Weak acids (HA) react with strong bases (OH–) to form a salt and water.
One of two things can happen:
- HA is in excess. The product mixture is acidic due to HA and H+ remaining.
- OH– is in excess. The product mixture is alkaline due to OH– remaining.
An early step in any calculation will therefore be to use an ICE table (or an equivalent method) to determine which of the reactants is in excess.
NB: If there is an error in this step, it can be very hard to score further marks if no ECF is given for a wrong / missing method. It is vital that you get this step correct to make progress in the question.
Have a look at the first example:
| Question: 0.40 g of NaOH is mixed with 100 cm3 of 0.2 mol dm-3 CH3COOH (KA = 2 x 10-5). Calculate the pH of the final mixture. Answer: First, work out the moles of CH3COOH and OH– in the mixture. n NaOH = m / Mr = 0.40 / 40 = 0.01 mol –> n OH– = 0.01 mol n CH3COOH = cv = 0.2 x (100/1000) = 0.02 mol Once we have the initial amounts of CH3COOH and OH–, we can write an ICE table for the neutralisation to predict the final amounts of CH3COOH and OH–. In this case, the HA is in excess.
The final volume is 100 cm3 so final [HA] = 0.1 mol dm-3. We can then re-arrange the KA expression to make [H+] the subject:
Substituting values gives: [H+] = (2 x 10-5) x (0.01/0.01) = 2 x 10-5 mol dm-3. This means that pH = -log (2 x 10-5) = 4.70
This is a very interesting special case. It shows that at half-neutralisation, when exactly half of the acid is neutralised, and where [HA] = [A-], the values of KA and [H+] are also equal (in this example 2 x 10-5). This can also be written as pKA = pH at half-neutralisation, where pKa = -log10KA |
Here is another example:
| Question: 0.80 g of NaOH are mixed with 50 cm3 of 0.1 mol dm-3 CH3COOH (KA = 2 x 10-5). Calculate the pH of the final mixture at 298 K. Kw = 1.00 x 10-14 mol2 dm-6. Answer: First, work out the moles of CH3COOH and OH– in the mixture. n NaOH = m / Mr = 0.80 / 40 = 0.02 mol –> n OH– = 0.02 mol n CH3COOH = cv = 0.1 x (50/1000) = 0.005 mol Once we have the initial amounts of CH3COOH and OH–, we can write an ICE table for the neutralisation to predict the final amounts of CH3COOH and OH–. In this case, the OH– is in excess.
The final volume is 50 cm3 so final [OH–] = 0.30. This means tha final [H+] = Kw / [OH–] = 3.3 x 10-14 mol dm-3 This means that pH = -log (3.3 x 10-14) = 13.48 This answer is EXACTLY the same as if we’d used a strong acid instead of a weak acid. Since all the acid has reacted, the pH doesn’t depend on whether the acid used was strong or weak. |
