Derivation of formula for Hydrolysis constant

While studying acid-base equilibrium I thought I would derive the formula for hydrolysis constant myself. But I ended up getting a weird result and I am not able to understand where I went wrong.

The equilibrium constants for reactions that tend to favour products are large. As the concentration of the reactants decrease the equilibrium constant slowly should move toward infinity. For example the ionisation of strong acids is taken to be a reaction that goes to completion. According to my understanding the equilibrium constant $K_\mathrm a=\infty$. The equilibrium constant for its reverse reaction would be zero. Similarly the equilibrium constant for the ionisation of salts should be $K_\mathrm s=\infty$.
Let us consider the reaction $\ce{HA + BOH <=> AB + H2O}$. This can be carried out in a number of steps:
$\ce{HA <=> A- + H+}$ with $K_\mathrm a$
$\ce{BOH <=> AB + H2}$ with $K_\mathrm b$
$\ce{H+ +OH- <=>H2O}$ with $1/K_\mathrm w$
$\ce{A- +OH1+<=> AB}$ with $1/K_\mathrm s=0$
So the hydrolysis constant should be
$1/K_\mathrm h=K_\mathrm aK_\mathrm b/(K_\mathrm wK_\mathrm s)=0\cdot K_\mathrm aK_\mathrm b/K_\mathrm w=0$
hence $K_\mathrm h=\infty$
I am sure there is something I havent understood well and hence I've ended up on this result. Can somebody explain where I am wrong?
Note: In most proofs the equilibrium constant of ionisation of salts is not considered but I see no reason to not consider it.

• Well, your reaction is wrong and there's no so thing as "infinite" pKa. – Mithoron Jul 14 '16 at 18:29
• @Mithoron Equilibrium of DNA burning in F2 is practically infinite. In other words, you will not be able to detect original DNA molecule once the system reaches equilibrium. – sixtytrees Jul 15 '16 at 1:23
• – Curt F. Oct 13 '16 at 12:26