How do I write a procedure for creating a buffer?

Question

I need to write an original procedure with exact numbers for creating a buffer:

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

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. 

We are given 0.1 M NaOH and 6 M acetic acid.

Using the HH equation, I found the ratio of NaCH3COOH to CH3COOH to be 0.10 : 1.

I think we can prepare a solution containing  0.10 M NaCH3COOH and 1.0 M CH3COOH. But with what we're given  (0.10 M NaCh3COOH and  6 M acetic acid), I don't know how to proceed and find exact numbers for a procedure.

Answer 1

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Answer 2

I answered this previously on this forum. Here is what I wrote:

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.

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Additional information for the current question:

If you want 50 ml (0.05L) and it is to be 0.1M NaCH3COO and 1.0 M CH3COOH, you can calculate moles and then mass and/or volume. Thus, 0.05 L x 0.1 mol/L = 0.005 moles sodium acetate is required

0.05 L x 1.0 mol/L = 0.05 moles acetic acid is required

82.0 g/mole NaAc x 0.005 moles = 0.41 g sodium acetate

(x L)(6 mol/L) = 0.05 moles and x = 0.0083 L = 8.3 mls

So weigh out 0.41 g NaAc, combine with 8.3 mls of 6 M HAc and bring final volume to 50 ml