18.1 – Calculations Involving Acids and Bases
18.1.1 – State the expression for the ionic product constant of water
The ionisation of water happens according to the equation:
Using this, we can find the expression of the equilibrium constant:
However, the concentration of water is relatively constant, so it can be removed from the equation. This is because the equilibrium lies so far to the left that the change in concentration does not have a significant effect on the value of Kc. Therefore, it can be written as:
The dissociation of water is an endothermic reaction. As a result, the Kw value increases as the temperature increases.
18.1.2 – Deduce [H+(aq)] and [OH–(aq)] for water at different temperatures given Kw values
Given a temperature and the value of Kw, we can calculate the value of [H+], using the equation:
Since OH- and H+ are in equal ratios of 1:1, in pure water, [H+] = [OH–], so:
[OH–] will be equal to [H+].
18.1.3 – Solve problems involving [H+(aq)], [OH-(aq)], pH and pOH
If the pH of a solution is <7, it is acidic. If the pH is >7, then it is alkaline. The pH of a solution is found according to the equation:
Therefore, given the pH, the [H+] can be found using:
Likewise, the pOH is found according to the equation:
Given the pOH, the [OH–] can be determined:
If you are given the concentration of a basic solution and need to find the pH, this can be done by rearranging the equation for Kw:
The [H+] can then be used to find the pH.
Alternatively, if you are given the [OH–] and calculate the pOH, then pH can be found by:
When you are given the equation for the reaction of a strong acid or base with water and the initial concentration of acid or base, then you can assume complete dissociation. This means that the [H+] or [OH–] will be equal to the initial concentration of acid or base. However, it is essential to check the stoichiometric ratio of acid-H+ or base-OH–, as this will need to be taken into account. For example, looking at a diprotic acid:
The concentration of [H+] (or [H3O+]) will be double the initial concentration of H2SO4.
Also, the pKw can be calculated, given the Kw:
18.1.4 – State the equation for the reaction of any weak acid or weak base with water, and hence deduce the expressions for Ka and Kb
When a weak acid reacts with water, the reaction will favour the left-hand side of the equation and it will have a small equilibrium constant.
The Ka is simply the Kc for the reaction multiplied by [H2O], so that it given the expression:
Likewise, if we are given a reaction for a weak base with water:
The Kb is found the same way, giving the expression:
You will only ever be asked about reactions involving the transfer of one proton.
18.1.5 – Solve problems involving solutions of weak acids and bases using the expressions:
In reactions of weak acids and bases, we cannot make the same assumptions as for the dissociation of strong acids and bases. Instead, we must use the relationships below.
For a weak acid and its conjugate base, the equation below can be used:
By finding the –log10 of these, this means that:
We can assume in these reactions that [acid]initial ≈ [acid]equilibrium since there is so little dissociation of the acid. Also, given that the acid is monoprotic, then [A–] = [H3O+]. Using these relationships, we can say that:
Therefore:
which can then be used to calculate pH. Remember to state the assumptions:
- [acid]initial ≈ [acid]equilibrium
- [A–] = [H3O+]
- There is negligible self-ionisation of the water so the [H3O+] is not affected
- The temperature is 25oC
Another relationship that can be used is:
Remember the IB will not ask you to use quadratic equations.
18.1.6 – Identify the relative strengths of acids and bases using values of Ka, Kb, pKa and pKb
The pKa or pKb is found using the equations:
Since they have a logarithmic relationship, a weaker acid will have a small Ka value but a larger pKa value. Likewise, a weak base will have a small Kb but a large pKb.
You will often be asked to compare between a number of acids or base to determine which is strongest or weakest.