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vande060
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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 dosen'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.
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