acids and bases: overview common uses of acids and bases

1

Acids and Bases: Overview

• Definitions of acids and bases

• Equilibria involving acids and bases

• Conjugate acid‐base pairs

• Autoionisation of water

• The “p” scale (pH, pOH, pKa, pKb)

• Weak acids and bases

• Calculations involving pKa and pKb• Strong acids and bases

• Polyprotic acids

• Salts of acids and bases

• Buffers

• Indicators

• Titrations

Chem 1102 1B 1

Common Uses of Acids and Bases

Chem 1102 1B 2

Common Types of Acids

•Many acids are oxy acids where the proton is attached to an oxygen atom.

Chem 1102 1B 3

Organic acids Organic acids have a carboxyl group

Hydrohalic Acids H-F, H-Cl, H-Br and H-I

Oxy Acids

Common Types of Bases

•Organic bases have a nitrogen atom with a lone pair

Chem 1102 1B 4

Organic bases

LiOH, NaOH, K2O etc.Hydroxides and oxides

Acids and Bases

•Acids were originally recognised by their sour taste, e.g., lemon (citric acid).

•Bases usually by their bitter taste and slippery feel, e.g., drain uncloggers (caustic soda or sodium hydroxide).

Chem 1102 1B 5

Arrhenius was the first to recognisethe nature of acids and bases.

Strong acids and bases:

HCl (aq) + NaOH (aq) ∆H0 = -56 kJ mol-1

HI (aq) + NaOH (aq) ∆H0 = -56 kJ mol-1

HNO3 (aq) + NaOH (aq) ∆H0 = -56 kJ mol-1... etc

Why do these all have the same heat of reaction? Because the only species that actually react are:

H+ (aq) + OH- (aq) = H2O(l) ∆H0 = -56 kJ mol-1

Definitions•Arrhenius: H+ (aq) + OH‐ (aq) ⇌ H2O (l)

– ACID: H+ producer in aqueous solution i.e. AH– BASE: OH‐ producer i.e. MOH

•Brønsted ‐ Lowry:•H+ + A‐⇌ HA

– ACID: proton donor (H+) e.g. HCl– BASE: proton acceptor e.g. NH3

•Lewis•A + :B⇌ A:B

•ACID: electron pair acceptor e.g. BF3– BASE: electron pair donor e.g. NH3

Chem 1102 1B 6

2

NaOH and Dry Ice

•1 M NaOH pH ≈ 12

CO2 (s) ⇌ CO2 (g) ⇌ CO2 (aq) ⇌ H+ (aq) + HCO3‐ (aq)

•H+ (aq) + OH‐ (aq) → H2O (l)

Chem 1102 1B 7

Electron-pair acceptor (Lewis acid)

DEMO: 5.10 dry ice in water plus 1M HCL and NaOH + litmus paper

Brønsted‐Lowry Model•acid = proton donor

•base = proton acceptor

Chem 1102 1B 8Note that water can act as an acid or a base - it is amphoteric

Acids, Bases & Equilibrium

•In water: an acid (e.g., HCl) ionises to produce H+ (aq)

– NB: Actually H3O+ (aq), but we usually just write H+ (aq)

Chem 1102 1B 9

H3O+

+

Generally exists as [H(H2O)n]+ (aq)


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

•A STRONG acid has equilibrium to the right (HA mostly dissociated)•A WEAK acid has equilibrium to the left (HA mostly intact)

•Equilibrium Equation:

•Ka is the ACID DISSOCIATION CONSTANT

Note: it has the same form as if reaction was a simple dissociation: HA (aq) ⇌ H+ (aq) + A- (aq)

Chem 1102 1B 10

Conjugate Acid‐Base Pairs

•NH4+ is the conjugate acid of NH3

•NH3 is the conjugate base of NH4+

•A conjugate base has one less proton than its conjugate acid.

•HSO4–: conjugate base is SO4

2–

conjugate acid is H2SO4

•H2SO4 is a dibasic or diprotic acid:

Chem 1102 1B 11

H2SO4 (aq) + H2O (l) ⇌ H3O+ (aq) + HSO4– (aq)

HSO4– (aq) + H2O (l) ⇌ H3O+ (aq) + SO4

2– (aq)

(lies ~100% to right)

Questions:

Chem 1102 1B 12

Write the formula of the Write the formula of the conjugate bases conjugate acids

H3O+ OH-

H2SO4 H2O

HClO4 CN-

CH3COOH NH3

HPO42– HPO4

2–

3

Acid‐Base Reactions

Chem 1102 1B 13

The positions of these equilibria determine the strengths of the acids.

Autoionisation of Water

•H2O (l) ⇌ H+ (aq) + OH– (aq)

•Equilibrium constant given special symbol:Kw = [H+][OH–]

NB: [H2O (l)] = constant

At 25 °C: Kw = 1.0 × 10‐14 ← REMEMBER THIS!

– Neutral solution: [H+] = [OH– ] = 10‐7 M– Acidic solution: [H+] > 10‐7 M– Basic : [H+] < 10‐7 M

Chem 1102 1B 14

This is why pure water is a weak electrolyte

The pH Scale

Because the concentrations of acids and bases can vary over many orders of magnitude, it is convenient to define a logarithmic scale to compare them:

pH = −log10[H+]

e.g. If [H+] = 1.0 × 10‐6 M then pH = ‐log(10‐6) = ‐ (‐6) = 6.0

Chem 1102 1B 15

The ‘p’ ConventionpH = – log10[H

+]pOH = – log10[OH

– ]pKw= – log10[Kw] = 14 at 25 °C

Acid : pH < 7Neutral: pH = 7Basic: pH > 7

Since Kw = [H+][OH– ]:log10 Kw = log10 [H

+] + log10 [OH– ]

– log10 [H+] – log10 [OH

– ] = – log10 Kw

pH + pOH = 14pOH = 14 – pH

Chem 1102 1B 16

Note that an acidic solution contains some OH-, and a

basic solution contains someH3O+

Temperature Dependence of pH

Kw = 1.0 × 10‐14 only at 25 °C

For T > 25 °C, Kw > 10‐14 ⇒ pH + pOH < 14 if T > 25 °C

For T < 25 °C, Kw < 10‐14 ⇒ neutral pH > 7 if T > 25 °CChem 1102 1B 17

H2O

H+ + OH-

ΔH0 = 56 kJ mol-1Ea

The pH Scale

Chem 1102 1B 18

Remember, this is a log scale -so 0.1 M HCl is ~100 × more acidic than 0.1 M acetic acid.

4

Autoionisation of Water

•H2O (l) ⇌ H+ (aq) + OH– (aq)

Equilibrium constant given special symbol:

Kw = [H+][OH–]NB: [H2O (l)] = constant

At 25 °C: Kw = 1.0 × 10-14

– Neutral solution: [H+] = [OH– ] =10-7 M– Acidic solution: [H+] > 10-7 M– Basic : [H+] < 10-7 M

Chem 1102 1B 19

This is why pure water is a weak electrolyte


•HA (aq) + H2O (l) ⇌ H3O+ (aq) + A‐ (aq)

A STRONG acid has equilibrium to the right (HAmostly dissociated)

A WEAK acid has equilibrium to the left (HAmostly intact)

Equilibrium Equation:

Ka is the ACID DISSOCIATION CONSTANT

– Note: it has the same form as if reaction was a simple dissociation:

HA (aq) ⇌ H+ (aq) + A‐ (aq)

Chem 1102 1B 20

Weak Acids

Most acids and bases are weak ‐ they do not completely ionise in water.

•The more positive the pKa, the weaker the acid (and the stronger the conjugate base).

Chem 1102 1B 21

pKa = – log10Ka

HA(aq) H+(aq) + A–(aq)

Acid dissociation constant

Acetic Acid: Vinegar

acetic (ethanoic) acid, CH3COOH (HAc for short):

Ka = 10‐4.7 = 1.99 x 10‐5

pKa = 4.7

pH of 0.1 M solution of acetic acid > 1

(pH would be –log(0.1) = 1

only if it were completely ionised)

Chem 1102 1B 22

Relationship Between Ka and pKa

The larger the value of Ka the stronger the acid and the lower the value of pKa.

•Ka = 1.02 × 10‐2 then pKa = ‐log10(1.02 × 10‐2) = 1.991

•pKa = 1.991 then Ka = 10‐1.991 = 1.02 × 10‐2

Chem 1102 1B 23

Examples of Ka Values

Chem 1102 1B 24

Note effect of changing the organic

part of carboxylic acids

5

Example

Find the pH of 0.1 M acetic acid, CH3COOH (HAc)

DATA: pKa = 4.7, Ka = 10‐4.7

Chem 1102 1B 25

HAc (aq) ⇌ H+ (aq) + Ac– (aq) Ka = 10-4.7

H2O (l) ⇌ H+ (aq) + OH– (aq) Ka = 10-14

Concns. (M) HAc (aq) ⇌ H+ (aq) + Ac– (aq)before eqm: 0.1 0 0

after: 0.1 – x x xmake x either

H+ or Ac–

Example (continued)

Chem 1102 1B 26

Since the equilibrium constant is very small we assume

x << 0.1, i.e. ( 0.1 – x ) ≈ 0.1 10-4.7 ≈ x 2 / 0.1

x2 ≈ 0.1 × 10-4.7 = 10-5.7

pH = – log10 [H+ ] = – log10 x = – log10 √10-5.7

= 2.9Check: x = 10-2.9 = 1.4 × 10-3 << 0.1,

( 0.1 – x ) = 0.0986 M

which is less than 5 % different to our assumption that 0.1 - x = 0.1, so our assumption was valid.

[Note that the exact answer found by solving a quadratic would be x = 0.098579]

Demo: Strong Acids versusWeak Acids

Chem 1102 1B 27DEMO: 5.9 strong and weak acid with metal

ExampleFind % ionisation of 0.50 M HF (pKa = 3.1).

Chem 1102 1B 28

Conc (M) HF (aq) ⇌ H+ (aq) + F– (aq)

before eqm.: 0.50 0 0

after: 0.50 – x x x

Hence Ka = 10-3.1 ≈ x2 / (0.50 - x)

If x << 0.50, 10-3.1 ≈ x2 / 0.50

x2 = 0.50×10-3.1

x = 2×10– 2 M (check: indeed x << 0.50 M)

% ionisation = x / 0.50 × 100 = 4 % (indeed a weak acid!)

Weak Bases

Ionisation of a weak base:NH3 (aq) + H2O (l) ⇌ NH4

+ (aq) + OH– (aq)

Equilibrium constant is called base ionisation constant, Kb :

•We can calculate pOH and hence pH, given Kb.

Chem 1102 1B 29

Weak Bases

Chem 1102 1B 30

Note effect of changing what the N is attached

to

6

pKa and pKbFor conjugate systems (Brønsted‐Lowry acid‐base pairs)

As acid (HA):HA (aq) ⇌ H+ (aq) + A– (aq)

As conjugate base (A– ):A– (aq) + H2O (l) ⇌ HA (aq) + OH– (aq)

•Hence only need values of pKa , since pKb = 14 – pKa

Chem 1102 1B 31

pKa + pKb = 14.0

ExampleFind the pH of 10–2 M NaHCO2 (pKa of formic acid (HCO2H) is 4.1)(check: it will be basic so expect pH > 7)

Chem 1102 1B 32

HCO2- (aq) + H2O (l) ⇌ OH- (aq) + HCO2H (aq)

before: 10-2 0 0

after: 10-2 – x x x

pKb = 14 – pKa = 14 – 4.1 = 9.9

Example (continued)

Chem 1102 1B 33

(indeed x << 10-2 )

x = [OH-] so pOH = 5.95 ; pH = 14.0 – 5.95 = 8.05

= 8.1 (to one significant figure)

When we say 10-2 M, experimentally we mean

0.95 × 10-2 M < conc < 1.05 × 10-2 M

Q. What would the pH be for 1.05 × 10-2 M NaHCO2 ?0.95 × 10-2 M NaHCO2 ?

Weak Acids and BasesHA (aq) ⇌ H+ (aq) + A– (aq)

Acid dissociation constantpKa = – log10Ka

More positive pKa⇒ weaker acid (and stronger conjugate base)

Chem 1102 1B 34

B (aq) + H2O (aq) ⇌ HB+ (aq) + OH– (aq) Base ionisation constant

pKb = – log10KbMore positive pKb⇒ weaker base (and stronger

conjugate acid)

Relative Strengths of Acids and Bases

Chem 1102 1B 35


Strongest acids lose their protons easily:For H-X bonds, larger X → weaker H-X bonds → stronger acids.For H-O-X bonds, more electronegative X → weaker H-O bonds → stronger acid

Chem 1102 1B 36

7


Strongest bases hold on to protons strongly:

An acid-base equilibrium always lies in the direction of the weaker acid and weaker base.

Chem 1102 1B 37

HCN (aq) + CH3COO– (aq) ⇌ CH3COOH (aq) + CN– (aq)

weaker acid & weaker base ← stronger acid & stronger base

pKa = 9.2 pKb = 9.3 pKa = 4.7 pKb = 4.8

Strong Acids and Bases

Completely ionise in water: e.g. HCl (aq) ⇌ H+ (aq) + Cl– (aq)

‐ equilibrium lies completely to right, Ka ≈ ∞Strong acids:

H2SO4 , HCl, HBr, HI, HNO3, HClO4

Strong bases:All hydroxides of Groups 1 & 2 (except Be):NaOH, Ca(OH)2, …

Chem 1102 1B 38

Learn these!

Example

What is the pH of a 0.1 M HCl solution?

Chem 1102 1B 39

What are the sources of [H+]?HCl (aq) → H+ (aq) + Cl- (aq) [H+] = 0.1 MH2O (l) ⇌ H+ (aq) + OH- (aq) [H+] = 10-7 M

The water equilibrium will be driven to the left by the high value of H+ from the HCl hence [H+] from auto-ionisation of water will be negligible and similarly [OH-] is negligible.

Thus [H+] = 0.1 M and pH = – log10 [H+] = 1.0

Example:

What is the pH of a 0.002 M NaOH solution?

Sodium hydroxide is fully soluble thus totally ionised

Chem 1102 1B 40

Thus [OH– ] = 0.002 M

pOH = – log10 [OH– ] = – log10 (0.002) = 2.7

pH = 14 – 2.7 = 11.3

More Examples

Calculate the pH of:

1) 0.001 M HNO3

2) 0.001 M NaOH

3) 0.001 M Ca(OH)2

Chem 1102 1B 41

Solutions

1) HNO3 is a strong acid thus fully dissociated.

HNO3 + H2O H3O+ + NO3

‐

[H3O+] = 10‐3 M Thus pH = ‐log10 10

‐3 = 3.0

2) Sodium hydroxide is fully soluble

[OH‐] = 10‐3 M and pOH = ‐log1010‐3 = 3.0

pH + pOH = 14 therefore pH = 14‐3.0 = 11

3) Ca(OH)2 gives 2 mole of OH‐ for each mole thus [OH‐] = 0.002M and pOH = ‐log100.002 =

Chem 1102 1B 42

8

Alternate question?

1) What is the pH of a solution formed by mixing 400 mL of 0.05 M HCl with 600 mL of 0.05 M NaOH

Chem 1102 1B 43

Solutions:1) First what is the reaction? Neutralization thus is only

H+ + OH‐ H2OWe are given concentrations and volumes thus when the

solutions are added final volume and concentrations must be calculated.

n= C x V from HCl mol of H+ = 0.05 M x 0.40L = 0.02molFrom NaOH mol of OH‐ = 0.05 M x 0.60L = 0.03molVolume = 1.0 L Excess OH‐ thus [OH‐] = (0.03‐0.02)/1.0L = 0.01MpOH = ‐log100.01 = 2.0 and pH = 12

Chem 1102 1B 44

Alternate question?

2) What is the [H+] of a solution with a pH of 4.5?

Chem 1102 1B 45

Solution:

If pH= 4.5 then [H+] = 10‐4.5 or 3.2 x 10‐5

Harder Example

What is the pH of a 2.0 × 10‐7 M HCl solution?

Chem 1102 1B 46

What are the sources of [H+]?

HCl (aq) → H+ (aq) + Cl‐ (aq) [H+] = 2 × 10‐7 MH2O (l ) ⇌ H+ (aq) + OH‐ (aq) [H+] = 10‐7 M

So is [H+] = 2 × 10‐7 M ??

NO ‐ the water equilibrium will be driven to the left by the acid.

Solution:

0cbxaxwhere

a2

ac4bbx

2

2

Chem 1102 1B 47

Conc (M) H2O ⇌ H+ + OH‐

Initial 2 × 10‐7 0

H2O autoionisation: (2 × 10‐7 + x) x

Kw = [H+][OH‐] = (2 × 10‐7 + x)(x) = 10‐14

Solve quadratic equation x = 4.14 × 10‐8 M,

giving pH = 6.62

Polyprotic Acids

H3PO4 (aq) ⇌ H+ (aq) + H2PO4– (aq)Ka1 = 10

‐2.2

H2PO4– (aq) ⇌ H+ (aq) + HPO4

2– (aq) Ka2 = 10‐7.2

HPO42‐ (aq) ⇌ H+ (aq) + PO4

3– (aq) Ka3 = 10‐12.4

Ka1 > Ka2 >Ka3

Harder to remove +ve charge against increasing ‐vecharge.

consider one equilibrium at a time

Chem 1102 1B 48

9

Polyprotic Acids

Chem 1102 1B 49

Ionisableprotons a long way apart

Ionisableprotons close together

Salts of Weak Acids and Bases

Is a solution of NaNO2 acidic or basic?

Take the ions apart

Na+ is from the strong base NaOH

NO2‐ is from the weak acid and HNO2 .

The base “wins”

pH > 7.

Chem 1102 1B 50

Overall reaction

H2O (aq) + NO2– (aq) ⇌ OH– (aq) + HNO2 (aq)

We can discount the other reaction:

H2O (aq) + Na+ (aq) ⇌ NaOH (aq) + H+ (aq)

Sodium hydroxide will not reform

NaOH is a strong base and cannot coexist with H+

Chem 1102 1B 51


Does a solution of NH4Cl have pH > 7 or < 7?

Separate the salt ions

NH4+ comes from the weak base NH4OH

Cl‐ comes from the strong acid HCl.

The acid “wins”

pH < 7.

Chem 1102 1B 52DEMO: ammonium chloride in water

Overall reaction

H2O (aq) + NH4+ (aq) ⇌ NH3 (aq) + H3O

+ (aq)

We can discount the other reaction:

H2O (aq) + Cl‐ (aq) ⇌ HCl (aq) + OH‐ (aq)

because HCl is a strong acid and cannot coexist

with OH‐

Chem 1102 1B 53


What is the pH of ammonium acetate at 25 °C ?2 possible reactions are:

CH3COO– (aq) + H2O (l) ⇌ CH3COOH (aq) + OH

– (aq)

Kb = 10–9.24

Anions of weak acids hydrolyse→ OH–

NH4+ (aq) + H2O (l) ⇌ NH3 (aq) + H3O

+ (aq)

Ka = 10–9.24

Cations of weak bases hydrolyse→ H3O+

In this case Ka = Kb so salt is neutral!

Chem 1102 1B 54

LOOK at the Ka and Kb how different are theyDEMO 5.13 + ammonium acetate in water

10

Salts of Strong Acids and Bases

Does a solution of NaCl have pH > 7 or < 7?We can discount the reaction:

H2O (aq) + Na+ (aq) → NaOH (aq) + H+ (aq)

because NaOH is a STRONG BASE and cannot coexist with H+, and we can discount the reaction:

H2O (aq) + Cl‐ (aq) → HCl (aq) + OH‐ (aq)

because HCl is a STRONG ACID and cannot coexist with OH‐

Salts of strong acids and strong bases are always neutral

Chem 1102 1B 55

pKa of Weak Acid and its Salt

What is the pH of a solution made up as 0.1 M in acetic acid (HAc) and 0.1 M in sodium acetate (pKa of HAc = 4.7)?

Reactions are:

CH3COOH (aq) + H2O (l) ⇌ CH3COO‐ (aq) + H3O

+ (aq)

NaCH3COO (aq) + H2O (l) ⇌ Na+ (aq) + CH3COO‐ (aq)

We have COMMON ION

Chem 1102 1B 56

Solution:

take 0.1 mol of CH3COOH and 0.1 mol NaCH3COO and dilute to 1 L

Can work either from acid or base dissociation constants. Take acid this time:

HAc (aq) ⇌ H+ (aq) + Ac‐ (aq)

Before: 0.1 0 0.1

After: 0.1 – x x 0.1 + x

(xmol HAc dissociates and forms xmol of Ac–)

Chem 1102 1B 57

could solve as quadratic, but make usual assumption that x << 0.1

pH = – log x = 4.7 = pKa

The Common Ion EffectIf you add the salt of an acid to a solution of the same acid then the equilibrium will shift towards neutral.

CH3COOH (aq) + H2O (l) ⇌ CH3COO‐ (aq) + H3O

+ (aq)

Addition of CH3COONa will boost [CH3COO‐]

By Le Châtelier’s principle the first equilibrium will shift to the left to remove CH3COO

‐ and therefore decrease [H3O

+].Same for a baseIMPORTANT FOR BUFFER SOLUTIONS

Chem 1102 1B 58

BuffersA solution containing both:

a weak acid + its salt

OR

a weak base + its salt

withstands pH changes when (limited) amounts of acid or base are added.

Reason: Le Châtelier’s principle.– if we add acid, then reaction HA ⇌ H+ + A– goes to

left to absorb change;

– vice‐versa if we add base

Chem 1102 1B 59

Sodium bicarbonate is used as a buffer in swimming pools

DEMO 5.11

Buffers

Consider the change in pH of pure water (pH = 7) if we add an equal amount of 10–3 M HCl:

Take 500 mL of water pH = 7.0 ([H+] = 10‐7M

Add 500 mL of 10‐3 M HCl

[H+] = 5 × 10–4 M (can neglect amount already present in water), so pH goes from 7 to 3.3!

Huge change!

What about buffers?

Chem 1102 1B 60

11

ExampleConsider a buffer solution with 0.1 M each of sodium acetate (NaAc) & acetic acid (HAc):

(found earlier that pH = 4.7)

What is the pH when 10–3 M HCl is added?

Chem 1102 1B 61

HAc ⇌ H+ + Ac–

initially: 0.1 10–3 0.1eqm: 0.1+10–3–x x 0.1–10–3 + x

i.e., suppose all but x of the added H+ forms HAc, but x will be very small even in comparison to 10‐3.

Chem 1102 1B 62

x = 1.02 × Ka = 0.000020 << 0.001

pH = – log x = 4.69

the pH hardly changes from 4.7!

Solution is buffered against pH change

Example (continued)

Chem 1102 1B 63

Henderson ‐ Hasselbalch Equation

For a buffer solution, which contains similar concentrations of a conjugate acid/base pair of a weak acid:

Chem 1102 1B 64

Since the dissociation of HA or

protonation of A‐ doesn’t lead to a significant change in the concentrations of these species.

Chem 1102 1B 65

Buffer Preparation

If the pH of a required buffer is the same as pKa of an available acid then use equimolaramounts of acid and conjugate base.

If the required pH differs from the pKa then use the Henderson‐Hasselbalch equation.

Chem 1102 1B 66

12

Buffer Preparation

]HA[

]A[logpKpH 10a

Chem 1102 1B 67

1) Choose conjugate acid‐base pair –chosen mostly by the desired pH. The buffer is most effective when the ratio of the acid/base is close to 1 pH close to pKa.

2) Calculate the ratio of the buffer component concentrations. Find the ratio of [A‐]/[HA] that gives the desired pH. Use the Henderson‐Hasselbalch equation

Example 1In the H3PO4 / NaH2PO4 / Na2HPO4 / Na3PO4

system, how could you make up a buffer with a pH of 7.40?

DATA: Ka1 = 7.2×10‐3, Ka2 = 6.3×10

‐8, Ka3 = 4.2×10‐13

Chem 1102 1B 68

H3PO4 ⇌ H2PO4‐⇌ HPO4

2‐ ⇌ PO43‐

38.12]10x2.4[logp

20.7]10x3.6[logp

14.2]10x2.7[logp

1310

810

310

a3

a2

a1

K

K

K

58.110

log20.740.7

logpKpH

20.0

10

10a

]PO[H

][HPO

]PO[H

][HPO

]PO[H

][HPO

42

24

42

24

42

24

Chem 1102 1B 69

pKa1 = 2.14 pKa2 = 7.20 pKa3 = 12.38

Must use mixture of H2PO4‐ and HPO4

‐. Could go

through whole procedure, or simply use

Henderson‐Hasselbalch equation

so the required ratio of Na2HPO4 to NaH2PO4

1.58:1

Example 2 ]HA[

]A[logpKpH 10a

Chem 1102 1B 70

Prepare a buffer solution with a pH = 3.90.

4.110]HCOOH[

]HCOO[

16.0]HCOOH[

]HCOO[log

]HCOOH[

]HCOO[log74.390.3

16.0

For every 1.0 mol of HCOOH we need 1.4 mol of HCOONa

The pKa of formic acid is 3.74 or Ka = 1.8 x 10‐4. The buffer

components can be formic acid HCOOH and the formateion HCOO‐ supplied by a soluble salt such as sodium

formate, HCOONa.

To calculate the component ratios use the equation

Buffer Capacity

Buffer capacity is related to the amount of strong acid or base that can be added without causing significant pH change.

Depends on amount of acid & conjugate base in solution:

‐ highest when [HA] and [A–] are large‐ highest when [HA] ≈ [A–] (most effective buffers have acid/base ratio less than 10 and more than 0.1)

pH range is ±1

Chem 1102 1B 71

Buffers in Natural SystemsBiological systems, e.g. blood, contain buffers: pH control essential because biochemical reactions are very sensitive to pH

Human blood is slightly basic, pH ≈ 7.39 – 7.45

In a healthy person, blood pH is never more than 0.2 pH units from its average value

pH < 7.2, “acidosis”; pH > 7.6, “alkalosis”

Death if pH < 6.8 or > 7.8

Chem 1102 1B 72

13

Buffers in Natural SystemsTo hold the pH of the blood close to 7.4 the body uses 3 buffer systems:

• Carbonate the most important

• Phosphate

• proteins

Chem 1102 1B 73

The acid is carbonic acid, H2CO3

the base is the carbonate HCO3‐.

The pKa of H2CO3 is 6.37.

Since the pH of an equimolar mixture of acid and conjugate base is equal to its pKa, a buffer made of equal

concentrations of H2CO3 and HCO3‐ has a pH of 6.37

Blood Buffering

10]CO[H

][HCO

32

3

Chem 1102 1B 74

• Blood however has a pH of 7.4. The carbonate buffer can maintain this pH only if ‐

[H2CO3] is not equal to [HCO3‐]

In Fact

The normal concentrations in blood are about 0.0025 M H2CO3

BLOOD 10]CO[H

][HCO

32

3

Chem 1102 1B 75

This means that it is a better buffer system for acids which lowers the ratio than for bases which increases the ratio.

This is in harmony with the functioning of the body as more acidic than basic substances enter the blood.

This ratio can be easily maintained as the body can readily alter the amount of CO2

entering the blood

Buffer System in Blood

“Extracellular” buffer (outside cell)

H+ (aq) + HCO3– (aq) ⇌ H2CO3 (aq)

H2CO3 (aq) ⇌ H2O (l) + CO2 (g)

Removal of CO2 shifts equilibria to right, reducing [H+], i.e., raising the pH

The pH can be reduced (made acidic) by:

H2CO3 (aq) + OH– (aq) ⇌ HCO3

– (aq) + H2O (l)

Chem 1102 1B 76

How much acid or base is in a given amount of solution?

• If solution is ACIDIC – add base until the acid is neutralised

• If solution is BASIC – add acid until base is neutralised

1. The unknown solution goes in the conical flask 2. The solution of known CONCENTRATION goes in the burette

3. The INDICATOR is chosen to change colour at the appropriate pH.

Chem 1102 1B 77

TitrationsEquivalence Point:

•When number of moles of

added base = original no. of moles of acid

– Strong acid/strong base pH = 7

– Weak acid/strong base pH > 7

– Strong acid/weak base pH < 7

•End Point:

•When a colour change in the indicator is observed

•⇒ Choose an indicator that changes colourclose to the equivalence point

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Indicatorsweak acid ⇌ weak base

•Each form has a different colour.

•The pH at which acid → base depends on the pKa of the indicator.

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Salts of Acids and Bases

Chem 1102 1B 80

Salt acidic/basic/neutral

NaOH BASIC – STRONGLY

NaHCO3 SLIGHTLY BASIC

Na2CO3 BASIC

NaCl NEUTRAL

NH4Cl ACIDIC

CH3COONH4 NEUTRAL

Al2(SO4)3 ACIDIC

KHSO4 ACIDIC

CH3COOH ACIDIC

HCl ACIDIC –STRONGLY

Salts of Acids and Bases

ReasoningNaHCO3 Na+ +HCO3

‐ HCO3‐+ H2O ?

Na2CO3 Na+ +CO32‐ CO3

2‐+ H2O HCO3‐+ OH‐

NH4Cl NH4+ + Cl‐ NH4

+ + H2O NH3 + H3O+

Al2(SO4)3 SO42‐ + Al(H2O)6

3+ + H2O Al(H2O)5(OH‐)2+ + H3O

+

KHSO4 K+ + HSO4‐ + H2O K+ + H3O

+ + SO42‐

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Titrations: Strong Acid / Strong Base

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25.0 mL of HCl solution is titrated with NaOH

Titrations: Weak Acid / Strong Base

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Equivalence point pH > 7 (value depends on starting concentrations)

Change is more gradual

Titrations: Weak Base / Strong Acid

Chem 1102 1B 84

Equivalence point pH < 7 (value depends on starting concentrations)

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5. Acids and Bases: Summary

•You should understand:

– The Brønsted‐Lowry concept of an acid as a proton donor and a base as a proton acceptor.

– The difference between strong acid (or base) and a weak acid (or base); and the role of the conjugate base (or acid) in equilibrium.

– The temperature dependence of Kw ∝ exp(‐∆H0/RT) and therefore pH.

– Trends in the relative strengths of acids and bases due to bond strength and electronegativity.

– How a buffer solution works due to the presence of a weak acid (or base) and its salt.

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• You should know:– The definition of the acid dissociation constant

Ka = [H+][A‐]/[HA].– The “p” convention, e.g.,

pH = ‐log10[H+], pOH = ‐log10[OH‐], pKa = ‐log10[Ka].– That pH + pOH = pKw = 14, and that pKa + pKb = pKw = 14.

You should be able to determine– The conjugate base of a given acid, and the conjugate acid of a

given base.– The pH of a strong or weak acid or base in solution.– The extent of ionisation of a weak acid or base in solution.– The pH of a buffer using Henderson‐Hasselbalch pH ≈ pKa +

log([added base]/[added acid]).– The equivalence point of a titration.

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acids and bases: overview common uses of acids and bases

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