Another method of making a buffer is to react a limiting amount of strong base (e.g., NaOH) with an excess of weak acid (e.g., CH3COOH). This results in partial neutralisation of the weak acid, and a solution with a large reservoir of both HA and A-.
The word react is emphasised above because this is IS a reaction so DOES require an ICE table.
Again – in this situation [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.
There are a number of possible questions that could be written:
- Finding the pH of a given buffer solution. This is quite a challenging question and commonly worth 5-6 marks at A Level.
| Question: 1.0 g of NaOH is added to a 400 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 NaOH and CH3COOH: n NaOH = m / Mr = 1 / 40 = 0.025 mol n CH3COOH = cv = 0.2 x (400/1000) = 0.080 mol We then can put this information into an ICE table:
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 (0.055/0.025) = 4.4 x 10-5 mol dm-3 This gives a pH of -log(4.4 x 10-5) = 4.36 |
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 strong base that needs to be added to make a buffer with a specific pH. This is VERY challenging and beyond what you’d be asked to do A Level. Just for fun.
| Question: X g of NaOH is added to a 400 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 initial [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. For this question, it is slightly more convenient to use concentrations in the ICE table rather than molar amounts, but both approaches are valid. In the ICE table below, “x” is used to denote the initial concentration of OH– in the mixture:
We can then write the KA expression and substitute in everything that we know:
This shows that initial [OH–] = 0.133 mol The volume of the solution is 400 cm3 so we can now work out the initial moles of OH- added and, thus, the mass of NaOH required: n OH- = cv = 0.133 x (400/1000) = 0.0533 mol m NaOH = n x Mr = 0.0533 x 40 = 2.13 g
Checking the answer: We have shown that 2.13 g NaOH added to 400 cm3 solution of 0.20 mol dm-3 CH3COOH gives a buffer with pH 5.00. We can check this calculation by doing the much easier calculation in the forward direction. initial CH3COOH = cv = 0.2 x (400/1000) = 0.08 mol initial NaOH = m / Mr = 2.13 / 40 = 0.0533 mol Putting this into an ICE table:
Using the re-arrangement from before and substituting in the values, with KA = 2 x 10-5 mol dm-3.
gives [H+] =(2 x 10-5) x 0.0267 / 0.0533 = 1.00 x 10-5 mol dm-3, which has a pH 5.00 (as required). |
We now have an equation in terms of x, so we can do some algebra to determine the value of x:
