One method of making a buffer solution is to mix a weak acid (e.g., CH3COOH) with a soluble salt of its conjugate base (e.g., CH3COONa). This results in partial neutralisation of the weak acid, and a solution with a large reservoir of both HA and A-.
The word mix is emphasised above because this is NOT a reaction so DOES NOT require an ICE table.
Also – importantly – the [H+] does not equal [A–] so if you start writing [H+]2 on the top of the KA expression…. STOP! This is wrong and will cost you many marks.
Using the algebra from before…
There are a number of possible questions that could be written:
- Finding the pH of a given buffer solution. This is relatively straightforward.
| Question: 1.0 g of CH3COONa is added to a 100 cm3 solution of 0.20 mol dm-3 CH3COOH. Calculate the pH of the mixture. KA = 2 x 10-5 mol dm-3 Answer: The first step is to work out the moles of CH3COONa and CH3COOH: n CH3COONa = m / Mr = 1 / 82 = 0.012 mol n CH3COOH = cv = 0.2 x (100/1000) = 0.020 mol Then we do some algebra:
We know KA and the molar amounts of HA and A-, so we can now work out the [H+] [H+] = (2 x 10-5) x (0.020/0.012) = 3.3 x 10-5 mol dm-3 This gives a pH of -log(3.3 x 10-5) = 4.48 |
Note that the [HA]/[A-] is a ratio of concentrations. However, this will be equal to simply a ratio of amounts HA/A- because the volume cancels:
2. Finding the amount of soluble salt that needs to be added to make a buffer with a specific pH. This is more challenging.
| Question: X g of CH3COONa is added to a 100 cm3 solution of 0.20 mol dm-3 CH3COOH. The pH of the mixture is 5.00, calculate X. KA = 2 x 10-5 mol dm-3 Answer: The question tells us that [HA] = 0.20 mol dm-3. The next step is work out [H+] by re-arranging the pH equation. The pH of the buffer needs to be 5.00, so substituting into [H+] = 10-pH equation gives us [H+] = 1 x 10-5 mol dm-3. We now know KA, [H+] and [HA] so the only unknown is [A–] and we re-arrange the KA equation to make [A–] the subject:
We can now work out [A-]: [A–] = (2 x 10-5) x (0.2 / 1 x 10-5) = 0.4 mol dm-3 The volume of the solution is 100 cm3 so: n A– = cv = 0.4 x (100/1000) = 0.04 mol If there were 0.04 mol A– added to the acid, then there must have been 0.04 mol CH3COONa added. We can use this to find the mass of CH3COONa. m CH3COONa = moles x Mr = 0.04 x 82 = 3.28 g |
Comparing the two examples, you can see that the more soluble salt added, the more the pH increases. This is really useful because we can vary the amount of soluble salt added to make a buffer with the desired pH.