Making Buffers
- Buffers can be made by a direct or indirect method
- The direct method involves measuring out the exact quantities needed of a solution of a weak acid and its salt
- The indirect method uses sodium hydroxide solution to partially neutralise the weak acid so that a mixture of the acid and salt are obtained
- To make an ethanoic acid / ethanoate buffer, 50 cm3 of 0.1 mol dm-3 ethanoic acid solution is measured into a beaker
- 2.05 g of sodium ethanoate is weighed out and added to the beaker and stirred until it has dissolved
- A funnel is used to transfer the solution to a 250 cm3 volumetric flask and the rinsings are included
- The solution is made up to the scratch mark with more 0.1 mol dm-3 ethanoic acid solution
- A calibrated pH probe or meter is used to measure the pH of the buffer solution
- To test the buffer 1cm3 of 1.0 mol dm-3 HCl is added to 100cm3 of the buffer in a beaker and the pH is measured
- This is repeated using 1cm3 of 1.0 mol dm-3 NaOH
- The pH should change very little on addition of the acid and alkali
Indirect method
- The indirect method involves using 0.1 mol dm-3 sodium hydroxide solution in a burette and adding the desired volume of sodium hydroxide to a fixed volume of the weak acid to make up the buffer
- In this example, 25 cm3 of 0.1 mol dm-3 sodium hydroxide would be needed to make up the buffer of the same pH
Analysis
- 2.05 g of sodium ethanoate represents 0.025 mol (molar mass 82.03 g mol-1)
- When this dissolves in 250 cm3 of 0.1 mol dm-3 ethanoic acid solution, the concentration is 0.1 mol dm-3 in sodium ethanoate
- Using the Henderson- Hasselbalch equation,
- When the ratio of salt to acid is 1:1, log10 1 becomes 0, so pH = pKa
- The pKa of ethanoic acid is 4.75, so this buffer should have pH of 4.75
- You can make any buffer by adjusting the ratio of salt to acid to match the difference between the pKa and the desired pH
- For example, suppose you wanted to make a buffer of pH = 4.25,
pH = 4.25
4.25 = 4.75 -0.50
[salt]/[acid] = 0.50
- The ratio of the salt to acid has to be 0.50, so following the same procedure, you would need to dissolve half the mass of sodium ethanoate in the same volume of ethanoic acid solution, i.e. 1.025 g
