Problem 20 - Chemistry: The Central Science

Question

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

Step-by-Step Solution

Answer

(a) In $\mathrm{HBrO}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{BrO}^{-}(a q)$, acid: $\mathrm{HBrO}$ , base: $\mathrm{H}_{2} \mathrm{O}$ , conjugate acid: $\mathrm{H}_{3} \mathrm{O}^{+}$ , conjugate base: $\mathrm{BrO}^{-}$. (b) In $\mathrm{HSO}_{4}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{SO}_{4}{\underline{\phantom{xx}}}^{2-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)$, acid: $\mathrm{HSO}_{4}^{-}$ , base: $\mathrm{HCO}_{3}^{-}$ , conjugate acid: $\mathrm{H}_{2} \mathrm{CO}_{3}$ , conjugate base: $\mathrm{SO}_{4}^{2-}$. (c) In $\mathrm{HSO}_{3}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$, acid: $\mathrm{H}_{3} \mathrm{O}^{+}$ , base: $\mathrm{HSO}_{3}^{-}$ , conjugate acid: $\mathrm{H}_{2} \mathrm{SO}_{3}$ , conjugate base: $\mathrm{H}_{2} \mathrm{O}$.

Step 1: Identify the Acid and Base

In the equation, $\mathrm{HBrO}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{BrO}^{-}(a q)\( , the \)\mathrm{HBrO}\( donates a proton to \)\mathrm{H}_{2} \mathrm{O}\(. Therefore, \)\mathrm{HBrO}\( is the acid, and \)\mathrm{H}_{2} \mathrm{O}$ is the base.

Step 2: Identify the Conjugate Acid and Conjugate Base

Upon losing a proton, \(\mathrm{HBrO}\) forms \(\mathrm{BrO}^{-}\), which is the conjugate base. \(\mathrm{H_{2}O}\) gains a proton to form \(\mathrm{H_{3}O^{+}}\), which is the conjugate acid. (b)

Step 3: Identify the Acid and Base

In the equation, $\mathrm{HSO}_{4}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{SO}_{4}{\underline{\phantom{xx}}}^{2-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q)\( , the \)\mathrm{HSO_{4}^{-}}\( donates a proton to \)\mathrm{HCO_{3}^{-}}\(. Therefore, \)\mathrm{HSO_{4}^{-}}\( is the acid, and \)\mathrm{HCO_{3}^{-}}$ is the base.

Step 4: {Identify the Conjugate Acid and Conjugate Base}

Upon losing a proton, \(\mathrm{HSO_{4}^{-}}\) forms \(\mathrm{SO_{4}^{2-}}\), which is the conjugate base. \(\mathrm{HCO_{3}^{-}}\) gains a proton to form \(\mathrm{H_{2}CO_{3}}\), which is the conjugate acid. (c)

Step 5: Identify the Acid and Base

In the equation, $\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)\(, the \)\mathrm{HSO}_{3}^{-}\( accepts a proton from \)\mathrm{H}_{3} \mathrm{O}^{+}\(. Therefore, \)\mathrm{HSO}_{3}^{-}\( is the base, and \)\mathrm{H}_{3} \mathrm{O}^{+}$ is the acid.

Step 6: Identify the Conjugate Acid and Conjugate Base

Upon gaining a proton, \(\mathrm{HSO}_{3}^{-}\) forms \(\mathrm{H}_{2} \mathrm{SO}_{3}\), which is the conjugate acid. \(\mathrm{H_{3}O^{+}}\) loses a proton to form \(\mathrm{H_{2}O}\), which is the conjugate base.

Key Concepts

Understanding Conjugate Acids

In Brønsted-Lowry acid-base theory, a conjugate acid is the species formed when a base gains a proton (H⁺). To picture this, imagine the base as something thirsty for a proton, and once it grabs one, it becomes a conjugate acid.
For example, in the reaction: \[\mathrm{HBrO}(aq) + \mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{BrO^-}(aq)\]Here, when the water molecule \((\mathrm{H_2O})\) gains a proton, it becomes hydronium \((\mathrm{H_3O^+})\), the conjugate acid. Seeing this in reactions:

• A base accepts a proton and turns into a conjugate acid.
•  This conjugate acid can potentially donate a proton back, making its role reversible.

Conjugate acids create a dynamic balance with their base counterparts in chemical equilibrium. This balance plays a critical role in maintaining chemical stability in various systems.

Exploring Conjugate Bases

A conjugate base is what’s left after an acid donates a proton. It’s like an acid after a "proton breakup." The conjugate base now has the capacity to re-accumulate the lost proton.Let's look at this reaction:\[\mathrm{HSO_4^-}(aq) + \mathrm{HCO_3^-}(aq) \rightleftharpoons \mathrm{SO_4^{2-}}(aq) + \mathrm{H_2CO_3}(aq)\]Here, after the \(\mathrm{HSO_4^-}\) donates a proton, it transforms into \(\mathrm{SO_4^{2-}}\), the conjugate base.
Key features about conjugate bases include:

• An acid loses a proton to become a conjugate base.
• This base can regain a proton to revert to its original acid form, creating equilibrium in the reaction.

The interplay of acids and their conjugate bases is a central theme in predicting the direction of proton transfer reactions. This concept underpins buffer solutions, which resist changes in pH levels.

Proton Transfer Reactions Explained

Proton transfer reactions are central to Brønsted-Lowry acid-base chemistry. They describe how a proton is relocated between different chemical species. These reactions can often look like a ‘proton bridge’ forming alleys between acids and bases.
Take the reaction:\[\mathrm{HSO_3^-}(aq) + \mathrm{H_3O^+}(aq) \rightleftharpoons \mathrm{H_2SO_3}(aq) + \mathrm{H_2O}(l)\]Here, \(\mathrm{H_3O^+}\) donates a proton to \(\mathrm{HSO_3^-}\), resulting in the formation of \(\mathrm{H_2SO_3}\) and \(\mathrm{H_2O}\). This signifies that:

• Acids are proton donors and make conjugate bases.
• Bases are proton acceptors and form conjugate acids.

Proton transfer reactions bidirectionally connect acids and bases, reinforcing their dynamic and reversible nature in achieving equilibrium. Understanding the directional nature and preference of proton transfer is key in predicting reaction outcomes and equilibria in various chemical processes.