Question

Designate the Brønsted-Lowry acid and the Brønsted-Lowry base on the left side of each equation, and also designate the conjugate acid and conjugate base of each on the right side. (a) \(\mathrm{HBrO}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{BrO}^{-}(a q)\) (b) \(\mathrm{HSO}_{4}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons\) \(\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\) (c) \(\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\)

Short answer

(a) Acid: HBrO, Base: H2O, Conjugate Acid: H3O+, Conjugate Base: BrO- (b) Acid: HSO4-, Base: HCO3-, Conjugate Acid: H2CO3, Conjugate Base: SO4^2- (c) Acid: H3O+, Base: HSO3-, Conjugate Acid: H2SO3, Conjugate Base: H2O

Step-by-step solution

Identify the acid and base on the left side

In this equation, HBrO donates a hydrogen ion (H+) to H2O. Therefore, HBrO is the Brønsted-Lowry acid and H2O is the Brønsted-Lowry base.

Identify the conjugate acid and conjugate base on the right side

On the right side of the equation, we have H3O+ and BrO-. Since H3O+ is the species formed after H2O accepted a hydrogen ion, it is the conjugate acid. Similarly, BrO- is the species left after HBrO donated a hydrogen ion, making it the conjugate base. (b) \(\mathrm{HSO}_{4}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons\) \(\mathrm{SO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\)

Identify the acid and base on the left side

In this equation, HSO4- donates a hydrogen ion (H+) to HCO3-. Therefore, HSO4- is the Brønsted-Lowry acid and HCO3- is the Brønsted-Lowry base.

Identify the conjugate acid and conjugate base on the right side

On the right side of the equation, we have SO4^2- and H2CO3. Since SO4^2- is the species left after HSO4- donated a hydrogen ion, it is the conjugate base. Similarly, H2CO3 is the species formed after HCO3- accepted a hydrogen ion, making it the conjugate acid. (c) $\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$

Identify the acid and base on the left side

In this equation, H3O+ donates a hydrogen ion (H+) to HSO3-. Therefore, H3O+ is the Brønsted-Lowry acid and HSO3- is the Brønsted-Lowry base.

Identify the conjugate acid and conjugate base on the right side

On the right side of the equation, we have H2SO3 and H2O. Since H2SO3 is the species formed after HSO3- accepted a hydrogen ion, it is the conjugate acid. Similarly, H2O is the species left after H3O+ donated a hydrogen ion, making it the conjugate base.

Key concepts

Acid-Base Reactions

Understanding acid-base reactions is crucial in chemistry, as they are among the most common types of chemical reactions. According to the Brønsted-Lowry acid-base theory, these reactions are all about the transfer of hydrogen ions, or protons. An acid is a substance that donates a proton, while a base is one that accepts it.

Let's take example (a), where \(\mathrm{HBrO}\) donates a hydrogen ion to \(\mathrm{H}_{2}\mathrm{O}\), forming \(\mathrm{H}_{3}\mathrm{O}^{+}\) and \(\mathrm{BrO}^{-}\). The reaction equation \(\mathrm{HBrO}(aq)+\mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+}(aq)+\mathrm{BrO}^{-}(aq)\) clearly shows this transfer. The direction of the reaction arrow indicates that the reaction can go both ways, meaning it can reach a state of equilibrium where the rate of the forward reaction equals that of the reverse reaction. It's important to note that in all acid-base reactions, there is a conservation of charge and mass.

To improve understanding, it's beneficial to visualize the reaction as a simple exchange — picture a hydrogen ion hopping from the acid to the base. This approach can demystify the concept and make the mechanism more approachable for students.

Conjugate Acid-Base Pairs

In every acid-base reaction, two conjugate acid-base pairs are formed. These pairs are two molecules or ions that are related by the loss or gain of a single hydrogen ion. After a Brønsted-Lowry acid donates a proton, it becomes a conjugate base. Similarly, after a Brønsted-Lowry base accepts a proton, it becomes a conjugate acid.

For instance, in example (b), \(\mathrm{HSO}_{4}^{-}\) (the acid) donates a proton to \(\mathrm{HCO}_{3}^{-}\) (the base), resulting in \(\mathrm{SO}_{4}^{2-}\) and \(\mathrm{H}_{2}\mathrm{CO}_{3}\). Here, \(\mathrm{SO}_{4}^{2-}\) and \(\mathrm{HCO}_{3}^{-}\) form one conjugate pair, while \(\mathrm{HSO}_{4}^{-}\) and \(\mathrm{H}_{2}\mathrm{CO}_{3}\) form the second pair.

To distinguish between the pairs and understand the reaction's dynamics, remember that the acid and its conjugate base differ by one hydrogen ion, as do the base and its conjugate acid. With this pattern, one can trace which species will act as an acid or base in a reverse reaction, providing a solid foundation for predicting the behavior of substances in different chemical contexts.

Hydrogen Ion Transfer

The hydrogen ion transfer is the defining process of Brønsted-Lowry acid-base reactions. A hydrogen ion (H+), which is simply a proton, is what gets transferred from the acid to the base. This transfer is the core event that determines the acidic or basic nature of the substances involved.

Looking at example (c), \(\mathrm{H}_{3}\mathrm{O}^{+}\), a hydronium ion, donates a proton to \(\mathrm{HSO}_{3}^{-}\), resulting in \(\mathrm{H}_{2}\mathrm{SO}_{3}\) and water \(\mathrm{H}_{2}\mathrm{O}\). The hydronium ion \(\mathrm{H}_{3}\mathrm{O}^{+}\) acts as the acid because it donates the proton, and \(\mathrm{HSO}_{3}^{-}\) acts as the base because it accepts the proton. The transfer of the proton is what transforms the reactants into the products — from \(\mathrm{H}_{3}\mathrm{O}^{+}\) and \(\mathrm{HSO}_{3}^{-}\) to \(\mathrm{H}_{2}\mathrm{SO}_{3}\) and \(\mathrm{H}_{2}\mathrm{O}\).

To make this concept easier to digest, imagine the proton as a 'hot potato' being passed between molecules. Once a molecule receives the 'hot potato,' it becomes charged positively (as in the formation of a conjugate acid) and becomes eager to pass it on, illustrating the propensity for the reaction to proceed in both directions, towards equilibrium.