[noiseordinance]

I'm working on a buffer problem. We have Formic acid (Ka =  1.7e-4), and we want a pH of 3.50 using 10 mL of 0.10 M solution of Formic acid (HFor) using the following bases: 0.10 M NaFor, and 0.10 M NaOH.

For the first reaction, I did the following:

pH = pKa + log (For-/HFor)
3.5 = -log(1.7e-4) + log (For-/HFor)
3.5 = 3.77 + log (For-/HFor)
-0.27 = log (For-/HFor)
10e-0.27 = (For-/HFor)
0.538 = (For-/HFor)   <--- mole ratio

10mL HFor (0.1 mol HCOOH / 1000mL) = 0.001 moles HFor
0.538 = [For-]/0.001 mol HFor, For- = 0.000538 moles (1000mL / 0.1 mol) = 5.38mL For-

So basically, 10mL of Formic acid and 5.38mL of Sodium Formate must be added to get our pH 3.50.

The problem I'm having is using NaOH as the base instead. So the first thing I'm assuming is that the reaction is like so:

HFor + NaOH  HCOONa + H₂O

When I run through the same exact steps as above, I end up with 5.38 mL of NaOH. This can't be right, since NaOH is a stronger base than the NaFor...

Is there some type of extra step required to use NaOH as the new base? Does it's Kb come into play at all?

[Borek]

Obviously you forgot where does the formate comes from when adding NaOH.