Acids and Bases: Reactions of strong acids with strong bases

Category Interesting QuestionsPosted on:

Strong acids (H+) react with strong bases (OH–) to form a salt and water. 

One of two things can happen:

  1. H+ is in excess. The product mixture is acidic due to H+ remaining.
  2. 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 are mixed with 100 cm3 of 0.2 mol dm-3 H2SO4. Calculate the pH of the final mixture.

 

Answer:

First, work out the moles of H+ and OH– in the mixture.

n NaOH = m / Mr = 0.40 / 40 = 0.01 mol

–> n OH– = 0.01 mol

 

n H2SO4 = cv = 0.2 x (100/1000) = 0.02 mol

–> n H+ = 0.04 mol

 

Once we have the initial amounts of H+ and OH–, we can write an ICE table for the neutralisation to predict the final amounts of H+ and OH–. In this case, the H+ is in excess.

The final volume is 100 cm3 so final [H+] = 0.30. 

This means that pH = -log (0.03) = 0.52

Here is the second example:

Question:

0.80 g of NaOH are mixed with 50 cm3 of 0.1 mol dm-3 HCl. 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 H+ and OH– in the mixture.

n NaOH = m / Mr = 0.80 / 40 = 0.02 mol

–> n OH– = 0.02 mol

 

n HCl = cv = 0.1 x (50/1000) = 0.005 mol

–> n H+ = 0.005 mol

 

Once we have the initial amounts of H+ and OH–, we can write an ICE table for the neutralisation to predict the final amounts of H+ 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

 

NB – if there was a small error in the ICE table and you actually found an H+ excess, you would not have any calculation using Kw so lose further marks. This means that one error might cost 2 or 3 marks.