Question about sig figs in buffer problem

[vande060]

Homework Statement

prepare 1L of a .1M acetate buffer with a pH of 5.22

Homework Equations

pK of acetate is 4.70

The Attempt at a Solution

this is what my professor did:

ph = pK + log(base/acid)

5.22 = 4.70 + log(base/acid)

antilog(.52) = (base/acid)

3 = (acetate/acetic acid) = 3.3/1

you therefore require 3 moles of acetate for every mole of acetic acid.
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I don't understand why the .3 was just ignored, it doesn't seem like she should have rounded down...any suggestions? I understand that eventually we will have to use .1M, which will reduce the sig figs to one, but that is not until the next step, and I thought you were not supposed to drop figures early.

 

[Borek]

I don't understand it either, especially if the question asks for pH 5.22 - that's quite specific (and difficult to prepare in the standard lab, 5.2 looks more reasonable; most pH meters are not able to reliably measure last digit, so you won't be even able to test it). I would go for 3.3.

Please note - sig figs are a faulty concept, something like "poor mans accuracy". Don't treat them too seriously, accuracy should be evaluated and reported using statistical methods. In the case of solution preparation they can be used as a rule of thumb, but even then you should be vigilant. If 1 means anything between 0.5 and 1.5, and 9 means anything between 8.5 and 9.5, the same one significant digit means about 50% error in the first case and about 5% error in the second case - so something is obviously wrong.