If a solution of a weak acid contains more HA than A-, then
The Attempt at a Solution
the answer is pH > pKa.
Okay my question is why is this the answer. I'm trying to understand in what situation you would have the pH greater or less than the pKa and what that would mean in terms of the amount of A- and HA. The Henderson-Hasselbach equation is
pH = pka + log [A-]\[HA]
So let's say the pH is 6 and pka is 5. Then your ratio of A-A will be 10:1. That's pretty easy. This example would contain more A- than HA. So here, pH > pKa correct? Why is that then the answer to the original question was pH > pKa when you have more HA than A-??
Like if your pH was 4 and your pKa was 5, then the ratio would be 0.1:1 for A-A and mean that there is more HA? Then that would mean pKa>pH. Maybe I'm misinterpreting this!