My question relates to dissolving $\ce{NH3}$ in water and how it produces $\ce{OH-}$ for the reaction $\ce{NH3 + H2O -> NH4+ + OH-}$?

I know that the $\mathrm{p}K_b$ of $\ce{NH3}$ is $4.8$ and $\mathrm{p}K_a$ of $\ce{NH4+}$ is $9.2$ so $\ce{NH3}$ is a stronger base than $\ce{NH4+}$ is an acid. But why the free $\ce{OH-}$? I am now taking organic chemistry and all I am reading is how strong $\ce{OH-}$ is and how it can attack anything slightly positive. So why is the above reaction favorable towards creating $\ce{OH-}$?

Actually any acid-base reaction that is creating $\ce{OH-}$ as a free species?


2 Answers 2


The shortest answer is "because the change in free energy is negative," but maybe you want to know how the mechanism works?

Ammonia is a Lewis base, and hydroxide is a stronger Lewis base. However, both are stronger Lewis bases than water is. Pure water always has some amount of "free" $\ce{H+}$ (it's actually hydronium ion) and $\ce{OH-}$ due to self-ionization ($pK_W \approx 14$):

$$ K_W = \frac{[\ce{H3O+}][\ce{OH-}]}{[\ce{H2O}]^2} $$

When you add ammonia to the system, some of those hydronium ions (which are very strong acids) react with the ammonia to form ammonium. Since the concentration of hydronium decreases, the concentration of hydroxide must increase to maintain equilibrium. Those hydroxide ions are stronger bases than ammmonia is, and so they will "win" the tug-of-war over protons. However, in the meantime, some other hydronium reacts with some other ammonia, and the net result is a dynamic equilibrium in which there is more hydroxide then there would be in pure water.

The same mechanism is responsible for any other acid/base reaction where a base accepts protons from water and leaves hydroxide as a free species, but the net concentration of hydroxide will be different for different strengths of bases.

  • $\begingroup$ Even if the change in free energy is negative, there are still processes that won't "go" under certain conditions without an input of energy. $\endgroup$
    – Dissenter
    Aug 12, 2014 at 15:01
  • $\begingroup$ Also, your statement about ammonia and hydroxide being Lewis bases is off base. Hydroxide ion is not necessarily a stronger Lewis base than ammonia. Consider the stabilities of metal ammonia and metal hydroxide complexes. Metal hydroxide complexes tend to be less stable than their metal ammonia counterparts. $\endgroup$
    – Dissenter
    Aug 12, 2014 at 15:05
  • $\begingroup$ "hydronium ions (which are very strong Lewis acids)" ... Hydronium ions are in no way Lewis acids. $\endgroup$
    – Dissenter
    Aug 12, 2014 at 15:05
  • 2
    $\begingroup$ @Dissenter - 1) Yes, there are activation energies. How does that relate to this question? 2) In this case, hydroxide ion is a stronger base than ammonia. 3) You are technically right - I originally had written free protons (which are Lewis acids) instead of hydronium, then went back and changed it in an attempt to avoid these kinds of pedantic arguments. I see that I failed by forgetting to remove the Lewis acid statement. I'll make you a deal - I'll remove that statement, and you work on keeping your comments polite and on sending them in smaller barrages. $\endgroup$
    – thomij
    Aug 12, 2014 at 19:23
  • $\begingroup$ 1) if the activation energy were high for the rxn there would be no hydroxide ion present. $\endgroup$
    – Dissenter
    Aug 12, 2014 at 19:25

Simply put, this reaction involves the formation of hydroxide, because it is relatively more stable than the alternative products: $\ce{{}^{-}NH2}$ and $\ce{H3O+}$. You're right; hydroxide is generally a reactive species, so you probably wouldn't expect it to be the product of the reaction. However, when you have two reactants, in this case $\ce{NH3}$ and $\ce{H2O}$ in an acid-base reaction, the stronger base dictates the outcome of the reaction, which of course is $\ce{NH3}$.

Also, this reaction is in equilibrium and there are more reactants than the products. Thus, this solution is hardly composed of hydroxides.


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