Acids and Bases

They’re Everywhere

Definitions

• Arrhenius Acids– Acids produce hydrogen ions in solution– Bases produce hydroxide ions in solution– Only have one type of base, hydroxide– First description of what an acid/base was

Definitions• Bronsted-Lowry Model

– Acid is a proton donor– Base is a proton acceptor

H---Cl + H---O---H (H---O---H)+1 + Cl-1

H

Water is the base (proton acceptor)HCl is the acid (proton donor)

+ +1 -1

Conjugate Pairs• The reaction that occurs is really an equilibrium• The H3O+1 is called the hydronium ion

H---Cl + H---O---H (H---O---H)+1 + Cl-1

H

HCl + H2O Acid Base

H3O+1 + + Cl-1

Acid Base

These are conjugate acid base pairs.Water is a base on one side and a acid on the other, this is called amphoteric

Quick Practice

• Strong acidHNO3 + H2O ↔ H3O+ + NO3

-

– What are the conjugate pairs?• Weak acid

HF + H2O ↔ H3O+ + F-– What are the conjugate pairs?

Weak Acids• Weak acids partially dissociate in water:• HA(aq) H+(aq) + A- (aq)

– Bronsted-Lowry: HA + H2O H3O+ + A-

• What is Kc for this dissociation?

• Kc in a weak acid case is better known as Ka, or the acid-dissociation constant.

• The larger the value of Ka, the stronger the acid.

Acid Dissociation Constant• HA + H2O H3O+1 + A-

• Generalized expression– The equilibrium expression for this process

Ka = [H3O+1 ] [A-] = acid dissociation constant [HA]Note: the [water] is a constant (concentration of a

pure solid or pure liquid is not included in an equilibrium expression)

Water does play an important part in the dissociation of the acid!

Acid Strength

• HA(aq) + H2O H3O+1(aq) + A-

(aq)

– A strong acid lies to the right Ka >>1• [H+1] [HA]0

– A weak acid lies to the left Ka<<1• [H+1] << [HA]0

– Typical Ka (dissociation) constants for weak acids• HF 7.2 x 10-7

• HC2H3O2 1.8 x 10-5

Conjugate Pairs and Strength

• HA(aq) + H2O H3O+1(aq) + A-

(aq)

– Strong acids have weak conjugate bases– Strong bases have weak conjugate acids

http://www.chem.ubc.ca/courseware/pH/index.html

Weak Acid

Acid Strength• In an acid, the strength of the bond between

the acidic hydrogen and the other atom (H-X) determines how strong the acid is.

• In general, the strength of an H-X bond weakens as atoms get bigger.– So, going down a group, the strength of an acid

increases.– HF < HCl < HBr < HI

Acid Strength• Going across a row, bond strengths don’t

change all that much. So, bond polarity is the major factor – the more polar the bond, the stronger the acid

• Period 2: CH4 < NH3 < H2O < HF

Acid-Base Behavior and Chemical StructureBinary Acids

Acid-Base Behavior and Chemical StructureFactors That Affect Acid StrengthConsider H-X. For this substance to be an acid we need:

• H-X bond to be polar with H+ and X- (if X is a metal then the bond polarity is H-, X+ and the substance is a base).

• the H-X bond must be weak enough to be broken,• the conjugate base, X-, must be stable.

Acid-Base Behavior and Chemical StructureBinary Acids• Acid strength increases across a period and down a group. • Conversely, base strength decreases across a period and

down a group.• What differences in atomic structure account for these

variations?

• The kernel charge increases across a period. Therefore, the nonmetallic element has a stronger pull on the shared electron pair and H is more easily ionized.

• As you go down a group, the strength of the H-X bond weakens as the X has more shells and its size increases. Therefore, H is more easily ionized

Acid-Base Behavior and Chemical StructureOxyacids• Oxyacids contain O-H bonds.• All oxyacids have the general structure Y-O-H.

• The strength of the acid depends on Y and the atoms attached to Y.– If Y is a metal (low electronegativity), then the substances

are bases.– If Y has intermediate electronegativity (e.g. I, EN = 2.5), the

electrons are between Y and O and the substance is a weak oxyacid.

Acid-Base Behavior and Chemical Structure

Oxyacids– If Y has a large electronegativity (e.g. Cl, EN = 3.0), the

electrons are located closer to Y than O and the O-H bond is polarized to lose H+.

– The number of O atoms attached to Y increase the O-H bond polarity and the strength of the acid increases (e.g. HOCl is a weaker acid than HClO2 which is weaker than HClO3 which is weaker than HClO4 which is a strong acid).

Acid-Base Behavior and Chemical StructureOxyacids

Carboxylic Acids• On a similar note,

carboxylic acids contain –OH groups, but are acids, because of the additional attached oxygen “aldehyde” group on the final carbon in the chain.

• Carboxylic acids are also stabilized by resonance once the hydrogen goes away.

B-l Acids and Bases

• Weak acids and Bases• Know the difference between a Bronsted-

Lowry acid/base and an Arrhenius acid and base

Strong Acid

• Example 0.10 M HNO3 • H2O H3O+ + OH-

• There are two sources of H+

– The nitric acid and water• Since [H+] >>[OH-] in 0.1M nitric

– Autoionization of water is insignificant– All the H+ is from HNO3

• [H+] = 0.10M pH = -log 0.1 = 1.0

Strong Acid

• The acid contributes all the [H+] • Example 1.0 x 10-10 HCl

– The [H+] from autoionization (1 x 10-7M) is much higher.

pH = - log 1 x 10-7 = 7

Weak Acids

• Treat like any equilibrium problem• What is pH of a 1.00M HF solution• Kc = 7.2 x 10 –4 • Kw = 1.0 x 10-14

– Since the Kc is so much bigger than the Kw,– HF is the major source of H+

pH of 1.0 M HF SolutionHF + H2O H3O+ + F-

I 1.0 0(1 x 10-7) 0C -x +x +xE 1.0 – x x x

Ka = 7.2 x 10-4 = [F- ] [H+] = x • x = x2 [HF] 1 – x 1

Assume x is small (5% rule) X = 2.7 x 10-2 = [H+]

Check for 5% Rule• Ka = x2 x2

[HA]o – x [HA]

Ka • [HA] x2

(Ka • [HA])1/2 x

X x 100 5% [HA]0

• X = 2.7 x 10-2 x 100 5%

1

Weak Acid Equilibrium• List the Major species in solution

– Don’t forget water as an acid source!• Choose the species that can produce H+

– Write balanced equation• Using the K values, choose dominate source H+ • Write the equilibrium expression for dominate H+ • Do “ICE” and solve using 5% rule• Verify 5%• Calculate [H+ ] and pH

Polyprotic Acids• When acids are polyprotic, like the triprotic H3PO4,

where all three protons are weak acids, different Ka values are used.

• H3PO4 H2PO4- + H+ Ka1 = 7.1 x 10-3

• H2PO4- HPO4

2- + H+ Ka2 = 6.3 x 10-8

• HPO42- PO4

3- + H+ Ka3 = 4.5 x 10-13

• If Ka1 is more than 103 larger than Ka2, you can ignore Ka2 and treat it like a monoprotic acid.

Calculate pH of 0.100 M HOCl

• Ka = 3.5 x 10-8 • You Calculate the pH

– Pg 675 if you need book• pH = 4.23

pH of Weak Acid Mixture

• Calculate the pH of a mixture of 1.00 M HCN, 5.00 M HNO2 and the equilibrium concentration of [CN-1]

• HCN Ka = 6.2 x 10-10

• HNO2 Ka = 4.0 x 10-4 • H20 Kw = 1 x 10-14 How do you approach this?

Calculate pH

• Ka = 4.0 x 10-4 = [H+][NO2-]

[HNO2] HNO2 H+ + NO2

-

I 5.00 0 0C -x +x +xE 5.00 – x x x

Calculate pH

• Ka = 4.0 x 10-4 = x2 = x2

5.00 - x 5 x = [H+] = 4.5 x 10-2 M

pH = - log [H+] = 1.35

Now calculate [CN-], you now [HCN] and [H+]

Calculate [CN-]

• Ka = 6.2 x 10 –10

6.2 x 10 –10 = [CN-][H+] = [CN-][4.5 x 10-2] [CN] 1.00Solve for [CN-] = 1.4 x 10-8 M

Percent Dissociation

• % dissociation = [amount disassociated] x 100 [initial concentration]

In the HF example [H+] = 1.27 x 10-2 M x 100 = 1.27% [HF] 1.00 M

Calculate the Percent Dissociation

• 1.00 M HC2H3O2 • Left side of room

– Write on board

• 0.100 M HC2H3O2 • Right side of room

– Write on board

Percent Dissociation• For solutions of weak acids:

– the more dilute the solution– The greater the percent dissociationGeneral ProofSuppose have acid HA, with [HA]0 Dilute it to 1/10 th initial concentrationQ = (x/10)(x/10) = x2 = 1/10 Ka

[HA]/10 10 [HA]Since Q < Ka, the reaction moves to the rightAnd you get a greater percent dissociation

Ka from % Dissociation• Lactic acid is 3.7% dissociated @ 0.100 M• HC3H5O H+ + C3H5O-

• Ka = [H+][C3H5O -] [HC3H5O]3.7% = x x 100 x = 3.7 x 10 -3

[HC3H5O]

Ka = [H+][C3H5O -] = (3.7 x 10 –3) (3.7 x 10 –3) [HC3H5O] 0.100Ka = 1.4 x 10 -4

Weak acid equilibriaExample

Determine the pH of a 0.10 M benzoic acid solution at 25 oC if Ka = 6.5 x 10-5

HBz(aq) + H2O(l) H+(aq) + Bz-(aq)

[H+] [Bz-][HBz]

The first step is to write the equilibrium expression.

Ka =

Weak acid equilibria HBz H+ Bz-

Initial conc., M 0.10 0.00 0.00

Change, DM -x +x +x

Eq. Conc., M 0.10 - x x x

[H+] = [Bz-] = x

We’ll assume that [Bz-] and [H+] are negligible compared to [HBz], since the value of the Ka<< [HBz].

(6.5 x 10-5 << 0.10 M)

Weak acid equilibria

Solve the equilibrium equation in terms of x

Ka = 6.5 x 10-5 =

x = (6.5 x 10-5 )(0.10)

= 0.00255 M H+

pH = - log (0.0025 M) = 2.6

x2

0.10

Weak acid exampleNow, lets go for the exact solution

Earlier, we found that for 0.10 M benzoic acid

Ka = 6.5 x 10-5 =

X2 + 6.5 x 10-5 X - 6.5 x 10-6 = 0

Use the quadratic equation to solve for x.

X = -b + b2 - 4ac2a

x2

0.10-x

Weak acid example

X = 0.00252 M H+

pH = - log (0.00252 M) = 2.599

versus pH = - log (0.00255 M) = 2.593

In this case, there is no significant difference between our two answers. If the Ka value is more than 2 powers of 10 different than the [acid], you can ignore the change in [acid].

X = -6.5 x 10-5 + [(6.5 x 10-5)2 +4 x 6.5 x 10-6]1/2

2

Dissociation of bases, Kb The ionization of a weak base can also be

expressed as an equilibrium.

B (aq) + H2O(l) BH+(aq) +OH- (aq)

The strength of a weak base is related to its equilibrium constant, Kb.

Kb =[OH-] [BH+]

[B]

Weak base equilibriaExample

The Bz-(aq) formed in the benzoic acid solution is a weak conjugate base. Determine the pH of a 0.10 M sodium benzoate solution NaBz(aq), at 25 oC

Bz-(aq) + H2O(l) HBz(aq) + OH-(aq)

[OH-] [HBz][Bz-]

The Na+(aq) are spectator ions, and are not part of the equilibrium expression.

Kb =

Weak base equilibriaExample

The Kb value is related to the Ka value by the equation Ka x Kb = Kw = 1.0 x 10-14

[H+] [Bz-] [OH-] [HBz] = [H+] [OH-][HBz] [Bz-]

Kb = Kw / Ka = 1.0 x 10-14 / 6.5 x 10-5

= 1.5 x 10-10

Weak base equilibria Bz - OH- HBz

Initial conc., M 0.10 0.00 0.00

Change, DM -x +x +x

Eq. Conc., M 0.10 - x x x

[OH-] = [HBz] = x

We’ll assume that [HBz] and [OH-] are negligible compared to [Bz -], since the value of the Ka << [Bz -].

(1.5 x 10-10 << 0.10 M)

Weak base equilibriaSolve the equilibrium equation in terms of x

Kb = 1.5 x 10-10 =

x = (1.5 x 10-10 )(0.10)

= 3.9 x 10-6 M

pOH = - log (3.9 x 10-6 M) = 5.4

pH = 14 - pOH = 8.6

x2

0.10

Relationship Between Ka and Kb

• What happens when you multiply Ka and Kb together?

• Ka x Kb = [H+][OH-] = Kw = 1.0 x 10-14

• And, just like pH + pOH = 14.00 for strong acids/bases at standard temperature…

• pKa + pKb = pKw = 14.00

Acidic and Basic Salts• Certain ions in solution can exhibit acid/base

properties.• For example, consider the weak base, ammonia:

– NH3 + H2O NH4+ + OH-

• What if you dissolve the salt, ammonium sulfate, in water?– (NH4)2SO4 2 NH4

+ + SO42-

• Because NH4+ is the conjugate acid of NH3, when

this salt dissolves in water, the solution will become slightly acidic.

Anions and Water

• An anion that is a conjugate base of a weak acid raises the pH of a solution:– X- + H2O HX + OH-

• Example – sodium acetate in water– Ionic: Na+ + CH3COO- + H2O

Na+ + CH3COOH + OH- – Net ionic: CH3COO- + H2O CH3COOH + OH-

• The Kb of this reaction can be found using the Ka of acetic acid (Ka x Kb = Kw)

Cations and Water

• An cation that is a conjugate acid of a weak base that contains hydrogen lowers the pH of a solution:– HX+ + H2O X + H3O+

• Example – ammonium chloride– Ionic: NH4

+ + Cl- + H2O NH3 + Cl- + H3O+

– Net ionic: NH4+ + H2O NH3 + H3O+

• The Ka of this reaction can be found using the Kb of ammonia (Ka x Kb = Kw)

Some Rules

• An anion that is the conjugate base of a strong acid will not affect the pH of a solution. (Ex: Br- from HBr)

• An anion that is the conjugate base of a weak acid will cause an increase in pH (CN- from HCN)

• A cation that is the conjugate acid of a weak base will cause a decrease in pH (NH4

+ from NH3)

Some Rules

• Alkali metal cations and Ca2+, Sr2+, and Ba2+ will not affect pH (they are conjugate acids of strong bases).

• Other metals (Al3+, etc.) will cause a decrease in pH.

• When a solution contains both a cation and anion that will affect pH, the ion with the larger equilibrium constant (Ka or Kb will have the greater influence on pH).

Lewis Acids and Bases• There is yet a third definition of acids and

bases – the Lewis definition.• A Lewis base is defined to be an electron pair

donor.• A Lewis acid is defined to be an electron pair

receiver.• Understanding Lewis acids and bases requires

the use of Lewis diagrams.

Lewis Acid-Base Reactions

• Example 1: H+ and NH3

– Empty s orbital on hydrogen• Example 2: NH3 and BF3

– Empty p orbital on boron

• In order for a Lewis acid to receive an electron pair, there must be an empty orbital in the electron configuration.

Coordination Complexes• Certain transition metal cations can act as

multiple Lewis acids (where the empty orbitals are is beyond the scope of AP Chemistry)

• Fe3+ has the ability to attract six electron pairs to itself:– Reaction of Fe3+ and CN-

• This kind of compound is known as a coordination complex.

Brief Preview of Organic Reactions

• Lots of organic chemistry relies on “bonding sites” – or determining and predicting reaction mechanisms based on chemical structures.

• Example – Lewis acid/base reaction between CO2 and H2O to create H2CO3

Hydrolysis of Metal Ions• When salts dissolve in water, the metal ions

become hydrolyzed, or the water acts like a Lewis base and forms coordination compounds with the metal ions.

• This is the mechanism behind metal ions acting like acids:

• [Fe(H2O)6]3+ Fe(H2O)5(OH)2+ + H+

Lewis Acid/Base Review• Which of the following compounds can act

as Lewis acids?• NH3

• H2O• H+

• SO42-

• BCl3

Hydride Bases• We are used to seeing hydrogen ions (H+) as

acids… but there are a class of compounds called hydrides (H-), which act as bases.– Ex: NaH (ionic bond w/ H-)