IB syllabus > acids & bases (hl) > 18.3 

These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

18.1 - Calculations involving acids and bases


18.1.1 State the expression for the ionic product constant of water (Kw).


Water equilibrium

Water is in equilibrium with its dissociated ions (hydrogen and hydroxide).

The equilibrium:

H2O H+ + OH-

Can be expressed according to the equilibrium law:

Kc = [H+][OH-]
[H2O]

However, as the concentration of the water effectively remains constant on both sides of the equilibrium then the [H2O] term can be removed to a very close approximation and the equilibrium constant denoted as Kw (sometimes called the ionic product of water).

This gives:

Kw = [H+][OH-]

The constant Kw remains unchanged at constant temperature (as all good constants should!).

At 25ºC the value of Kw = 1 x 10-14 mol2 dm-6

As the concentration of the hydrogen ions equals the concentration of the hydroxide ions (see note 1) then the concentration of hydrogen ions in pure water at 25ºC = the square root of the ionic product of water:

= 1 x 10-7 mol dm-3

All equilibrium constants are temperature dependent (and this one is no exception):

The dissociation of water molecules into ions is bond breaking and is therefore an endothermic process (energy must be absorbed to break the bonds). Endothermic processes are favoured by an increase in temperature and so as the temperature rises the equilibrium moves further to the right hand side and Kw gets larger.

As Kw gets larger so do the values of the hydrogen ion concentration and the hydroxide ion concentration.

As pH is a measure of the hydrogen ion concentration (pH = -log[H+]) then as the temperature increases the pH gets lower - i.e. the water becomes more acidic.

This is calculated in the following section.


18.1.2: Deduce [H+(aq)] and [OH-(aq)] for water at different temperatures given Kw values.


Variation of Kw with temperature

The equilibrium

H2O H+ + OH-

involves the breaking of bonds and is therefore endothermic - energy must be applied to break one of the the H-O-H bonds to give the ions. Consequently, according to Le Chatelier, an increase in temperature favours the forward reaction - i.e. the position of equilibrium shifts towards the right hand side and Kw becomes larger.

However, as the ratio of hydrogen ions to hydroxide ions in pure water must remain 1:1, then if we know the value of Kw, it is a simple matter to calculate the value of either H+ or/and OH- to obtain the concentrations and hence the values of pH and pOH.

Example: Calculate the pH when Kw = 6,5 x 10-14 mol2 dm-6

As...

Kw = [H+][OH-]

and...

[H+] =[OH-]

Then...

Kw = [H+]2

Therefore...

[H+] = √ Kw

[H+] = √ 6.5 x 10-14

[H+] =2.55 x 10-7

pH = 6.59

The pOH value will also be the same as [H+] =[OH-]


18.1.3: Solve problems involving [H+(aq)], [OH-(aq)], pH and pOH


Definition of pH

pH is defined as the negative of the logarithm (base 10) of the hydrogen ion concentration

For example at 25ºC the hydrogen ion concentration of pure water is 1 x 10-7 mol dm-3

The logarithm of 1 x 10-7mol dm-3 = -7

The negative of -7 = +7

Therefore the pH of pure water at 25ºC is 7

Definition of pOH

This is basically (no pun intended) the same as pH but from the point of view of the OH- ions

Thus the negative logarithm of the OH- ions gives the pOH

Note that at 25ºC ... pH + pOH = 14 it is therefore a simple matter to obtain one from the other.

Definition of pKw

As you can probably guess from the previous two definitons, pKw is the negative logarithm of Kw

And as... Kw = [H+][OH-]

Then... pKw = pH + pOH = 14 (at 25ºC)


18.1.4: State the equation for the reaction of any weak acid or weak base with water, and hence deduce the expressions Ka and Kb. Only examples involving the transfer of one proton will be assessed


Calculation of [H+(aq)]

If we are dealing with a strong acid then this is straightforward. It is simply a matter of treating the hydrogen ion concentration as the molar concentration of the acid for monobasic acids (such as nitric acid) and double the acid concentration for dibasic acids (such as sulphuric acid).

Example: Calculate the [H+(aq)] of 0.25 M sulphuric acid

As sulphuric acid dissociates 100% according to the equation

H2SO4 2H+ + SO42-

Then 0.2 M sulphuric acid gives 0.25 x 2 M hydrogen ions solution = 0.5M

The pH of this solution is:

pH = - log 0.5 = 0.3

If we are dealing with a weak acid (or base) then the Ka (or pKa) of the acid must be known

Example: Calculate the [H+(aq)] of 0.2 M ethanoic acid (Ka = 1.78 x 10-5)

As ethanoic acid is a weak acid it only partially dissociates according to the equation:

CH3COOH CH3COO- + H+

Applying the equilibrium law:

Ka = [H+][CH3COO-]
[CH3COOH]

We can assume that as the acid only slightly dissociates then the concentration of the acid at equilibrium is the same (to a close approximation) as the concentration of the original acid (in this case = 0,2 M)

Therefore:

1.78 x 10-5 = [H+][CH3COO-]
[0.2]

And as the hydrogen ion concentration equals the ethanoate ion concentration then:

0.2 x 1.78 x 10-5 = [H+]2

[H+] = √ 3.56 x 10-6

[H+] =1.89 x 10-3

The pH of this solution is:

pH = -log 1.89 x 10-3 = 2.7


18.1.5: Solve problems involving solutions of weak acids and bases using the expressions: Ka x Kb = Kw, pKa + pKb =pKw, pH + pOH = pKw. Students should state when approximations are used in equilibrium calculations. The use of quadratic equations will not be assessed


Acid dissociation equations

It is important to identify the acidic hydrogen(s) in order to be able to write the equation representing dissociation. In most acids the hydrogen that is released causing acidity is fairly obvious:

H2SO4

HNO3

HCl

However, in organic acids this is not always the case:

CH3COOH

(COOH)2

HCOOH

Once the acidic hydrogens are identified, it is a case of writing the equation showing the ion resulting from removal of the H+ ion(s)

H2SO4 2H+ + SO42-

HNO3 H+ + NO3-

HCl H+ + Cl-

CH3COOH CH3COO- + H+

(COOH)2 (COO)2- + 2H+

Base equations

In these cases the base removes an ion of hydrogen from the water molecule. The base is hydrolysing (breaking apart) the water to produce hydroxide ions. As the base gains a hydrogen ion, it itself will produce a species with a positive charge (positive ion)

NH3 + H2O NH4+ + OH-

The acid (base) equilibrium expression

Once the equation is written down the equilibrium law states that the acid (base) equilibrium constant is equal to the concentrations of the products raised to their stoichiometries divided by the concentration of the reactant(s) raised to the stoichiometry

Example: For the equilibrium: CH3COOH CH3COO- + H+

The equilibrium law gives:

Ka = [H+][CH3COO-]
[CH3COOH]

18.1.6: Identify the relative strenghs of acids and bases using values of Ka, Kb, pKa, and pKb.


Derivation of Ka x Kb = Kw

For the equation: CH3COOH CH3COO- + H+

CH3COOH is the acid and CH3COO- is its conjugate base.

Ka = [H+][CH3COO-]
[CH3COOH]

And for the conjugate base reaction: CH3COO- + H2O CH3COOH + OH-

Kb = [OH-][CH3COOH]
[CH3COO-][H2O]

As water is in vast excess on both sides of the equilibrium it can be safely eliminated to give

Kb = [OH-][CH3COOH]
[CH3COO-]

Combining Ka and Kb gives:

 

Ka x Kb = [H+][CH3COO-] x [OH-][CH3COOH]
[CH3COOH] [CH3COO-]

Cancelling out terms from top and bottom gives:

Ka x Kb = [H+] x [OH-]

And as: Kw = [H+][OH-]

Then:

Ka x Kb = Kw


Ka is the acid equilibrium constant - i.e. the equilibrium constant of the products of acid dissociation divided by the acid concentration at equilibrium (however the approximation that the acid concentration at equilibrium is the same as the original acid concentation is usually used for convenience).

Ka is usually a very small number (for example 1.78 x 10-5 for ethanoic acid). It is more convenient to use the logarithm of this Ka value to give number that are handled more easily. However taking logs of very small number produces a negative value. To avoid this the negative of the logarithm is used and called the pKa value.

Hence:

-log Ka = pKa

If we are dealing with bases then Kb again is very small and so pKb is used to define base strength where:

-log Kb = pKb

As shown in section 18.3.6 above:

Ka x Kb = [H+] x [OH-]

Consequently at 25ºC

Ka x Kb = 1 x 10-14

And:

pKa + pKb =14


Ka values

Using the typical weak acid (HA) equation, this is represented by the equilibrium

HA H+ + A-

From which, by the equilibrium law:

It may be seen that an increase in the components of the right hand side of the equilibrium will give rise to a greater value for Ka.

Hence the stronger the acid the larger the value of Ka


pKa values

The relationship between pKa and Ka is one of an inverse log and so the larger the value of Ka the smaller the value of pKa.

Hence the stronger the acid the smaller the value of pKa

This may be illustrated by some Ka and pKa values

Acid or base Ka pKa acid strength
Trichloroethanoic acid 5.10 x 10-2 1.29 decreasing acid strength
Chloroethanoic acid 1.38 x 10-3 2.86
Methanoic acid 1.77 x 10-4 3.75
Ethanoic acid 1.78 x 10-5 4.75
Propanoic acid 1.26 x 10-5 4.90
Carbonic acid 3.98 x 10-7 6.40
Water 1.00 x 10-7 7.00
Ammonia 5.26 x 10-10 9.25
Methylamine 2.24 x 10-11 10.65

Remember that Ka + Kb = Kw

And so, pKb = 14 - pKa for the bases


Example:: Calculate the pH of 0,25M ethanoic acid (pKa = 4.75)

For the equilibrium: CH3COOH CH3COO- + H+

The equilibrium law gives:

Ka = [H+][CH3COO-]
[CH3COOH]

pKa = 4.75 therefore Ka = 1.78 x 10-5

and [H+] = [CH3COO-]

Therefore:

1.78 x 10-5 x [CH3COOH] = [H+]2

1.78 x 10-5 x 0.25 = [H+]2

Therefore: [H+] = √1.78 x 10-5 x 0.25

Therefore: [H+] = 2.11 x 10-3

pH = -log [H+]

Therefore pH = 2.68


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