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When a neutralisation reaction happens, for example, $ \pu{100 mol l^-1}$ of $\ce{HCl}$ with $\pu{100 mol l^-1}$ $\ce{NH_3}$, why does all of the base and acid get converted to salt? Why isn't there an equilibrium established between salt and acid-base?

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Acids and bases don't necessarily completely react with each other. Consider your hypothetical reaction between hydrochloric acid and ammonia.

\begin{align} \ce{HCl + H3N &-> H4N+ + Cl-}& \text{Extent:}& \approx100\% \end{align}

The products are ammonium ion and chloride ion - the conjugate acid and conjugate base, respectively. So if the forward (above) reaction is complete, how can the reverse reaction between the conjugate acid and conjugate base be anything but insignificant? I.e.

\begin{align} \ce{H4N+ + Cl- &<<=> HCl +H3N}& \text{Extent:}& \approx0\% \end{align}

An equilibrium is established whenever the extent of reaction is not complete. Another good example would be an acetic acid solution. Acetic acid and water exist in equilibrium with acetate anion and hydronium ion. So no not all acid-base reactions go to an extent of 100%. Things to consider: stoichiometry and acid and base strengths.

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  • $\begingroup$ Thanks for the answer. The reason I took ammonia was that it was a weak base, and for weak bases, dissociation is not complete. My question then was how ammonia gets incompletely dissociated when alone in water, but then completely reacts with HCl. So, what you've essentialy pointed out is that there is an equilibrium for strong acid-weak base reactions. Right ? $\endgroup$ – Gaurav Dec 30 '14 at 19:43
  • $\begingroup$ Ammonia being a base doesn't disassociate as do acids. Ammonia does react incompletely with water because water is a weak acid and incapable of protonating significant concentrations of ammonia. HCl however is a good few billion times stronger as an acid than water. $\endgroup$ – Dissenter Dec 30 '14 at 19:48
  • $\begingroup$ And no generally there is no equilibrium involving the reaction of a strong acid with a weak base because the base with be completely protonated. $\endgroup$ – Dissenter Dec 30 '14 at 19:49

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