[gicky]

k i have 20 ml of CH3COOH and 80 CH3OONa the PH of the buffer mixture is 5.36. my question is this. what is the ph if 10 ml of this buffer is diluted with water to a final volume of 100 ml. would it still be 5.36?

[mike]

As long as the acid to salt ratio stays the same the pH should stay the same.

Example log[A-]/[HA] = log80/20 = log8/2 = 0.6

I think you have to assume that the initial concentrations of the acid and salt are the same though.

[gicky]

gotcha, just want to make sure i did it right. thanks man!

[Borek]

As long as the acid to salt ratio stays the same the pH should stay the same.

While you are right it is a dangerous way of stating it with regard to dilution.

Let's take a look at the pH of acetic buffer 50/50:

0.1M	4.76
0.01M	4.76
10-3M	4.79
10-4M	4.95
10-5M	5.47
10-6M	6.31
10-7M	6.89

What is wrong? Well, when the solutions gets diluted ration salt/acid changes as the more and more acid dissociates. That's logical - very diluted solution is pure water, so it must have pH=7.00.

Does it mean H-H equation doesn't hold? It holds, but only if it contains real equilibrium concentrations of the acid and base. In not so diluted solutions (like 0.01M) equilibrium concentration and stoichiometric concentration (calculated using amount of substances put into solution) are for all practical purposes identical. If the acid is too strong, or the buffer is too diluted - that's no longer true.

See my lecture on pH of buffers for example with the not so weak acid.

Mitch~The forums are having trouble making a correct link for the above, so formatting has been removed from it. I don't know why this is.

Borek~Seems it is OK now. Perhaps lacking // caused problems?

[mike]

Oh great I have another misconception about pH

Only joking So as long as the concentration is not very, very low you could assume that diluting it is not going to change the pH?

Thanks once again Borek