Topic 8 Acids and Bases

8.2 Properties of Acids and Bases (SL/HL)

Acids = Substances that are able to produce H+ ions when dissolved in water. They cause litmus to go red, phenolphthalein to go colourless and methyl orange indicator to go red.

Bases = Substances that can neutralise acids.

Alkalis = Soluble bases that produce OH- ions when in water. They cause litmus to go blue, phenolphthalein to go pink and methyl orange to go yellow.

Typical Reactions

Give a fully balanced equation as an example to illustrate each type of neutralisation reaction.

1.  Neutralisation with different types of base such as metal oxides or hydroxides.

HNO3 + KOH

HCl + NH4OH

H2SO4 + CaO

2.  Reaction with both soluble and insoluble carbonates and hydrogencarbonates.

HCl + NaHCO3

3.  Reaction with metals.

H2SO4 + Mg

8.1 Bronsted-Lowry Acids and Bases (SL/HL)
Acids are substances which have a hydrogen ion concentration greater than 1 x 10-7.

In order to predict if a substance will be an acid or a base, we need a definition in terms of structure.

Acids are proton H+ donars
Bases are proton H+ acceptors

For a species to act as an acid it must therefore contain a hydrogen atom attached by a bond which can be easily broken.

HCl H+ + Cl-

Will HF and HI be stronger or weaker acids than HCl?

Draw ionic equations to illustrate the action of these acids.

For a species to act as a base, it must contain a non-bonding e- pair which it can use to form a bond with a H+ ion.

H+ + OH- H2O

Donates H+ ion = acid Accepts H+ ion = base

2H+ + O2- H2O

2H+ + CO32- H2O + CO2

Conjugate Acids and Bases

When an acid donates H+, the species produced is called a conjugate base as it now has the ability to accept a H+.

When a base gains H+ the species produced is called a conjugate acid, as it now has the ability to donate a H+.

H2SO4 + Cl- HSO4- + HCl

Acid Base conjugate base conjugate acid

HCl + H2O H3O+ + Cl-

Acid Base conjugate acid conjugate base

NH3 + H2O NH4+ + OH-

Base Acid conjugate acid conjugate base

The stronger the acid, the weaker the conjugate base it produces. (HCl to Cl-)

The weaker the acid, the stronger the conjugate base it producers. (H2O to OH-)

CH3COOH + H2O H3O+ + CH3COO-

What are the conjugate bases of the following acids?….H2SO4, HNO3, H2CO3, C2H5OH, C6H5OH, OH-, H2O.

What are the conjugate acids of the following bases?…OH-, H2O, NH3, CO32-, HCO3- HNO3, C2H5NH2.

Label the Bronsted Lowry acid bases and conjugate acid bases in the following equation.

H2SO4 + HNO3 HSO4- + H2NO3+

Amphiprotic (amphoteric)

Substances (such as water) which are able to either donate or accept a H+ and so act as either an acid or a base is said to be amphiprotic.

H3O+ H2O OH-

Accepts H+ and so acts as a base donates H+ and so acts as an acid.

8.1 Lewis Acids and Bases (SL/HL)

Ø  Lewis extended the definition of acids and bases to include substances which do not contain hydrogen ions but which can still act as an acid or a base.

An acid is an electron pair acceptor
A base is an electron pair donor.

E.g. 2 HCl + Mg MgCl2 + H2

E.g. BF3 + NH3 (BF3 NH3)

·  Where H+/ BF3 have empty orbitals and so can act as a Lewis acids by accepting pairs of electrons.

·  Mg/NH3 have non-bonded electron pairs that can donate and so can act as a Lewis bases.

·  AlCl3 can also act as a Lewis acid as it has an empty bonding orbital.

·  Bronsted lowry acids and bases can also be considered using the Lewis definition.

E.g. H+ (can accept e- pair = acid) + OH- / O2- / CO32- (all have un-bonded e- pair that can be donated = base)

Use ‘curly arrows to show the movement of electron pairs to illustrate the lewis acid base nature of these reactions.

HCl + NaOH

H2SO4 + CuO

HNO3 + CaCO3

·  All transition metal compounds can act as Lewis acids since they all have empty orbitals.

·  Any molecule that can form a co-ordinate bond by donating a spare electron pair can act as a Lewis base. E.g. Cl-, H2O, CN-, OH-, NH3 etc. (called ligands)

·  This means that all metal complexes which form (see Topic 3) are actually Lewis acid base reaction, where the metal is the acid and the ligand (H2O etc) is the base.

E.g. Cu2+ + 6H2O [Cu(H2O)6]2+

Lewis acid Lewis base

Use ‘curly’ arrows to show which of the following are lewis acids and bases.

Fe3+ + 6 CN-

Cu2+ + 4 Cl-

Ag+ + 2 NH3

TOK: Which of the two acid and base theories is the easiest to use?

Why is there a need for two theories?

8.3 Strong and Weak Acids and Bases. (SL/HL)

Strong acids dissociate (ionise) completely in aqueous solution

This means that all the molecules split up into ions in water.

Examples of strong acids include hydrochloric, sulphuric and nitric.

HCl H+ + Cl-

H2SO4 2H+ + SO42-

HNO3 H+ + NO3-

Strong acids therefore have large concentrations of hydrogen ions in aqueous solution.
Weak acids do not completely dissociate (ionise) in aqueous solution.

This means that some of the molecules remain intact and do not split up into ions.

The undissociated acid molecules and the dissociated ions form an equilibrium.

Examples of weak acids include ethanoic and carbonic acids.

CH3CO2H H+ + CH3CO2-

H2CO3 H+ + HCO3-

Weak acids therefore have lower concentrations of hydrogen ions in aqueous solution.

Strong and Weak Bases.

Strong bases completely dissociate. Examples are group 1 metal hydroxides.

E.g. NaOH Na+ + OH-

Weak bases only partially dissociate. Examples are ammonia solution, carbonate solutions and amines (see Topic 10 organic chemistry).

E.g. NH3 + H2O NH4+ + OH-

CO32- + 2H2O CO2 + 2 OH-

Experimental Determination of Strength of Acids. (SL/HL)

Ø  Varying concentrations of H+ ions can be tested experimentally to see if the acid is strong or weak.

1.  Conductivity: Since electrical conductivity relies on the movement of charges, strong acids have higher concentrations of H+ ions and so have more charges available. Strong acids are able to conduct electricity much better than weaker acids. This can be tested with a simple circuit containing an ammeter.

2.  By Reaction with Carbonates: A simple indication of the strength of an acid can be given by comparing how violently it reacts with a carbonate. If equal amounts of two acids of the same concentration are added to calcium carbonate, the one that produces carbon dioxide at the greatest rate is the strongest.

3.  pH: This is the measurement of H+ concentration. The lower pH value, the higher the concentration of H+, the stronger the acid. If two acids of equal concentration are tested with a pH meter the stronger one will give the lower value.

Note: Weak acids are relatively safe, yet over a long period of time their effects become apparent. This happens in atmospheric acid rain. Weakly acidic rain slowly reacts with limestone and marble in buildings, and causes the leaching of important metal ions out of soil so that plants are unable to photosynthesise and grow.

8.4 pH Scale (SL/HL)

Ø  pH stands for power of hydrogen.

Ø  The pH scale = 0 (very acidic, U.I red, stomach acid)

3-4 (weak acid, U.I orange, vinegar, lemon or orange juice)

7 (Neutral, U.I green, pure water)

9-10 (Weak alkali, U.I blue, washing powder)

14 (Strong alkali, U.I purple, oven cleaner)

Where U.I is universal indicator used as a mix of indicators giving a wide range of colours.

pH = - log10 [H+]

Therefore if [H+] = 1 x 10-1 pH = 1

= 1 x 10-2 pH = 2

= 1 x 10 –5 pH = 5

= 1 x 10-7 (pure water) pH = 7

= 1 x 10-14 pH = 14

Ø  If the pH value changes by 1, this then actually means that the hydrogen ion concentration has changed by a factor of 10.

A volume of 10cm3 of 0.1 mol dm-3 solution of HCl is diluted with water to 1000 cm3, what are the initial and final pH’s of the acid?

What is the pH of a 5 cm3 sample of 0.01 mo dm-3 HNO3?

Describe how the pH could be changed to pH = 6 by dilution.

TOK: Why is it necessary to use a logarithmic scale to represent pH?

What are the advantages?

18.1 Calculations involving acids and bases (HL ONLY)

pH Calculations

pH = - log10[H+]

The strength of a strong acid or base will be the same as its hydrogen ion concentration since both will completely ionise in solution

So a 0.1 mol dm-3 solution of hydrochloric acid will have [H+] = 0.1 mol dm-3

Calculate the pH of the following a) 0.25 mol dm-3 HCl

b) 2.5 HNO3

c) 1.5 x 10-2 HCl

d) 1.4 x 10-5 HNO3

Calculate the hydrogen ion concentration of the following

e) pH = 2.5

f) pH = 6.4

g) pH = 4.5

18.1 The Ionic Product of Water (Kw) (HL ONLY)

H2O H+ + OH-

Equilibrium constant Kc = [H]+ [OH]- / [H2O]

Since the equilibrium position is so far to the LEFT, the [H2O] concentration can be taken as being constant and so can be incorporated into the equilibrium constant.

Ionic product of water Kw = [H]+ [OH]-

(units = mol2 dm-6)

Kw = 1 x 10-14 mol2 dm-6

at 298 K (25 degrees) room temperature

Using Kw

This can be used to determine the pH of alkaline solutions.

E.g. Calculate the pH of a 0.1 mol dm-3 NaOH solution.

[OH-] = 0.1 mol dm-3 since this is a strong alkali and fully ionises.

Since Kw = 1 x 10-14 = [H+] [OH-]

Using this to find the concentration of hydrogen ions

[H+] = 1 x 10-14 / [ 0.1 ]

[H+] = 1 x 10-13

pH = 13

E.g. 2 Calculate the pH of a 0.001 mol dm-3 solution of KOH

E.g. 3 Calculate the pH of a 0.15 mol dm-3 solution of NaOH.

E.g. 4 Calculate the pH of a 0.04 mol dm-3 solution of LiOH.

Answer 2. pH 11 3. pH 13.2 4. pH 12.6

18.1 pH, pOH and pKw (HL ONLY)

pOH is a measure of the hydroxide ion concentration on a log10 scale

pOH = -log10 [OH-]

pKw is a measure of the combined hydrogen and hydroxide concentrations on a log10 scale.

pKw = -log10 Kw

Since Kw = 1 x 10-14 at room temperature

pKw = 14

(at room temperature)

This gives a useful and easily learnt relationship between pH and pOH
pH + pOH = 14

E.g. Calculate the pOH of a solution of monoprotic acid of pH 4.

pOH = 14 - 4 = 10

E.g. 2 Calculate the pOH of a monoprotic acid of concentration 0.02 mol dm-3

E.g. 3 Calculate the pOH and pH of a 0.03 mol dm-3 solution of LiOH.

Answers 2)12.3 3) pH = 12.48 pOH = 1.52

18.1 Kw and Temperature (HL ONLY)

HEAT + H2O H+ + OH-

·  This is an endothermic reaction.

·  Increasing the temperature will therefore a) increase the amount of ionisation

b)  move the equilibrium to the RHS

c)  Increase both [H+] and [OH-].

Kw increases as the temperature increases

·  Since [H+] increases, the pH will go down below 7, but the [OH-] will increases as well. This causes the pH to change, but the water to remain neutral.

Water at a temperature above 25 degrees has a pH less than 7

Water at a temperature below 25 degrees has a pH greater than 7

·  E.g. Water at 70 degrees has a pH of 6.6 even though it remains neutral.

·  Calculate the hydrogen ion concentration and pH of water if at a particular temperature Kw = 5 x 10-13

(Answer = pH 6.15)

18.1 Equilibrium constants for weak acids and bases (HL ONLY)

·  Weak acids and bases are only partially dissociated in water.

Weak Acids

HA(aq) H+(aq) + A-(aq)

HA = acid molecule

A-  = salt ion

Ka = [H+] [A-] / [HA]

Ka = acid dissociation constant (the equilibrium constant for the dissociation of a weak acid.)

·  If Ka is large = more dissociation

= equilibrium further to RHS

= stronger acid

·  If Ka is small = less dissociation

= equilibrium further to LHS

= weaker acid

·  Since the values for Ka can have such a large range, the numbers are usually given as log10.

pKa = - log10 [Ka]

·  A pKa of 1 or 2 would therefore be a very strong acid.

·  A pKa of 4 or 5 would be a weaker acid such as ethanoic acid.

·  A pKa of 13 or 14 would be a very weak acid such as ethanol.

Weak Bases (HL ONLY)

B + H2O BH+ + OH-

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

And

pKb = - log10 Kb

·  So that strong bases have a pKb of 1 or 2.

·  Weaker bases (such as ammonia) have a pKb of 4 or 5.

·  Very weak bases have pKb values of 9 or 10.

Use the data booklet to:

a) Calculate the Ka for ethanoic acid.

b) Calculate the Kb for ammonia.

18.1 Calculations Involving Weak Acids and Bases (HL ONLY)

HA(aq) H+(aq) + A-(aq)

Ka = [H+] [A-] / [HA]

If the initial concentration of the weak acid HA = a

The equilibrium concentration of H+ and OH- = x

Then the equilibrium concentration of HA = (a-x)

This gives the equilibrium expression Ka = x2 / (a – x)

Since x is so much smaller than a for a weak acid, we can assume that (a – x) = a

This gives:

Ka = x2 / a

Where x is the equilibrium concentration of H+ calculated from the pH of the solution and a is the initial concentration of the weak acid.