[mandy9008]

A compound has a pKa of 7.4. To 100 mL of a 1.0 M solution of this compound at pH 8.0 is added 30 mL of 1.0 HCl. What is the pH of the resulting solution?

This is what we did in class:

pH=pKa + log (A-/HA)
8.0 = 7.4 + log (A-/HA)
(A-/HA) = 3.98
(A-/HA) = 4

1.0 M = mol/100 mL
=0.1 mol

A-=0.08 mol
HA=0.02 mol

1.0 M HCl = mol/30 mL
=0.03 mol HCl

x  0.08 - 0.03 = 0.05
y  0.02 + 0.03 = 0.05

pH = 7.4 + log (0.05/0.05)
pH= 7.4

I don't understand how my prof got the 0.08 and 0.02 from the 0.1 mols. When I asked him, he told me it was because the ratio was calculated, using the H-H equation first, to be 4, which meant that x was 4 times what y was. Could someone please explain to me how that is relevant. Also, why would he subtract and add the 0.03 in order to get the 0.05 mol? I am very confused. Any help will be greatly appreciated.

[Borek]

Unfortunately, it is not clear to me what you don't understand - as all you wrote is exactly correct.

Henderson-Hasselbalch equation can be used in two different ways. First, it can be used to calculate pH of the solution if you know concentrations of the acid and conjugate base (or just their ratio). Second, it can be used to calculate ratio if the pH is given. Here, you know total concentration of acid and conjugate base, but you don't know the ratio. However, you can calculate ratio from a given pH, then, knowing sum of concentrations and ratio of concentrations you can calculate concentrations themselves.

After that - you add strong acid to the buffer. That means weak (conjugate) base present in the solution gets protonated (converted to acid form). You can assume this reaction proceeds quantitatively to the end. That means amount of conjugate base left is "initial-acid added" and amount of conjugate acid is "initial+acid added". That's where the -0.03 and +0.03 come from.