Okay, here's the deal. You got da skillz wit da strong/strong. But what if the acid isn't strong? What if de acid is some flabby, puny little girlie man? (insert Austrian accent as appropriate) Well, then things get a little more complex. | |

To start with, a weak acid or base won't dissociate in water to completion, so we invite someone else over to come and help us out: The acid dissociation constant. This guy, who also goes by the alias K _{a}, is our key to figuring out how much H^{+} or OH¯ is left over after the acid dissociates. | |

So where do we start? We go through the same first steps as before. First we look at the problem, and seeing that the acid isn't a strong acid (Why? Take a wild guess), we determine that it must be a weak acid. Next, we look for our good friend Ka, and he introduces us to his buddy K _{b}, who happens to be equal to K_{w}/K_{a} (K_{w} = 1.0*10^{-14}). | |

So what next? Now comes the fun part: we do an equilibrium problem! For every single part! But wait! There's good news. A couple of guys named Henderson and Hasselbach came up with this crazy shortcut that cuts off about, oh, say 10% - 90% of the work! They figured out that when you're between 10% and 90% of the equivalence point, you can fudge the math with a simple formula: pH = pK (by the way, pK_{a} - log ([HA]/[A¯])_{a} = -log(K_{a}))Also, since the volume is the same when an acid and its conjugate base are in solution: pH = pK_{a} - log (mols HA / moles A¯)We won't bother with it right now, but if you're really curious, you can see the derivation. | |

So what do we do now? Well, we know that Henderson/Hasselbach (heretoforth referred to as H/H) works within 10% and 90% of the equivalence point, so wouldn't it be nice if we had some way of finding out what that point was? Oh yeah! So, we use M _{A}V_{A} = M_{B}V_{B} to find the equivalence point, and then we find the range in which H/H can help us out. | |

But guess what? There's another shortcut! If you're too lazy to use H/H, then at the max buffer point you can use the formula pH = pK_{a}. Why? Because at the max buffer point (which happens to be half of the equivalence point), the quantities of the acid and its conjugate base are equal. So pH = pK_{a} - log ([Ha]/[A]) is really pH = pK_{a} - log (1) = pK_{a}! | |

So there you have it. Our steps are: - Find K
_{a}and K_{b} - Find equivalence point and max buffer point (M
_{A}V_{A}= M_{B}V_{B}) - Find H/H range
- Do the math, baby!
| |

Oh, yeah, one more thing. We haven't talked about this before, but we think by now you're ready for it. Everything that we've done up to this point involving titrations, including everything we're going to do, can be done by titrating a base with an acid! That's right, you just switch it all around, as you'll see in the example. And by the way, the Henderson/Hasselbach equation for pOH is pOH = pK._{b} - log ([B]/[HB^{+}]) | |

Howz about we do a little problem, eh? |