+ in water c) Solutions of these combined ion salts have pH = 7.00

B) Basic Salts1) Salts containing the conjugate base of a weak acid produce basic solutions2) The conjugate base must be strong if the acid is weak, so it must have a

strong affinity for H+, which will affect the pH of a solution

3) Sodium Acetate ExampleNaC2H3O2 Na+ + C2H3O2

-

C2H3O2- + H2O HC2H3O2 + OH-

???]OH[C

]][OHOH[HCK -232

232b

4) How do we find Kb for the conjugate base of a weak acid?

5) For any weak acid and its conjugate base, Ka x Kb = KW

Kb = KW / Ka = 1 x 10-14 / 1.8 x 10-5 = 5.6 x 10-10 for acetate

6) Example: pH of 0.30 M NaF Ka for HF = 7.2 x 10-4

C. Base Strength in Water1) Any base in water must compete with the hydroxide anion for protons2) Hydrocyanic acid example:

HCN + H2O H3O+ + CN- Ka = 6.2 x 10-10

weak acid strong base (compared to H2O)

CN- + H2O HCN + OH- Kb = KW / Ka = 1.6 x 10-5

weak base strong acid

]][OH[H]OH[C

]][OHOH[HC]OH[HC

]OHC][H[K x K -232

232

232

-232

ba

-14W 10 x 1K]][OH[H

OH- > CN- > H2O(compared to OH-)

D. Acidic Salts1) Salts having the conjugate acid of a weak base produce acidic solutions2) Ammonium chloride = NH4Cl NH4

+ + Cl- NH4

+ NH3 + H+

3) Since ammonia is a weak base, ammonium is “strong” acid and will effect pH

4) Example: pH of 0.1 M NH4Cl Kb = 1.8 x 10-5 a) Find Ka from KW b) Make sure the conjugate acid is stronger than water or it will have no effect

5) Highly charged metal ions can also be acidica) AlCl3 + 6 H2O Al(H2O)6

3+ + 3 Cl- b) Al(H2O)6

3+ + H2O Al(H2O)5(OH)2+ + H3O+

c) The higher the metal’s charge the more acidicExample: pH of 0.01 M AlCl3 Ka = 1.4 x 10-5

E. Salts containing both acidic and basic components1) NH4C2H3O2 NH4

+ + C2H3O2-

2) The calculations for these compounds are complex3) We can, however, at least decide if the solution is acidic or basic

a) If Ka > Kb, the solution will be acidic

b) If Kb > Ka, the solution will be basic

c) If Ka = Kb, the solution will be neutral

4) Example: Will the following solutions be acidic or basic?NH4C2H3O2 NH4CN Al2(SO4)3

IV. Acid-Base Properties and Molecular StructureA. Polarity

1) Not all H containing molecules are acidica) CHCl3 H+ + CCl3

-

b) Strong, nonpolar bonds don’t dissociate easily

2) Polar X—H bonds are easily dissociated (acidic)a) H—Cl is a polar bondb) H—OH is a polar bond

3) Bond strength also plays a part in aciditya) Hydrohalide Polarity: H—F > H—Cl > H—Br > H—I b) Bond Strength (kJ.mol) 565 427 363 295c) Acidity weak strong strong strong

4) Oxyacids: the more O on the central atom, the stronger the acid in a seriesa) Electronegative Oxygens remove electrons from the center atomb) This polarizes and weakens the O—H bond even morec) HClO4 > HClO3 > HClO2 > HClO

d) H2SO4 > H2SO3

5) H—O—X Moleculesa) The more electronegative X is, the more acidic the molecule isb) Electronegative X removes electrons from the H—O bondc) Acidity: H—O—Cl > H—O—Br > H—O—I > H—O—CH3

d) Electronegativity: 3.0 2.8 2.5 2.3

B. Acid-Base Properties of Oxides1) A generic oxide can be represented as X—O 2) If X—O is a strong and covalent bond, the oxide will be acidic in water

a) H—O—X H+ + -O—X b) If X is electronegative, like O, it should form a strong covalent O—X bond

3) If X—O is weak and ionic, the oxide will be basic in watera) H—O—X X+ + OH- b) NaOH Na+ + OH- c) The X atom in these oxides is usually not electronegative (Na+)

4) Examples of Acidic Oxidesa) SO2 + H2O H2SO3 H+ + HSO3

-

b) CO2 + H2O H2CO3 H+ + HCO3-

5) Examples of Basic Oxidesa) CaO + H2O Ca(OH)2 Ca2+ + 2 OH-

b) K2O + H2O KOH K+ + OH-

V. Lewis Acid-Base DefinitionA. Definitions

1) Lewis Acid = an electron pair acceptor2) Lewis Base = an electron pair donor3) Example: H+ + :NH3 NH4

+

4) This model includes the other acid-base concepts

5) This model accounts for many other chemical reactions that the others don’ta) BF3 + :NH3 H3N:BF3

Lewis acid Lewis Base Lewis Acid-Base complex

Lewis acid Lewis Base Lewis Acid-Base complex

b) AlCl3 + 6 H2O Al(H2O)63+ + 3 Cl-

c) Example: Identify Lewis Acid and Lewis Basei. Ni2+ + 6 NH3 Ni(NH3)6

2+

ii. H+ + H2O H3O+

Summary of Acid-Base Problem Solving1. List major species in solution2. Look for reactions that go to completion a. concentration of product b. major species left

3. Identify acids and bases4. Solve the equilibrium problem, check the approximation, find pH