# Does Brønsted-Lowry theory apply to neutralization reactions?

Does the Brønsted-Lowry theory apply to neutralization reactions such as: $$\ce{HCl + NaOH -> H2O + NaCl}$$

and would there be a conjugate acid/base?

## 2 Answers

The Brønsted-Lowry acid/base theory is based on a proton donor (acid), and a proton acceptor (base).

In the reaction: $$\ce{HCl + NaOH -> H2O + NaCl}$$ the $\ce{HCl}$ will dissolve into $\ce{H^+}$ and $\ce{Cl^-}$ because it is a strong acid, and likewise $\ce{NaOH}$ will dissolve into $\ce{Na^+}$ and $\ce{OH^-}$.

Considering that $\ce{H^+}$ is a free proton (neglecting the hydrogen isotopes), then in the left side of the reaction, $\ce{HCl}$ is an acid because it donating a proton in form of $\ce{H^+}$ to $\ce{NaOH}$ according to the following mechanism:

$$\ce{HCl -> H^+ + Cl^-}$$ $$\ce{H^+ + NaOH -> Na^+ + H_2O}$$ $$\ce{Na^+ + Cl^- -> NaCl}$$ This means that $\ce{HCl}$ is an acid because it's donating a proton, while $\ce{NaOH}$ is a base because it's accepting the proton.

The same is true for the conjugate: $\ce{H_2O}$ is the conjugate acid while $\ce{NaCl}$ is the conjugate base. $$\ce{H_2O -> H^+ + OH^-}$$ $$\ce{H^+ + NaCl -> HCl + Na^+}$$ $$\ce{Na^+ + Cl^- -> NaCl}$$

• – Jan Dec 25 '17 at 10:37

The Brønsted-Lowry theory describes acid-base reactions as proton transfers. The acid is the species which donates a proton while the base is the species that accepts a proton.

Usually, $\ce{HCl}$ and $\ce{NaOH}$ are considered dissolved in water in the context of acid base reactions. In the case of sodium hydroxide, this means that it is better considered as $\ce{Na+ (aq) + OH- (aq)}$, while in the case of $\ce{HCl}$ both $\ce{HCl (aq)}$ and $\ce{H+ (aq) + Cl- (aq)}$ are valid descriptions; in the Brønsted-Lowry context things are easier to understand if $\ce{H+ (aq)}$ is replaced by $\ce{H3O+ (aq)}$ although the latter is neither more nor less correct than the former when dealing with acidic aqueous solutions. I will restrict myself to discussing the problem assuming $\ce{HCl (aq)}$; but note that hydrogen chloride is a much stronger acid than hydronium and thus will be practically completely dissociated.

To fully analyse the reaction in question, we need to determine where a proton has been transferred. I will write it slightly differently to emphasise the fact that the products remain in aqueous solution; compare $(1)$.

$$\ce{HCl (aq) + Na+ + OH- -> H2O + Na+ + Cl-}\tag{1}$$

If we look at the non-hydrogen elements, we notice that chlorine is bound to hydrogen on the reactant side but no longer on the product side. Likewise, oxygen is bound to one hydrogen on the reactant side but to two on the product side. We have thus determined the species that will donate and accept a proton; the reaction can be classified as a proton transfer reaction.

$$\ce{\underset{acid}{HCl (aq)} + Na+ + \underset{base}{OH-} -> \underset{{conj. acid}}{H2O} + Na+ + \underset{{conj. base}}{Cl-}}\tag{1'}$$

As expected, the neutralisation can be considered a Brønsted-Lowry acid-base reaction. The corresponding proof in the case of dissociated hydrochloric acid ($\ce{H3O+ (aq) + Cl- (aq)}$) is left to the reader.