1

Acids and Bases

L. Scheffler

Lincoln High School

2

Acids and Bases The concepts acids and bases were loosely defined as

substances that change some properties of water. One of the criteria that was often used was taste. Substances were classified

salty-tasting sour-tasting sweet-tasting bitter-tasting

Sour-tasting substances would give rise to the word 'acid', which is derived from the Greek word oxein, which mutated into the Latin verb acere, which means 'to make sour'

3

Acids• React with certain metals to produce

hydrogen gas.• React with carbonates and bicarbonates

to produce carbon • dioxide gas• Vinegar is a solution of acetic acid.

Citrus fruits contain citric acid.

• Have a bitter taste• Feel slippery. • Many soaps contain bases.

Bases

4

Properties of Acids Produce H+ (as H3O+) ions in water (the hydronium ion

is a hydrogen ion attached to a water molecule)

Taste sour

Corrode metals

Electrolytes

React with bases to form a salt and water

pH is less than 7

Turns blue litmus paper to red “Blue to Red A-CID”

5

Properties of Bases Generally produce OHGenerally produce OH-- ions in water ions in water

Taste bitter, chalkyTaste bitter, chalky

Are electrolytesAre electrolytes

Feel soapy, slipperyFeel soapy, slippery

React with acids to form salts and waterReact with acids to form salts and water

pH greater than 7pH greater than 7

Turns red litmus paper to blueTurns red litmus paper to blue “ “BBasicasic BBluelue””

6

Arrhenius DefinitionArrhenius

Acid - Substances in water that increase the concentration of hydrogen ions (H+).

Base - Substances in water that increase concentration of hydroxide ions (OH-).

Categorical definition – easy to sort substances into acids and bases

Problem – many bases do not actually contain hydroxides

7

Bronsted-Lowry Definition

Acid - neutral molecule, anion, or cation that donates a proton.

Base - neutral molecule, anion, or cation that accepts a proton. HA + :B HB+ + :A-

Ex HCl + H2O H3O+ + Cl-

Acid Base Conj Acid Conj Base

Operational definition - The classification depends on how the substance behaves in a chemical reaction

8

Conjugate Base - The species remaining after an acid has transferred its proton.

Conjugate Acid - The species produced after base has accepted a proton.

HA & :A- - conjugate acid/base pair

:A- - conjugate base of acid HA

:B & HB+ - conjugate acid/base pair

HB+ - conjugate acid of base :B

Conjugate Acid Base Pairs

9

Note: Water can act as acid or base

Acid Base Conjugate Acid Conjugate Base

HCl + H2O H3O+ + Cl-

H2PO4- + H2O

H3O+ + HPO4

2-

NH4+ + H2O

H3O+ + NH3

Base Acid Conjugate Acid Conjugate Base :NH3 + H2O

NH4+ + OH-

PO43- + H2O

HPO42- + OH-

Examples of Bronsted-Lowry Acid Base Systems

10

Lewis

Acid - an electron pair acceptor

Base - an electron pair donor

G.N. Lewis Definition

11

The pH Scale

pH [H3O+ ] [OH- ] pOH

12

pH and acidity1. Acidity or Acid Strength depends on Hydronium Ion

Concentration [H3O+]

2. The pH system is a logarithmic representation of the Hydrogen Ion concentration (or OH-) as a means of avoiding using large numbers and powers.

pH = - log [H3O+] = log(1 / [H3O+])

pOH = - log [OH-] = log(1 / OH-])

3. In pure water [H3O+] = 1 x 10-7 mol / L (at 20oC)

pH = - log(1 x 10-7) = - (0 - 7) = 7

4. pH range of solutions: 0 - 14

pH < 7 (Acidic) [H3O+] > 1 x 10-7 m / L

pH > 7 (Basic) [H3O+] < 1 x 10-7 m / L

13

pH and acidity

Kw = [H3O+] [OH-] = 1.0 x10-14

In water

[H3O+] = [OH-] = 1.0 x10-7

pH + pOH = 14

14

Calculating the pHpH = - log [H3O+]

Example 1: If [H3O+] = 1 X 10-10

pH = - log 1 X 10-10

pH = - (- 10)

pH = 10

Example 2: If [H3O+] = 1.8 X 10-5

pH = - log 1.8 X 10-5

pH = - (- 4.74)

pH = 4.74

15

Indicators

16

pH and acidity

The pH values of several common substances are shown at the right.

Many common foods are weak acids

Some medicines and many household cleaners are bases.

17

Neutralization An acid will neutralize a base,

giving a salt and water as products

Examples HCl + NaOH NaCl + H2O

H2SO4 + 2 NaOH Na2SO4 + 2 H2O

H3PO4 + 3 KOH K3PO4 + 3 H2O

2 HCl + Ca(OH) 2 CaCl2 + 2 H2O

18

Neutralization Calculations If the concentration of acid or base is

expressed in Molarity or mol dm-3

then:--The volume in dm3 multiplied by the

concentration yields moles. -- If the volume is expressed in cm3

the same product yields millimoles

19

Neutralization Problems If an acid and a base combine in a 1 to 1 ratio, then

the volume of the acid multiplied by the concentration of the acid is equal to the volume of the base multiplied by the concentration of the base

Vacid C acid = V base C base

If any three of the variables are known it is possible to determine the fourth

20

Neutralization Problems Example 1: Hydrochloric acid reacts with potassium hydroxide according to the following reaction:

HCl + KOH KCl + H2O If 15.00 cm3 of 0.500 M HCl exactly neutralizes 24.00 cm3 of

KOH solution, what is the concentration of the KOH solution?

Solution:Vacid Cacid = Vbase Cbase

(15.00 cm3 )(0.500 M) = (24.00 cm3 ) Cbase

Cbase = (15.00 cm3 )(0.500 M) (24.00 cm3 ) Cbase = 0.313 M

21

Neutralization Problems

Whenever an acid and a base do not combine in a 1 to 1 ratio, a mole factor must be added to the neutralization equation

n Vacid C acid = V base C base

The mole factor (n) is the number of times the moles the acid side of the above equation must be multiplied so as to equal the base side. (or vice versa)

Example

H2SO4 + 2 NaOH Na2SO4 + 2 H2O

The mole factor is 2 and goes on the acid side of the equation. The number of moles of H2SO4 is one half that of NaOH. Therefore the moles of H2SO4 are multiplied by 2 to equal the moles of NaOH.

22

Neutralization Problems Example 3: Phosphoric acid reacts with potassium hydroxide according to the following reaction:

H3PO4 + 3 KOH K3PO4 + 3 H2O If 30.00 cm3 of 0.300 M KOH exactly neutralizes 15.00 cm3 of

H3PO4 solution, what is the concentration of the H3PO4 solution?

Solution:In this case the mole factor is 3 and it goes on the acid side, since the mole ratio of acid to base is 1 to 2. Therefore

3 Vacid Cacid = Vbase Cbase

3 (15.00 cm3 )(Cacid) = (30.00 cm3 ) (0.300 M) Cacid = (30.00 cm3 )(0.300 M) (3) (15.00 cm3 ) Cacid = 0.200 M

23

Neutralization Problems Example 2: Sulfuric acid reacts with sodium hydroxide according to the following reaction:

H2SO4 + 2 NaOH Na2SO4 + 2 H2O If 20.00 cm3 of 0.400 M H2SO4 exactly neutralizes 32.00 cm3 of

NaOH solution, what is the concentration of the NaOH solution?

Solution:In this case the mole factor is 2 and it goes on the acid side, since the mole ratio of acid to base is 1 to 2. Therefore

2 Vacid Cacid = Vbase Cbase

2 (20.00 cm3 )(0.400 M) = (32.00 cm3 ) Cbase

Cbase = (2) (20.00 cm3 )(0.400 M) (32.00 cm3 )

Cbase = 0.500 M

24

Neutralization Problems Example 4: Hydrochloric acid reacts with calcium hydroxide according to the following reaction:

2 HCl + Ca(OH)2 CaCl2 + 2 H2O If 25.00 cm3 of 0.400 M HCl exactly neutralizes 20.00 cm3 of

Ca(OH)2 solution, what is the concentration of the Ca(OH)2 solution?

Solution:In this case the mole factor is 2 and it goes on the base side, since the mole ratio of acid to base is 2 to 1. Therefore

Vacid Cacid = 2 Vbase Cbase

(25.00 cm3) (0.400) = (2) (20.00 cm3) (Cbase)

Cbase = (25.00 cm3 ) (0.400 M) (2) (20.00 cm3 )

Cbase = 0.250 M

25

Acid Base DissociationAcid-base reactions are equilibrium processes.

The relationship between the relative concentrations of the reactants and products is a constant for a given temperature. It is known as the Acid or Base Dissociation Constant.

The stronger the acid or base, the larger the value of the dissociation constant.

]B[:

][OH [HB] K O][H K

[HA]

][H ]A[: K O][H K

constant. is solutions dilute in O][H

][H OH

:Note

O][H ]B:[

]OH][HB[ K

O][H HA][

]OH][ A[: K

waterin base a For waterin acid an For

-

-

b2eq

-

a2eq

2

3

2-

-

eq2

3-

eq

26

Acid Strength Strong Acid - Transfers all of its protons to water;

- Completely ionized; - Strong electrolyte; - The conjugate base is weaker and has a negligible tendency to be protonated.

Weak Acid - Transfers only a fraction of its protons to water;

- Partly ionized; - Weak electrolyte; - The conjugate base is stronger, readily accepting protons from water

As acid strength decreases, base strength increases. The stronger the acid, the weaker its conjugate base The weaker the acid, the stronger its conjugate base

27

Acid Dissociation ConstantsDissociation constants for some weak acids

28

Base Strength Strong Base - all molecules accept a proton; - completely ionizes; - strong electrolyte; - conjugate acid is very weak, negligible tendency to donate protons.

Weak Base - fraction of molecules accept proton; - partly ionized; - weak electrolyte; - the conjugate acid is stronger. It more readily donates protons.

As base strength decreases, acid strength increases. The stronger the base, the weaker its conjugate acid. The weaker the base the stronger its conjugate acid.

29

Common Strong Acids/Bases

Strong BasesStrong BasesSodium Hydroxide

Potassium Hydroxide

*Barium Hydroxide

*Calcium Hydroxide

*While strong bases they are not very soluble

Strong AcidsStrong AcidsHydrochloric Acid

Nitric Acid

Sulfuric Acid

Perchloric Acid

30

Water has the ability to act as either a Bronsted- Lowry acid or base.

Autoionization – spontaneous formation of low concentrations of [H+] and OH-] ions by proton transfer from one molecule to another.

Equilibrium Constant for Water

7--

o14--w

o14--3w

-3

22c

22

-3

c

10 x 0.1 ][OH ][H

: WaterPure In

C)25(at 10 x 0.1 ][OH ][H K

C)25(at 10 x 0.1 ][OH ]O[H K

][OH ]O[H O]H[K

O]H[

][OH ]O[H K

Water as an Equilibrium System

31

Weak Acid EquilibriaA weak acid is only partially ionized.Both the ion form and the unionized form exist at equilibrium HA + H2O H3O+ + A-

The acid equilibrium constant is

Ka = [H3O+ ] [A-] [HA]

Ka values are relatively small for most weak acids. The greatest part of the weak acid is in the unionized form

32

Weak Acid Equilibrium Constants

Sample problem . A certain weak acid dissociates in water as follows: HA + H2O H3O+ + A-

If the initial concentration of HA is 1.5 M and the equilibrium concentration of H3O+ is 0.0014 M. Calculate Ka for this acidSolutionKa = [H3O+ ] [A-]

[HA] I C E Substituting

[HA] 1.5 -x 1.5-x Ka = (0.0014)2 = 1.31 x 10-6 [A-] 0 +x x 1.4986[H3O+ ] 0 +x x

x = 0.0014 1.5-x = 1.4986

33

Weak Base EquilibriaWeak bases, like weak acids, are partially ionized. The degree to which ionization occurs depends on the value of the base dissociation constantGeneral form: B + H2O BH+ + OH-

Kb = [BH+][OH-] [B]

ExampleNH3 + H2O NH4

+ + OH-

Kb = [NH4+][ OH-]

[NH3]

34

Weak Base Equilibrium Constants

Sample problem . A certain weak base dissociates in water as follows: B + H2O BH+ + OH-

If the initial concentration of B is 1.2 M and the equilibrium concentration of OH- is 0.0011 M. Calculate Kb for this baseSolutionKb = [BH+ ] [OH-]

[B]

I C E Substituting[B] 1.2 -x 1.2-x Kb = (0.0011)2 = 1.01 x 10-6 [OH-] 0 +x x 1.1989[BH+ ] 0 +x x

x = 0.0011 1.2-x = 1.1989

35

Weak Acid Equilibria Concentration Problems

Problem 1. A certain weak acid dissociates in water as follows: HA + H2O H3O+ + A-

The Ka for this acid is 2.0 x 10-6. Calculate the [HA] [A-], [H3O+ ] and pH of a 2.0 M solutionSolutionKa = [H3O+ ] [A-] = 2.0 x 10-6

[HA] I C E Substituting

[HA] 2.0 -x 2.0-x Ka = x2 = 2.0 x 10-6

[A-] 0 +x x 2.0-x[H3O+ ] 0 +x x If x <<< 2.0 it can be dropped

from the denominator

The x2 = (2.0 x10-6)(2.0) = 4.0 x10-6 x = 2.0 x 10-3

[A-] = [H3O+ ] = 2.0 x10-3 [HA] = 2.0 - 0.002= 1.998

pH = - log [H3O+ ] =-log (2.0 x 10-3) = 2.7

36

Weak Acid Equilibria Concentration Problems

Problem 2. Acetic acid is a weak acid that dissociates in water as follows: CH3COOH + H2O H3O+ + CH3COO-

The Ka for this acid is 1.8 x 10-5. Calculate the [CH3COOH],[CH3COO-] [H3O+ ] and pH of a 0.100 M solutionSolution

Ka = [H3O+ ] [CH3COO- ] = 1.8 x 10-5

[CH3COOH] I C E Substituting

[CH3COOH] 0.100 -x 0.100-x Ka = x2 = 1.8 x 10-5

[CH3COO- ] 0 +x x 0.100-x[H3O+] 0 +x x If x <<< 0.100 it can be dropped

from the denominatorThe x2 = (1.8 x10-5)(0.100) = 1.8 x10-6 x = 1.3 x 10-3

[CH3COO--] = [H3O+ ] = 1.3 x10-3 [CH3COOH ] = 0.100 - 0.0013 = 0.0987

pH = - log [H3O+ ] =-log (1.3 x10-3) = 2.88

37

Weak Base EquilibriaExample1. Ammonia dissociates in water according to the following equilibrium

NH3 + H2O NH4+ + OH-

Kb = [NH4+][ OH-] = 1.8 x 10-5

[NH3]Calculate the concentration of [NH4

+][ OH-] [NH3 ]and the pH of a 2.0M solution.

I C E Substituting[NH3] 2.0 -x 2.0-x Kb = x2 = 1.8x 10-5

[OH-] 0 +x x 2.0-x[NH4

+] 0 +x x If x <<< 2.0 it can be dropped from the denominator

The x2 = (1.8 x10-5)(2.0) = 3.6 x10-5 x = 6.0 x 10-3 [OH-] = [NH4

+] = 6.0 x10-3 [NH3] = 2.0- 0.006= 1.994pOH = - log [OH-] =-log (6.0 x10-3) = 2.22pH = 14-pOH = 14-2.22 = 11.78

38

Amphoteric Solutions A chemical compound able to react with both an

acid or a base is amphoteric.    Water is amphoteric. The two acid-base couples

of water are H3O+/H2O and H2O/OH-

It behaves sometimes like an acid, for example

And sometimes like a base :

Hydrogen carbonate ion HCO3- is also amphoteric,

it belongs to the two acid-base couples H2CO3/HCO3

- and HCO3-/CO3

2-

39

Common Ion EffectThe common ion effect is a consequence of Le Chatelier’s Principle When the salt with the anion (i.e. the conjugate base) of a weak acid is added to that acid,

It reverses the dissociation of the acid.Lowers the percent dissociation of the

acid.A similar process happens when the salt with the cation (i.e, conjugate acid) is added to a weak base.These solutions are known as Buffer Solutions.

40

Buffer Solutions - Characteristics A solution that resists a change in pH.

It is pH stable. A weak acid and its conjugate base

form an acid buffer. A weak base and its conjugate acid

form a base buffer. We can make a buffer of any pH by

varying the concentrations of the acid/base and its conjugate.

41

Buffer Solution CalculationsCalculate the pH of a solution that is 0.50 M CH3COOH and 0.25 M NaCH3COO.

CH3COOH + H2O H3O+ + CH3COO- (Ka = 1.8 x 10-5)

Solution

Ka = [H3O+ ] [CH3COO- ] = 1.8 x 10-5

[CH3COOH] I C E . Substituting

[CH3COOH] 0.50 -x 0.50-x Ka = x (0.25+x) = 1.8 x 10-5

[CH3COO-] 0.25 +x 0.25+x (0.50-x)[H3O+] 0 +x x If x <<< 0.25 it can be dropped from both

expressions in ( ) since adding or subtracting a small amount will not significantly change the value of the ratio

Then the expression becomes x(0.25)/(0.50) = 1.8 x 10-5

x = 3.6 x 10-5 = [H3O+]

pH = - log [H3O+] =-log(3.6 x 10-5 ) = 4.44

42

Buffer Solutions - Equations

1. Ka = [H3O+] [A-] [HA]

2. [H3O+] = Ka [HA] [A-]

The [H3O+] depends on the ratio [HA]/[A-] Taking the negative log of both sides of equation

2 above

pH = -log(Ka [HA]/[A-])

pH = -log(Ka) - log([HA]/[A-])

pH = pKa + log([A-]/[HA])

43

Henderson Hasselbach Equation

pH = pKa + log([A-]/[HA]) pH = pKa + log(Base/Acid) This expression is known as the Henderson-

Hasselbach equation. It provides a shortcut from using the I.C.E. model for buffer solutions where the concentration of both [A-] and [HA] are significantly greater than zero.

44

Using the Henderson -Hasselbach Equation

pH = pKa + log([A-]/[HA]) Example

Calculate the pH of the following of a mixture that contains 0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)

HC3H5O3 + H2O H3O+ + C3H5O3-

Solution

Using the Henderson-Hasselbach equation

pH = - log (1.4 x 10-4) + log ( 0.25/0.75 )

= 3.85 + (-0.477) = 3.37

45

Henderson-Hasselbach Equation and Base BuffersFor a base a similar expression can be written

pOH = pKb + log ([BH+] / [B])

pOH = pKb + log ([Acid] / [Base])

Example: Calculate the pH of a solution that contains 0.25 M NH3 and 0.40 M NH4Cl

(Kb = 1.8 x 10-5)

Solution

pOH = - log(1.8 x 10-5) + log (0.40/0.25)

= 4.74 + 0.204 = 4.94

pH = 14 - pOH = 14 - 4.94 = 9.06

46

Henderson-Hasselbach Equation & Base Buffers Methyl amine is a weak base with a Kb or 4.38 x 10-4

CH3NH2 + H2O CH3NH3+ + OH-

Calculate the pH of a solution that is 0.10 M in methyl amine and 0.20 M in methylamine hydrochloride.

pOH = pKb + log ([BH+] / [B])

Solution pOH = -log (4.38 x 10-4) + log (0.20 / 0.10)

= 3.36 + 0.30 = 3.66

pH = 14- 3.66 = 10.34

47

Additional Buffer ProblemsHow many grams of sodium formate, NaCHOO, would have to

be dissolved in 1.0 dm3 of 0.12 M formic acid, CHOOH, to

make the solution a buffer of pH 3.80? Ka= 1.78 x 10-4

pH = pKa + Log ([A-]/[HA])

Solution 3.80 = -log (1.78 x 10-4) + Log [A-] - Log [0.12]

3.80 = 3.75 + Log [A-] - (-0.92)

Log [A-] = 3.80 - 3.75 - 0.92 = - 0.87

[A-] = 10-0.87 = 0.135 mol dm-3

The molar mass of NaCHOO = 23+12+1+2(16) = 58.0 gmol-1

So (0.135 mol dm-3)(58.0 gmol-1 ) = 7.8 grams per dm-3

48

Relationship of Ka, Kb & Kw HA weak acid. Its acid ionization is

A- is the conjugate base Its base ionization is

Multiplying Ka and Kb and canceling like terms

49

Titration Curves A graph showing pH vs volume of acid

or base added The pH shows a sudden change near

the equivalence point The Equivalence point is the point at

which the moles of OH- are equal to the moles of H3O+

50

Strong acid-strong base Titration Curve

At equivalence point, Veq:

Moles of H3O+ = Moles of OH-

There is a sharp rise in the pH as one approaches the equivalence point

With a strong acid and a strong base, the equivalence point is at pH =7

Neither the conjugate base or conjugate acid is strong enough to affect the pH

15_327

01.0

Vol NaOH added (mL)

50.0

7.0

13.0

pH

100.0

Equivalencepoint

pH

cm3 base added

51

Weak acid-strong base Titration Curve

The increase in pH is more gradual as one approaches the equivalence point

With a weak acid and a strong base, the equivalence point is higher than pH = 7

The strength of the conjugate base of the weak acid is strong enough to affect the pH of the equivalence point

52

Buffered Weak Acid-Strong Base Titration Curve

The initial pH is higher than the unbuffered acid

As with a weak acid and a strong base, the equivalence point for a buffered weak acid is higher than pH =7

The conjugate base is strong enough to affect the pH

53

Polyprotic Weak Acids Polyprotic acids have more than one hydrogen

that can be neutralized Phosphoric Acid has three hydrogen ions.

H3PO4 + H2O H3O+ + H2PO4-

H2PO4- +H2O H3O+ + HPO4

2-

HPO42- +H2O H3O+ + PO4

3-

At given pH only one acid form and one conjugate base predominate

pH 0-4.7: H3PO4 and H2PO4-

pH 4.7-9.7: H2PO4- and HPO4

2-

pH 9.7-14: HPO42- and PO4

3-

54

Polyprotic Weak Acid-Strong Base Titration Curve

Phosphoric Acid has three hydrogen ions.

There are three equivalence points

H3P04 + H2O H3O+ + H2PO4-

H2PO4- +H2O H3O+ + HPO4

2-

HPO42- +H2O H3O+ + PO4

3-

55

Salts A salt is the neutralization product of an acid and

a base. The anion comes from the acid and the cation

from the base. Examples

HCl + NaOH NaCl + H2O.

H2SO4 + 2 KOH K2SO4 + H2O.

56

Salts If a salt is the result of a

-- Strong acid and a strong base, the pH is near neutral.

HCl + NaOH NaCl + H2O.

-- Weak acid and a strong base, the pH will be greater than 7.

CH3COOH + NaOH NaCH3COO + H2O

-- Strong acid and a weak base, the pH will be lower than 7.

NH4OH + HCl NH4Cl +H2O

-- Weak acid and a weak base, the pH depends on whether the acid or the base is stronger.