Create a buffer using CH3COOH and NaCH3COO that has a pH of exactly 3.75.?

Question

a 50 mL sample of your buffered solution will have to be able to withstand the addition of 25.0 mL of 0.100 M NaOH solution. 

the buffered solution will break after the addition of no more than 35.0 mL of the 0.10 M NaOH. 

Can someone give me an overview of how to approach this? Thanks

Answer 1

Probably the easiest way to approach this is by using the Henderson Hasselbalch equation:

pH = pKa + log [salt]/[acid]

Before setting this up and doing the calculations, one should ask what is meant by "withstand" addition of 25 ml of 0.1 M NaOH? Also, what is meant by "break"? I'll assume "break" means that there is no longer any buffering capacity, ie all of the acetic acid has been converted to the salt. I'll also assume that "withstand" means there is still buffering capacity and that addition of this amount of NaOH can still be consumed and acid converted to salt.

Taking the pKa of acetic acid to be 4.75, and a desired pH of 3,75, we have...

3.75 = 4.75 + log [salt]/[acid]

log [salt]/[acid] = -1

[salt]/[acid] = 0.100 which means we want the ratio of the NaCH₃COO : CH₃COOH to be 0.100. To accomplish this we can prepare a solution containing 0.1M NaCH₃COO and 1.0M CH3COOH, or it could be 0.01M NaCH₃COO and 0.1M CH₃COOH or any other combination of salt and acid give the desired ratio.

However, if you want 50 ml of this buffer to "withstand" 25.0 ml of 0.1 M NaOH, we can determine moles NaOH added and hence the desired moles of salt and acid.

25 ml x 0.1 mmoles/ml = 2.5 mmoles NaOH added and the final volume will be 50 ml + 25 ml = 75 ml.

There must be at least 2.5 mmoles CH₃COOH present in order to react with the 2,5 mmoles NaOH. Thus, 2.5 mmoles CH₃COOH/50 ml = 0.05 mmoles/ml = 0.05 M CH₃COOH. This being the case, the NaCH₃COO concentration would be 0.1 x 0.05 M = 0.005 M. So, the initial buffer could be made up as 0.005 M NaCH₃COO + 0.05 M CH₃COOH.

You can do a similar calculation for "break" point by using 35 ml x 0.1 mmol/ml = 3.5 mmoles NaOH being added. In this case you'd calculate [salt] and [acid] so that there was fewer than 3.5 mmoles of CH₃COOH in the 50 mls of buffer.