Question

(a) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base in \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as an acid in \(\mathrm{H}_{2} \mathrm{O}(l) .\) (c) What is the conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q) ?\) What is its conjugate base?

Short answer

a) The equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base in \(\mathrm{H}_{2} \mathrm{O}(l)\) is: \[ \mathrm{H}_{2}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}(aq) + \mathrm{OH^{-}}(aq) \] b) The equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as an acid in \(\mathrm{H}_{2} \mathrm{O}(l)\) is: \[ \mathrm{H}_{2}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HC}_{6}\mathrm{H}_{6}\mathrm{O}_{5}^{2-}(aq) + \mathrm{H}_{3}\mathrm{O^{+}}(aq) \] c) The conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) is \(\mathrm{H}_{3}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}(aq)\), and its conjugate base is \(\mathrm{HC}_{6}\mathrm{H}_{6}\mathrm{O}_{5}^{2-}(aq)\).

Step-by-step solution

Write the equation of the reaction.

To write the equation where \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base in \(\mathrm{H}_{2} \mathrm{O}(l)\), consider the behavior of a base according to Bronsted-Lowry definition: a base accepts a proton \((\mathrm{H}^{+})\). In the reaction with water, the base will accept a proton from \(\mathrm{H}_{2}\mathrm{O}(l)\). The equation will look like this: \[ \mathrm{H}_{2}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}(aq) + \mathrm{OH^{-}}(aq) \] b) H2C6H7O5- acting as an acid in H2O(l)

Write the equation of the reaction.

To write the equation where \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as an acid in \(\mathrm{H}_{2} \mathrm{O}(l)\), consider the behavior of an acid according to Bronsted-Lowry definition: an acid donates a proton \((\mathrm{H}^{+})\). In the reaction with water, the acid will donate a proton to \(\mathrm{H}_{2}\mathrm{O}(l)\). The equation will look like this: \[ \mathrm{H}_{2}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}^{-}(aq) + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HC}_{6}\mathrm{H}_{6}\mathrm{O}_{5}^{2-}(aq) + \mathrm{H}_{3}\mathrm{O^{+}}(aq) \] c) Conjugate acid and conjugate base

Identify the conjugate acid.

The conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) is the species formed after accepting a proton. From the equation in (a), the conjugate acid is: \[ \mathrm{H}_{3}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}(aq) \]

Identify the conjugate base.

The conjugate base of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) is the species formed after donating a proton. From the equation in (b), the conjugate base is: \[ \mathrm{HC}_{6}\mathrm{H}_{6}\mathrm{O}_{5}^{2-}(aq) \]

Key concepts

Understanding Conjugate Acids

In the Bronsted-Lowry acid-base theory, a conjugate acid forms when a base accepts a proton (H^{+}). Let's explore this idea using our example with H_{2}C_{6}H_{7}O_{5}^{-}. In reaction with water, it acts as a base and accepts a proton, transforming into its conjugate acid:

• The water molecule donates a proton to H_{2}C_{6}H_{7}O_{5}^{-}.
• This results in the formation of H_{3}C_{6}H_{7}O_{5}, the conjugate acid.

This reaction highlights the transferable nature of hydrogen ions and their role in forming conjugate acid-base pairs, illustrated by the equation:\[H_{2}C_{6}H_{7}O_{5}^{-} + H_{2}O \rightleftharpoons H_{3}C_{6}H_{7}O_{5} + OH^{-}\]The concept of conjugate acids helps us see the dynamic balance in chemical reactions.

Identifying Conjugate Bases

A conjugate base emerges when an acid donates a proton. In our case, H_{2}C_{6}H_{7}O_{5}^{-} acts as an acid and donates a proton to water, producing its conjugate base:

• H_{2}C_{6}H_{7}O_{5}^{-} gives up an H^{+} to H_{2}O.
• This forms HC_{6}H_{6}O_{5}^{2-}, the conjugate base.

This reaction can be displayed as:\[H_{2}C_{6}H_{7}O_{5}^{-} + H_{2}O \rightleftharpoons HC_{6}H_{6}O_{5}^{2-} + H_{3}O^{+}\]Understanding conjugate bases is crucial, as it shows how substances can shift roles in chemical reactions, altering their identity in the process. This flexibility is key in studying equilibria in acid-base chemistry.

Exploring Acid-Base Reactions

Acid-base reactions are foundational in chemistry, particularly through the lens of the Bronsted-Lowry theory. This theory describes how protons are exchanged between substances:

• An acid is defined as a proton donor.
• A base is a proton acceptor.

In the context of H_{2}C_{6}H_{7}O_{5}^{-} , these roles illustrate how molecules can behave differently depending on their environment:

• In an acidic role, it donates a proton, generating hydronium ion H_{3}O^{+} and a conjugate base.
• In a basic role, it accepts a proton, producing OH^{-} and a conjugate acid.

These reactions highlight the reversible nature of many chemical processes, demonstrating the interconversion between acids and bases.