6b.6: Buffers
|
Definition of a buffer:
A solution of a weak acid mixed with its conjugate base What's the point of a buffer? When you add acid or base to it, the pH doesn't change very much Why does that work? If you add base, it reacts with the weak acid and makes a weak base. If you add acid, it reacts with the weak base and becomes a weak acid. How do you make a buffer? Option 1: take a weak acid, and add a lesser amount of strong base to turn some, but not all of it, into the conjugate base. Option 1a: Same thing but with a weak base plus strong acid Option 2: take a weak acid and directly add the conjugate base as an ionic compound. |
|
What's the math on a buffer?
If you set up the Ka ICE box and do a little rearrangement, you end up with the Henderson-Hasselbalch equation:
pH = pKa + log[A-]/[HA] (where pKa is -log(Ka) )
When does a buffer work best?
When concentrations of both HA and A- are high (the more weak acid and weak base you have to react, the better).
When the concentrations of both HA and A- are approximately equal (it will resist change in both directions equally well).
What's this half-equivalence point marked off on the graph?
It's halfway (on the x-axis) to the equivalence point. It's also the point where some math magic happens.
At that point, moles OH- added are exactly half moles of HA that we started with. Therefore:
[HA] = [A-]
So the Henderson-Hasselbalch equation simplifies to:
pH = pKa
If you set up the Ka ICE box and do a little rearrangement, you end up with the Henderson-Hasselbalch equation:
pH = pKa + log[A-]/[HA] (where pKa is -log(Ka) )
When does a buffer work best?
When concentrations of both HA and A- are high (the more weak acid and weak base you have to react, the better).
When the concentrations of both HA and A- are approximately equal (it will resist change in both directions equally well).
What's this half-equivalence point marked off on the graph?
It's halfway (on the x-axis) to the equivalence point. It's also the point where some math magic happens.
At that point, moles OH- added are exactly half moles of HA that we started with. Therefore:
[HA] = [A-]
So the Henderson-Hasselbalch equation simplifies to:
pH = pKa
