Topic 8 Acids and Bases

8.2Properties 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

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

HCl + NaHCO3

  1. 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.

HClH+ + 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 = acidAccepts 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

AcidBaseconjugate base conjugate acid

HCl+H2OH3O++Cl-

AcidBaseconjugate acidconjugate base

NH3+H2ONH4++OH-

BaseAcidconjugate acidconjugate 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+H2OH3O++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+H2OOH-

Accepts H+ and so acts as a basedonates H+ and so acts as an acid.

8.1 Lewis Acids and Bases (SL/HL)

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 + MgMgCl2 + 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 dative covalent 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?

8.3Strong 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.

HClH++Cl-

H2SO42H++SO42-

HNO3H++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.

CH3CO2HH++CH3CO2-

H2CO3H++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+H2ONH4++OH-

CO32- + 2H2O CO2 + 2 OH-

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

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.
  1. 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.
  1. 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 photosynthesize and grow.

8.4pH Scale (SL/HL)

pH stands for power of hydrogen.

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 10-2pH = 2

= 1 x 10 –5pH = 5

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

= 1 x 10-14pH = 14

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 mol 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 followinga)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)

H2OH++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 waterKw=[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. 2Calculate 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. 4Calculate the pH of a 0.04 mol dm-3 solution of LiOH.

Answer 2.pH 113.pH 13.24.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.

Answers2)12.33) pH = 12.48pOH = 1.52

18.1 Kw and Temperature (HL ONLY)

HEAT+H2OH++OH-

  • This is an endothermic reaction.
  • Increasing the temperature will thereforea) 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+H2OBH++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.

1. Calculating pH

Calculate the pH of a 0.1 mol dm-3 solution of a weak acid of pKa 4.2.

pKa = 4.2pKa = - log10 Ka

Ka = 6.31 x 10-5 mol dm-3

Ka = x2 / a

6.31x 10-5 = x2 / 0.1

x = 2.51 x 10-3 mol dm-3

pH = - log10 [2.51 x 10-3]

pH = 2.6

2. Calculating concentrations

Calculate the concentration of a weak acid HF of pH 2 where Ka = 6.76 x 10-4 mol dm-3

pH = 2 therefore2 = - log10 [H+]

[H+]=[F-]=1 x 10-2 mol dm-3=x

Ka=[H+] [F-] / [HF]

Ka=x2 / a

6.76 x 10-4=(1 x 10-2)2 / [HF]

therefore[HF]=0.148 mol dm-3

3. Calculating Ka

Calculate the dissociation constant Ka for a weak acid HA if a 0.01 mol dm-3 solution has a pH of 3.1.

[HA]=0.01 mol dm-3=a

pH = 3.1therefore3.1 = - log10 [H+]

[H+]=7.94 x 10-4 mol dm-3= x

Ka=x2 / a

=(7.94 x 10-4)2 / (0.01)

=6.31 x 10-5 mol dm-3

What is the value for pKa? Use the data booklet to identify this organic acid.

Weak Bases (HL ONLY)

B+H2OBH++OH-

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

Note the [H2O] concentration is very large and therefore taken as constant. It is incorporated into the Kb value.

Kb = y2 / b

Same as acids!

Note: Kw = Ka x Kb = 1 x 10-14

so that:

pKa+pKb=14

Sneaky eh?

Questions

  1. Calculate the pH of a 0.1 mol dm-3 solution of ethanoic acid if Ka is 1.8 x 10-5 mol dm-3.
  1. Find Ka of a weak acid of concentration 0.02 mol dm-3 if its pH is 3.9.
  1. Find the concentration of a weak acid of pH 4.5 if Ka = 4.1 x 10-6 mol dm-3
  1. Calculate the pH of a weak base of concentration 0.01 mol dm-3if Kb = 1.8 x 10-5 mol dm-3.
  1. If the pH of a weak base is 10, and its concentration is 3 x 10-2, calculate the pKb of this base.
  1. If the pKa of methylamine (a weak base) is 10.64, and it’s pH is 10.8, what will be it’s concentration?

Answers = 1) pH 2.87 2) 7.92 x 10-7mol dm-3 3) 2.44 x 10-4 mol dm-3 4) pH 10.65. 5) 6.48 6) 9.13 x 10-4mol dm-3

18.2Buffer Solutions (HL ONLY)

Buffers resist changes in pH when small amounts of acids or bases are added.

Acidic Buffers:

pH remains constant and acidic with a pH of less than 7.

Prepared by mixing a weak acid with an aqueous solution of a salt of that weak acid.

E.g. Weak acid = ethanoic acid (CH3CO2H)

Mixed with salt of weak acid = sodium ethanoate (CH3CO2Na).

The weak acid partially dissociates and sets up an equilibrium that obeys Le Chateliers principle:

CH3COOH CH3COO- +H+

Addition of a small amount of acid H+ shifts the equilibrium position to the LHS so that the H+ concentration stays constant and the pH remains the same.

Addition of a small amount of base (OH-) reacts with the H+ ions and causes the equilibrium position to shift to the RHS. The H+ concentration and also the pH therefore remain constant.

Ideally there should be equal concentrations of weak acid (CH3COOH) and the salt of the weak acid (CH3COO-) so that the solution can buffer equally in both directions.

Note: A buffer can also be made by neutralising excess ethanoic acid with sodium hydroxide as the limiting reagent:

CH3CO2H + NaOHCH3CO2Na+ CH3CO2H + H2O

(excess) (limits)(salt) (weak acid)

0.2 moles 0.1 moles 0.1 moles 0.1 moles

Basic Buffers

pH remains constant and basic with a pH greater than 7.

Prepared by mixing a weak base with the salt of a weak base.

E.g. Weak base = Ammonia NH4OH

Salt of a weak base = Ammonium chloride NH4Cl

The weak base partially dissociates and sets up an equilibrium that obeys Le Chateliers principle:

NH4OH NH4+ +OH-

Small additions of acid will react with OH- and shift the equilibrium position to the RHSso that the pH does not change.

Small additions of OH- will shift the equilibrium to the LHS. pH stays constant.

Ideally there should be equal concentrations of weak base (NH4OH) and the salt of the weak acid (NH4+ ) so that the solution can buffer equally in both directions.

Note: The salt can be either added directly or made by neutralising excess ammonia with some hydrochloric acid:

NH3 + HCl NH4Cl + NH3

0.2 mole 0.1mole 0.1 mole 0.1 mole

Examples of Natural Buffers:Blood

Blood contains a very important buffer which makes it function properly.

In the lungs, [O2] is high, eq shifts to RHS causing O2 to combine with haemoglobin molecules.

HHb+O2H++HbO2-

In the muscles [O2] is low, eq shifts to LHS causing O2 to be released.

Changes in concentration of H+ will affect whether the blood is able to function correctly. The blood therefore has a buffer in it to make sure that that the pH cannot alter significantly.

18.2 Buffer Calculations (HL ONLY)

  • Since buffers are fundamentally weak acids and bases, the calculations are very similar

E.g.CH3COOHH++CH3COO-

  • The essential difference is that a large amount of the salt is also added to the solution and completely ionises, which makes the concentration of the salt ion (CH3COO-) different to the concentration of hydrogen ions.
  • The assumption made is that since the equilibrium of the weak acid/base is so far to the LHS, that the ionisation of the acid is so small that concentration of the salt ion comes only from the salt and not from the dissociation of the weak acid.

E.g. [CH3COO-] = from the concentration of salt solution

[H+] = from the pH of the weak acid

[CH3COOH] = From the initial concentration of the weak acid.

E.g.A buffer solution was prepared by adding2.05g of sodium ethanoate to 500cm3 of ethanoic acid of concentration 0.2 mol dm-3. Calculate the pH of the resulting buffer. pKa ethanoic acid = 4.76.

pKa = - log10 Ka = 1.74 x 10-5

Moles of CH3COONa = m / Mr = 2.05 / 82 = 0.025

[CH3COO-] = n / V = 0.025 / 0.5 dm3 = 0.05 mol dm-3

[CH3COOH] = 0.2 mol dm-3

Using eq equation: Ka=[H+] [CH3COO-] / [ CH3COOH]

1.74 x 10-5=[H+] [0.05] / [ 0.2]

[H+]=6.96 x 10 –5 mol dm-3

pH=4.16

Extension: How could this buffer solution be made more acidic?

E.g. 2If a buffer is made by mixing 0.2 moles of sodium ethanoate into 0.5 dm3 of ethanoic acid of concentration 0.1 mol dm-3, then what will be the resulting pH of the buffer?(pKa = 4.76)

E.g.3The weak acid propanoic acid (pKa = 4.87) can be used to make a buffer if it is mixed with it’s salt sodium propanoate (C2H5COONa). Calculate the mass of salt that must be dissolved in 0.25 dm3 of 1 mol dm-3 propanoic acid to give a buffer of pH 4.87.

Answers:2.pH = 5.363.24 g

Note: Most buffers are prepared so that they can shift equally to the LHS and to the RHS. This means that usually the concentration of the weak acid equals the concentration of the salt.

[weak acid] = [salt]

(E.g. [CH3COOH] = [CH3COO-] )

So that they cancel each other out.

This means that often

Ka = [H+]

And so (often) in an ideal buffer:

pKa = pH

E.g. If 40 cm3 of 0.1 mol dm-3 ethanoic acid is mixed with 20 cm3 of 0.1 mol dm-3 sodium hydroxide, what would the pH of the buffer be if Ka = 1.74 x 10-5 mol dm-3?

E.g. If 100cm3 of 0.5 mol dm HCl were mixed with 200cm3of ammonia solution. Calculate the pH of the resulting buffer. (pKa for ammonia = 4.75)

Answer 1. pH = 4.76 2. pH = 9.25

18.3 Salt Hydrolysis

Salts are derived from acids that have been neutralised by a base.

Salts made from strong acids and bases will be neutral.

If the acid and base involved are both strong, the resulting salt will be neutral.

E.g. Cl-, SO42-, NO3- are conjugate bases of strong acids and so will be only very weak bases.

E.g. Na+, K+ are conjugate acids of strong bases and so will be only very weak acids.

NaCl, KNO3 etc will therefore be neutral.

Salts made from weak acids and strong bases will be alkaline.

E.g. CH3COOH is a weak acid and therefore gives a conjugate base CH3COO- which is a strong.

CH3COONa will therefore be alkaline.

Salts made from strong acids and weak bases will be acidic.

E.g. NH3is a weak base and therefore gives a conjugate acid NH4+ which is a strong.

NH4Cl will therefore be acidic.

Salts made from weak acids and bases will be neutral.

This is because the conjugate acid and base produced will both be strong. However they will cancel each other out.

CH3COONH4 will therefore be neutral.

  1. KCl
  2. Na2SO4
  3. CH3CO2K
  4. NH4NO3
  5. Na2CO3
  6. (NH4)2CO3
  7. NaNO3
  8. (NH4)2SO4

Metal Salt Solutions:

Some metal salt solutions may be acidic due to the charge on the metal ion.

+1 metal ions in solution tend to be neutral, but due to the increased charge on the metal ion, +2 and then +3 metal ion solutions increase in acidity.

E.g. AlCl3 dissolved in water gives a complex ion Al(H2O)63+

The Al3+ is acting as Lewis acid due to its empty orbitals.

H2O is acting as a Lewis base due to it donating electron pairs to form 6 co-ordinate bonds.

Since the Al3+ is so highly charged it can cause the water molecules to break apart (called hydrolysis)