I. Acid TheoryA. Classifications

1) Acids

a) Have a sour taste

b) Can dissolve metals

2) Bases

a) Have a bitter taste

b) Feel slippery

3) Arrhenius Definition

a) Acids produce H+ in water

b) Bases produce OH- in water

c) Applies only to aqueous solutions

d) Allows for only one kind of base (OH-)

4) Bronsted-Lowery Definition

a) Acid is an H+ donor

b) Base is an H+ acceptor

c) HCl + H2O H3O+ + Cl-

5) General Acid Equation

HA(aq) + H2O(l) H3O+(aq) + A-

(aq)

a) Conjugate base = what is left after H+ leaves acid

b) Conjugate acid = base + H+

c) Conjugate acid-base pair are related by loss/gain of H+

d) Competition for H+ by A- and H2O; strongest base wins

acid base hydronium ion

H O

H

H Cl H O

H

H Cl+ +

acid baseconjugateacid

conjugatebase

“The differences between the various acid-baseconcepts are not concerned with which is right,but which is most convenient to use in a particular situation.” James E. Huheey

6) Ka = acid dissociation constant

7) Example: Write simple Ionizations for:

HCl, HC2H3O2, NH4+, C6H5NH3

+, Al(H2O)63+

8) Bronsted-Lowery theory allows for non-aqueous solutions

9) More examples:

[HA]

]][A[H

[HA]

]][AO[HK 3

a

H N

H

H

H Cl H N

H

H

H

Cl+ +

acidbase

conjugateacid

conjugatebase

B. Acid Strength

1) Acid strength describes the equilibrium position of the ionization reaction

HA + H2O H3O+ + A-

2) Strong Acid = equilibrium lies far to the right

a) Almost all HA has ionized to H+ and A- ([H+] = [HA]0)

b) A strong acid has a weak conjugate base

i. To ionize fully, the conjugate base must have low proton affinity

ii. The conjugate base must be weaker than water

3) Weak Acid = equilibrium lies far to the left

a) Almost all HA remains unionized ([H+] << [HA]0)

b) A weak acid has a strong conjugate base

c) The conjugate base is much stronger than water

Strong acid Weak Acid

4) Common Strong Acids (Ka = very large)

a) Sulfuric Acid = H2SO4 is a diprotic acid (has 2 ionizable protons)

H2SO4 H+ + HSO4- Ka = ∞

HSO4- H+ + SO4

2- Ka = 1.2 x 10-2

b) Hydrochloric Acid = HCl is a monoprotic acid (has 1 ionizable proton)

HCl H+ + Cl-

c) Nitric Acid = HNO3 d) Perchloric Acid = HClO4

5) Oxyacids = acids with the ionizable proton attached to oxygen

HClO4 HNO3 H2SO4 are all oxyacids

6) Hydrohalic Acids = acids with the ionizable proton attached to a halide

HCl, HBr, HF, HI are all hydrohalic acids

7) Common Weak Acids

a) Phosphoric Acid = H3PO4 (triprotic acid)

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

b) Nitrous Acid = HNO2 Ka = 4.0 x 10-4

c) Organic Acids have a carbon backbone and a carboxyl group

i. Acetic Acid = CH3COOH (HC2H3O2) Ka = 1.8 x10-5

ii. Benzoic Acid = C6H5COOH Ka = 6.4 x10-5

iii. Only the OH hydrogen is acidic

8) Example: Relative Basicity of H2O, F-, Cl-, NO2-, CN- (Table 15.5)

C

O

OH

C. Water as an Acid and Base

1) An amphoteric substance can behave as an acid or a base (water)

2) Autoionization of water (reaction with itself)

H2O + H2O H3O+ + OH-

3) Ionization constant for water = KW = [H3O+][OH-] = [H+][OH-]

a) For any water solution at 25 oC, [OH-] x [H+] = KW = 1 x 10-14

b) Neutral solutions (pure water) have [OH-] = [H+] = 1 x 10-7

c) Acidic solutions: [H+] > [OH-]

d) Basic solutions: [OH-] > [H+]

e) Example: Calculate [OH-] or [H+] for the following:

i. [OH-] = 1 x 10-5 M

ii. [OH-] = 1 x 10-7 M

iii. [H+] = 10 M

4) Kw is temperature dependent. At 60 oC, KW = 1 x 10-13

Example: Is water autoionization exothermic or endothermic? [H+]??

D. pH Scale

1) pH = -log[H+] (simplifies working with small numbers)

2) If [H+] = 1.0 x 10-7, pH = -log(1 x 10-7) = -(-7.00) = 7.00

3) Number of decimal places in a log:

The number of sig. fig’s in the number is

how many decimal places you keep when

you take the log

4) Other p Scales:

a) pOH = -log[OH-]

b) pKa = -logKa

5) pH changes by 1 unit for every power

of 10 change in [H+]

a) pH = 3 [H+] = 10 times the [H+] at pH = 4

b) pH decreases as [H+] increases

(pH = 2 more acidic than pH = 3)

6) Example: Find pH and pOH for [OH-] = 1 x 10-3 and [H+] = 1.0 M

7) Relationship of pH and pOH

a) KW = [H+][OH-]

b) -logKW = -log[H+] – log[OH-]

c) -log(1 x 10-14) = pH + pOH = 14

d) Example: Find [H+], [OH-], and pOH for sample of pH = 7.41

II. Acid-Base Problem SolvingA. Calculating pH of Strong Acid Solutions

1) Solution labels tell what went into the solution, not what is present now

a) 1.0 M HCl

b) 1.0 M H+ and 1.0 M Cl-

2) Major Species = those present in large amount

a) For 1.0 M HCl = H+, Cl-, H2O (not OH-)

b) Identifying these is the key to solving acid-base problems

3) Find pH of 1.0 M HCl

a) Since HCl is a strong acid, we assume [H+] = [HA]0 = 1.0 M

b) Ionization of water only produces [H+] = 1 x 10-7 M (ignore this)

c) pH = - log(1) = 0

4) Example: Find pH of 0.1 M HNO3 and 1 x 10-10 M HCl

B. pH of Weak Acid Solutions

1) pH of 1.00 M HF? Ka = 7.2 x 10-4

2) Write the major species: HF, H2O

3) Which can furnish H+?

a) Dominant = HF H+ + F- Ka = 7.2 x 10-4

b) Small amount = H2O H+ + OH- KW = 1 x 10-14

4) Do equilibrium calculation based on HF only

HF H+ + F-

Inititial 1 0 0

Change -x +x +x

Equil. 1-x x x

x

x

110 x 7.2

[HF]

]][F[HK

24

a

Assume x = small, then 1- x ≈ 1x2 = 7.2 x 10–4 x = 2.7 x 10–2

pH = 1.57

5) Is our approximation valid?

a) Most Ka values are only known to within 5% error

b) If x is really small, x/[HA]0 < 5% and our approximation is valid

c) If x/[HA]0 > 5%, we have to use the quadratic, no approximation

d) 2.7 x 10-2 / 1.00 x 100% = 2.7% so our approximation is ok

6) Example: Find pH of 0.1 M HOCl Ka = 3.5 x 10-8

7) Example: Find pH and [CN-] of a solution of

1.0 M HCN (Ka = 6.2 x 10-10) and 5.0 M HNO2 (Ka = 4.0 x 10-4)

8) Percent Dissociation = amount dissociated / initial conc. x 100%

a) For 1.00 M HF we found [H+] = 2.7 x 10-2

Percent Dissociated = (2.7 x 10-2 / 1.00) x 100% = 2.7%

b) Strong acids are 100% dissociated always

c) For weak acids, percent dissociation increases as acid is diluted

Example: %Dissociation of 1 M and 0.1 M Acetic acid Ka = 1.8 x 10-5

d) Why does dilution increase percent dissociation of weak acids?

If we dilute by 10 times:

Now if we calculate Q, we can see which way the equilibrium will shift

Since Q < Ka, the equilibrium will shift to the right; %Dissoc.Increases

e) Example: Find Ka for 0.1 M Lactic Acid when %Dissoc. = 3.7%

0

2

a [HA]

x

[HA]

]][A[HK

10

[HA][HA] and

10

x][A][H 0

a0

2

0

K10

1

[HA]

x

10

1

10[HA]

10x

10x

Q