Calculating pH of weak acids and bases

Weak acids and bases do not dissociate completely. That means their reactions with water are equilibrium reactions. Only a few of the acid or base molecules donate their protons to water molecules to form ions: the rest remain as molecules.

With strong acids and bases, knowing the concentration of the solution is sufficient to work out the pH.

With weak acids and bases we need to use the equilibrium constant for the dissociation reaction to find [H3O+].

Example 1: pH of weak acid

What is the pH of a 0.325 mol L-1 solution of methanoic acid, Ka(HCOOH) = 1.82 × 10-4.

1 Write the equilibrium equation for the reaction between methanoic acid and water, and hence write the expression for Ka.

2 Make the assumptions

Assumption 1

We assume that [HCOOH] is what we put in: 0.325 mol L-1.

We know that it isn’t, because that would mean that none of the acid had reacted with the water, which would make the pH equal to 7.

However, we also know that the proportion of acid which has dissociated is very small, so the amount of unreacted acid is almost the same as the original amount.

Assumption 2We assume that [HCOO-] = [H3O+]. That’s what the equation says, so why is that an assumption? Because there will also be a small amount of H3O+ present ‘naturally’ in the water. That extra H3O+ is very little though, which is why it is a valid assumption. 3 Put the numbers into the Ka expression.

4 Calculate the [H3O+].

5 Calculate the pH.

Example 2: pH of weak acid

What is the pH of a 0.200 mol L-1 solution of hydrofluoric acid ? pKa(HF) = 3.17.

1 Write the equilibrium equation for the reaction between HF and water, and hence write the expression for Ka.

2 Make the assumptions

Assumption 1: [HF] = 0.200 mol L-1

Assumption 2: [F-] = [H3O+]

Note: you’re not (normally) asked to justify your assumptions, but you do have to state them.

3 Put the numbers into the Ka expression.

Before we do that we’re going to have to convert the pKa given in the question into Ka.

It’s the same process as changing pH into [H+].

Now it’s the same as before:

4 Calculate the [H3O+].

5 Calculate the pH.

Example 3: Calculating the pH of a weak base.

Calculate the pH of a 0.100 mol L-1 solution of aminomethane, CH3NH2. Ka(CH3NH3

+) = 2.57 × 10-11.

In this calculation we will find the [OH-], and use that to find the [H3O+] and pH.

1 Write the equilibrium equation for the reaction between CH3NH2 and water, and hence write the expression for the dissociation constant.

Since we have produced OH–, not H3O+, we need to write an expression for Kb, not Ka:

2 Make the assumptions

Assume: 1 [CH3NH3+] = [OH–]

2 [CH3NH2] = 0.100 mol L–1

3 Put the numbers into the Kb expression and calculate [OH–]

First we will need to calculate Kb from Ka:

Now calculate [OH–]:

4 Use [OH–] to calculate [H3O+] and hence pH