Now when it comes to dichotic buffers, we're going to say that they deal with the presence of two equivalence points and two Henderson Hasselbeck equations as a result of 2K values. Remember, diprotic means you have two acidic hydrogens, so you can lose 2H pluses, each with its own unique K value. Here we're going to say the relationships between the equivalence point and equations are shown as.

So here we have our dissociation steps. First, remember a diprotic acid is in the common form of H_{2}A here. This is its form where it has both acidic hydrogens. So we're going to say that this is the acidic form. K_{A1} deals with losing that first acidic hydrogen. When it does, it's now HA^{-}. This is called the intermediate form. Now it still has another acidic hydrogen, so if it decides to lose that, that means we're dealing with K_{2} and we finally get to this last phase A^{2-} where it's lost both of acidic hydrogens. So it would be a basic form.

Now we could also think in the reverse. We could start out with the basic form and we can gain our first H to give us our intermediate form. Gaining that first H plus means we're dealing with AB_{1}. This intermediate form could then accept a second H plus to reform its acidic form. Accepting that second H plus would give us KB_{2}. Now we'd see that K_{1} and KB_{2} are paired and K_{2} and KB_{1} are paired. That takes us to the K_{a}KB equations. So here we say K_{a1} times KB_{2} gives us k_{W}, K_{A2} times KB_{1} gives us k_{W}.

Now the Henderson Hasselbeck equation. There's two of them because we have two acidic hydrogen. Now remember the general form, the regular form of our Henderson Hasselbeck equation is

pH = pK a 1 + log ( base acid )So if we take a look at this first Henderson Hasselbeck equation, it connects together the acidic form and the intermediate form. So here we're talking about losing the first acidic hydrogen. So it would be pK_{A1} plus log of base over acid here. Remember the base has one last H plus, so the base here would be the intermediate form and the acid would be the acidic form.

Now the second Henderson Hasselbeck equation deals with the intermediate form of the basic form. Here we're talking about losing the 2nd acidic hydrogen to create the basic form. So we're dealing with K_{2} and therefore pK_{A2} plus log of base over acid. Here again, the base is the one with one less H plus ion, so that will be the basic form, and the asset form here would actually be the intermediate. Remember, an intermediate can act as an asteroid base by donating an H or accepting an H. So these will be the two Henderson Hasselbeck equations associated with a typical diprotic buffer solution.