Acid Base Equilibria

Mullis 1

Acid Base Equilibria

Complications: Beyond pH of strong acids or strong bases

HCl(aq) H+(aq) + Cl- (aq) Ka > 1

HF(aq) H+(aq) + F- (aq) Ka = 7.2 x 10-4

Strong acid

Weak acid

Weak conjugate base

stronger conjugate base

Mullis 2

Conjugate Bases• Bronsted-Lowry Model• Conjugate base remains after acid loses its

proton• Strongest conjugate base = best at getting

available protons• Factors affecting conjugate base strength:

1. Electronegativity S- > Cl-

2. Size if acids were binary F- > I-

3. Number of oxygens if acids were oxyacids

ClO- > ClO2-

Mullis 3

pH of Mixture of Weak Acids• Find pH of a solution that contains 1.00M

HCN (Ka = 6.2 x 10-10) and 5.00 M HNO2

(Ka = 4.0 x 10-4).

• Because these are weak acids, dissociation is not complete and [H+] ≠ [HNO2].

• You must use ICE table strategy to solve pH problems with weak acids or bases.

Mullis 4

How to Determine [H+] in an Acid Mixture

• HCN H+ + CN- Ka = 6.2 x 10-10

• HNO2 H+ + NO2- Ka = 4.0 x 10-4

• H2O H+ + OH- Ka = 1.0 x 10-14

• Compare species and find the most likely contributor of protons. Here it is by far HNO2: Ka is 6 orders of magnitude greater than the closest possible H+ contributor.

Mullis 5

Equilibrium Method to Find [H+]

R HNO2 H+ NO2-

I 5.00 M 0 0

C -x +x +x

E 5.00 – x x x• Ka = x2 = 4.0 x 10-4

(5.00 –x)

x2 = 20. x 10-4 x = 4.5 x 10-2

Small: reactants favored

Change must be very small relative to initial conc.: Ignore x in this term

Mullis 6

Accuracy check

• When dealing with weak acids and bases, Ka values are typically valid to a ±5% accuracy. Verify that our assumption in ignoring x in the reduction of [HNO2]0 was valid:

• x x 100% = 4.5 x 10-2 (100%) = 0.90 %

[HNO2] 5.00

• Since 0.90% is less than 5%, the assumption is valid---in other words, our answer for x is good.

Mullis 7

What was the Original Question?

• Find pH of this solution:

• [H+] = 4.5 x 10-2

• pH = -log[H+] = -log(4.5 x 10-2) = 1.35

• Number of decimals in pH = Number of significant figures in concentration (2 here)

Mullis 8

Find [CN-] in that problem.• [H+] = 4.5 x 10-2 M• HCN H+ + CN- Ka = 6.2 x 10-10

• 6.2 x 10-10 = [4.5 x 10-2 ][CN-][HCN]

Remember our assumption that since Ka is small, we ignore the small reduction in initial concentration of product? Be consistent and make that assumption here, too. So:

• 6.2 x 10-10 = [4.5 x 10-2 ][CN-]1.00

[CN-] = 1.4 x 10-8 M (That’s a small number!)

Mullis 9

% DissociationEx: HNO2 H+ + NO2

-

[H+] x 100% = % dissociation

[HNO2]

Or…

[NO2-] x 100% = % dissociation

[HNO2]

If you know % dissociation and original concentration of acid [HNO2], can solve for [H+] or [NO2

-] (or x, the change in concentration).

Mullis 10

% Dissociation and DilutionMore Concentrated Diluted

Less % dissociation More % dissociation(more chance of

recombination)

If Q = K for this reaction Q < Ka for this one, so equilibrium shifts

right

Mullis 11

Ka from % DissociationA 0.100 M solution of lactic acid (HC3H5O3) is 3.7% dissociated. Calculate the value of Ka.

R HC3H5O3 H+ C3H5O3 -

I 0.100 M 0 0

C -x +x +x

E 0.100 – x x x

% dissociation = x x (100%) Ka = x2

0.10 (0.100 –x)

3.7% (0.10) = x Ka = 0.00372

x = 0.0037 M (0.100 - .0037)

Ka = 1.4 x 10-4

Mullis 12

Organic Acids(Acid = proton donor)

-OH phenol

Compound Formula Example Alcohol R-OH hydroxyl group 1-propanol Alkyl Halide -X X = any halide 1,2-dibromopropane Ether R-O-R’ one oxygen bonded to 2

hydrocarbon groups diethyl ether CH3-CH2 –O--CH2--CH3

Aldehyde O || R-C-H

Carbonyl group attached to end carbon

Ethanal O || CH3—C--H

Ketone O || R-C-R’

Carbonyl group attached to a middle carbon

2-propanone O || CH3—C-- CH3

Carboxylic Acid O || R-C-OH

Carboxyl group ethanoic acid O || CH3—C—OH

Ester O || R-C-O-R’

Carboxyl group without the H

methyl ethanoate O || CH3—C—O-- CH3

Mullis 13

Bases without OH-

• B (aq) + H2O BH+ (aq) + OH- (aq)

• NH3 (aq) + H2O NH4+ (aq) + OH- (aq)

• Kb = [BH+][OH-]

[B]

conjugate baseconjugate acidacidbase

Mullis 14

Amines RxNH(3-x)

• Ammonia with one or more of the H atoms replaced by another group.

• Many organic molecules involve amines.

N. .pyridine

ethylamine

. .N

HH

C2H5HO

HO

OH

CHCH2NHCH3

adrenaline

Mullis 15

Find the pH of 1.0M methylamine (Kb=4.38x10-4).

R CH3NH2 CH3NH3+ OH-

I 1.0 M 0 0

C -x +x +x

E 1.0 – x x x

CH3NH2 + H2O CH3NH3+ + OH-

Kb = x2 = 4.38x10-4 x2 ≈ 4.38x10-4

1.0-x x≈ 2.1x10-2

[OH-] = x = 2.1x10-2 M

pOH=-log[2.1x10-2 ] pOH = 1.68 pH= 14 -1.68= 12.32

Mullis 16

PolyProtic Acids• Lose one proton at a time: H3PO4 H2PO4

- HPO42- PO4

3-

• The reactions written to express this loss have successively smaller Ka values:

• Ka1 > Ka2 > Ka3

• …..makes sense: The loss of a second proton occurs less readily than the loss of the first one.In fact, Ka1 is usually so much larger than the

others, the loss of the first proton is the only reaction

that significantly contributes [H+].

Mullis 17

Sequential Loss of Protons

H3PO4 H+ + H2PO4- Ka1 = 7.5 x 10-3

H2PO4- H+ + HPO4

2- Ka2 = 6.2 x 10-8

HPO42- H+ + PO4

3- Ka3 = 4.8 x 10-13

Mullis 18

pH of a Polyprotic Acid• Find pH of 5.0M H3PO4 and the equilibrium concentrations of

each species: H3PO4, H2PO4-, HPO4

2-, PO43-

• Ka1 = [H+][H2PO4-] = x2 = 7.5 x 10-3

[H3PO4] 5.0-x (see previous problems’ RICE tables & pH calculations for details)

x2 = (5.0)7.5 x 10-3 x ≈ 0.19 M= [H2PO4-] = [H+] pH = 0.72

[H3PO4] = 5.0 – x = 4.8 M

Ka2 = [H+][HPO42-] = (0.19) [HPO4

2-] = 6.2 x 10-8

[H2PO4-] 0.19

[HPO42-]= 6.2 x 10-8 M

Ka3 = [H+][PO43-] = (0.19) [PO4

3-] = 4.8 x 10-13

[HPO42-] 6.2 x 10-8

[PO43-]= 1.6 x 10-19 M

Mullis 19

Sulfuric acid: Relatively High Conc.

• Find pH of 1.0 M H2SO4.

• H2SO4 H+ + HSO4- Ka1 > 1

• HSO4- H+ + SO4

2-Ka2 = 1.2 x 10-2

• In this case, does the 2nd reaction contribute significantly to [H+]?

• [H+] is at least 1.0 since the first step is total dissociation.

Mullis 20

Sulfuric acid: Relatively High Conc.

1.2 x 10-2= [H+][SO42-] = (1.0 + x)(x)

[HSO4-] (1.0 – x)

Assume change is small compared to 1.0, so: 1.0(x) 1.0

X = 1.2 x 10-2, or 1.2% of 1.0 M. If this is added to 1.0,result is 1.0012M, but with 2 sig. figs., [H+] is 1.0 M. (pH = 0.00)

R HSO4- H+ SO4

2-

I 1.0 M 1.0 0

C -x +x +x

E 1.0 – x 1.0 + x x

From 1st disso

ciation step

Mullis 21

0.01 M H2SO4 (Relatively Low Conc.)

1.2 x 10-2= [H+][SO42-] = (.01 + x)(x)[HSO4

-] (.01 – x)If assume change is small compared to .01, .01(x)

.01x = 1.2 x 10-2, or more than 0.01 M! Therefore, cannot ignore the subtraction

or addition of x—use the quadratic to find x = .0045. Total [H+] = 0.01(from 1st dissociation step) + .0045 = 0.0145.

pH = -log(0.0145) = 1.84

R HSO4- H+ SO4

2-

I 0.01 M 0.01 0

C -x +x +x

E 0.01 – x 0.01 + x x

From 1st disso

ciation step

Mullis 22

Salts of Weak Acids or Bases

• Cations of strong bases do not change pH of a solution. – (Ex: Na+ will not attract or contribute H+)

• Salts in which cations do not change pH and the anion is the conjugate base of a weak acid will produce basic solutions.– Ex. KC2H3O2 dissociates to produce C2H3O2

- and OH-

• Salts in which anions are not a base and the cation is a conjugate acid of a weak base will produce acidic solutions. – Ex. NH4Cl dissociates to produce NH3 and H+.

Mullis 23

Finding Kb for Conjugates

• Look up the Ka for the weak acid that makes the conjugate base.

• Ex: NaC2H3O2 (aq)

• C2H3O2- + H2O HC2H3O2 + OH-

• Ka(acetic acid) = 1.8 x 10-5

• KaKb = Kw

• Kb = 1x10-14 = 5.6 x 10-10

1.8 x 10-5

Mullis 24

Salt as a Weak Base• Find pH of a 0.30 M NaF solution.• Na+ does not change pH. • F- is the conjugate base of HF. Water has to

be the proton donor for this solution.

• F- (aq) + H2O (aq) HF (aq) + OH- (aq)

• Ka for HF is 7.2 x 10-4.

• Kb = 1.0 x 10-14 = 1.4 x 10-11

7.2 x 10-4

Mullis 25

Salt as a Weak Base

Kb = 1.4 x 10-11 = [HF][OH-] = x2

[F-] .30-xx2 =.30(1.4 x 10-11) x = 2.0 x 10-6 (<5% of .30)

[OH-] = 2.0x10-6

pOH = 5.69 pH= 14.00 – 5.69 = 8.31

R F- H2O (l) HF OH-

I 0.30 -- 0 ≈ 0

C -x -- +x +x

E 0.30 - x -- + x x

Mullis 26

Salt as a Weak Acid• Find pH of a 0.10 M NH4Cl solution.

• Cl- does not change pH.

• NH4+ produces H+ in water. It is the conjugate

acid of NH3.

• NH4+ (aq) NH3 (aq) + H+ (aq)

• Kb for NH3 is 1.8 x 10-5.

• Ka for NH+ = 1.0 x 10-14 = 5.6 x 10-10

1.8 x 10-5

Mullis 27

Salt as a Weak Acid

Kb = 5.6 x 10-10 = [NH3][H+] = x2

[NH4+] .10-xx2 =.10(5.6 x 10-10 ) x = 7.5 x 10-6 (<5% of .10)

[H+] = 7.5 x 10-6 pH = -log(7.5 x 10-6 ) = 5.13

R NH4+ NH3 + H+

I 0.10 0 ≈0

C -x +x +x

E 0.10 - x + x x

Mullis 28

Acid-Base Properties of SaltsSalt Example pH+ from strong base

- from strong acidNaCl neutral

+ from strong base

- from weak acidNaC2H3O2 Anion acts as

basebasic

+ conj. acid fm. weak base

- from strong acidNH4Cl Cation acts as

acidacidic

+ conj. acid fm. weak base

- conj. base fm. weak acidNH4C2H3O2 Anion acts as

base, cation as acid

Acidic if Ka> Kb

+ from metal ion

- from strong acidFeCl3

Hydrated cation acts as acid

acidic

Mullis 29

Practice1. A 0.15 M solution of a weak acid is 3.0% dissociated.

Calculate Ka.

2. What are the major species present in a 0.150 M NH3 solution? Calculate the [OH-] and the pH of this soln.

3. Calculate the pH of a 5.0 x 10-3 M solution of H2SO4.

4. Sodium azide is sometimes added to water to kill bacteria. Calculate the concentration of all species in a 0.010 M solution of NaN3. The Ka value for HN3 (hydrazoic acid) is 1.9 x 10-5.

5. Calculate the pH of 0.10 M CH3NH3Cl. Kb of CH3NH3+

is 4.38 x 10-4.