Brønsted–Lowry definition Knowledge-Based Study

Study notes

1. Expanding the Acid–Base Concept

   In 1884, Svante Arrhenius defined acids as substances that produce hydrogen ions ($\text{H}^+$) in aqueous solution, and bases as substances that produce hydroxide ions ($\text{OH}^-$). While useful, this definition was limited because it applied only to aqueous solutions and could not account for basic substances that do not contain hydroxide ions, such as ammonia ($\text{NH}_3$).

   In 1923, Danish chemist Johannes Brønsted and English chemist Thomas Lowry independently proposed a broader, more functional definition centered on proton transfer:

   • Brønsted–Lowry Acid: A chemical species (molecule or ion) that donates a proton ($\text{H}^+$) to another substance.
   • Brønsted–Lowry Base: A chemical species that accepts a proton ($\text{H}^+$) from another substance.

   Under this model, an acid–base reaction is essentially a proton-transfer process, which does not require water as a solvent or the explicit presence of hydroxide ions.

   Understanding Conjugate Pairs

   Because Brønsted–Lowry acid–base reactions involve the transfer of a proton, every reaction creates a linked pair of products capable of undergoing the reverse process. A conjugate acid–base pair consists of two species that differ from each other by exactly one hydrogen ion ($\text{H}^+$).

   • When a Brønsted–Lowry acid loses a proton, the remaining species becomes its conjugate base.
   • When a Brønsted–Lowry base gains a proton, the newly formed species becomes its conjugate acid.

   Example Reaction

   Consider the reaction between ammonia and water:

   $$\text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)}$$

   • $\text{H}_2\text{O}$ donates a proton to $\text{NH}_3$; therefore, $\text{H}_2\text{O}$ is the acid and $\text{OH}^-$ is its conjugate base.
   • $\text{NH}_3$ accepts a proton from $\text{H}_2\text{O}$; therefore, $\text{NH}_3$ is the base and $\text{NH}_4^+$ is its conjugate acid.

   The two conjugate pairs in this reaction are $\text{H}_2\text{O} / \text{OH}^-$ and $\text{NH}_4^+ / \text{NH}_3$.

   Dual Acid–Base Behavior

   Certain chemical species can either donate or accept a proton depending on the other reactant present. A substance capable of acting as either a Brønsted–Lowry acid or a Brønsted–Lowry base is termed amphiprotic (a specific category of amphoteric substances).

   Water as an Amphiprotic Molecule

   Water is the most common amphiprotic solvent. Its role depends entirely on the nature of the dissolved species:

   • Acting as a base: When reacting with hydrochloric acid, water accepts a proton:
     $$\text{HCl(aq)} + \text{H}_2\text{O(l)} \rightarrow \text{H}_3\text{O}^+\text{(aq)} + \text{Cl}^-\text{(aq)}$$
   • Acting as an acid: When reacting with ammonia, water donates a proton:
     $$\text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)}$$

   Other Common Amphiprotic Species

   Polyatomic anions containing ionizable hydrogens frequently display amphiprotic behavior, such as the hydrogen carbonate ion ($\text{HCO}_3^-$) and the dihydrogen phosphate ion ($\text{H}_2\text{PO}_4^-$).

   Systematic Approach to Acid–Base Analysis

   To identify the Brønsted–Lowry acid, base, and conjugate pairs in any chemical equation, follow these steps:

   1. Locate the proton transfer: Compare the reactant formulas to the product formulas to identify which species lost an $\text{H}^+$ and which gained an $\text{H}^+$.
   2. Identify the acid and base on the reactant side: The species that lost the proton is the acid; the species that gained the proton is the base.
   3. Match the conjugate partners on the product side: The acid becomes the conjugate base (one less $\text{H}^+$, charge decreases by 1). The base becomes the conjugate acid (one more $\text{H}^+$, charge increases by 1).

   Example for Practice

   In the reaction:$$\text{HSO}_4^-\text{(aq)} + \text{CO}_3^{2-}\text{(aq)} \rightleftharpoons \text{SO}_4^{2-}\text{(aq)} + \text{HCO}_3^-\text{(aq)}$$

   • $\text{HSO}_4^-$ donates $\text{H}^+$ $\rightarrow$ Acid (conjugate base: $\text{SO}_4^{2-}$)
   • $\text{CO}_3^{2-}$ accepts $\text{H}^+$ $\rightarrow$ Base (conjugate acid: $\text{HCO}_3^-$)