# 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

## 1 Answer

Sorry, I didn't get that. first of all, which hydrolysis are you talking about? I will talk about weak acid and strong base equilibrium. You have three concentrations [HA], [H+], [A-]. For instance, if they took 1L of 1 mM of HOAc (acetic acid) and added 0.3 mmoles of NaOH (concentrated, so no dilution) you get [A-]+[HA] = 1mM Then you say HA <-> H+ + A- has Ka. Then

[H+][A-]/[HA] = Ka. Then you say [HA] = x, so [A-] = concentration_total- x (in this case 1mM -x). Then you get an equation [H+]*[concentration_total-x]/x=Ka. Then you solve this square equation.

Regarding your question. I think you are mixing up to things: protonation of weak acid (stuff, relevant to analytical chemistry with a lot of drama, pH swings etc) and the behavior of other salts (they just dissociate completely and we forget about them). In stories like "to a solution of NaCl and K-acetat a drop of acid was added. What happen?" you should first forget about NaCl. It has no impact on the question. You should also forget about K+. Now you have "to a solution of Acetate- a drop of acid was added. What happen to pH?" Now you can easily calculate it using a square equation.

Run away from chemistry. Carcinogens, stinky chemicals, no jobs... It teaches you how to make poisons and drugs, but how useful is that?

• Thank you for your answer but I still don't understand what went wrong in my derivation... – oshhh Jul 16 '16 at 7:07