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Chapter 16: Problem 22

(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 \(\mathrm{H}_{2} \mathrm{O}(l) .(\mathrm{c})\) What is the conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-} ?\) What is its conjugate base?

Short Answer

Expert verified
(a) As a base, the reaction is: \(\mathrm{H}_{2}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}^{-} + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5} + \mathrm{OH}^{-}(a q)\). (b) As an acid, the reaction is: \(\mathrm{H}_{2}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}^{-} + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HC}_{6}\mathrm{H}_{6}\mathrm{O}_{5}^{2-} + \mathrm{H}_{3}\mathrm{O}^{+}(a q)\). (c) The conjugate acid is \(\mathrm{H}_{3}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}\), and the conjugate base is \(\mathrm{HC}_{6}\mathrm{H}_{6}\mathrm{O}_{5}^{2-}\).

Step by step solution

01

1. Analyze the given molecule as a base

When the given molecule acts as a base, it will accept a proton (\(\mathrm{H}^{+}\)) from the water molecules in the solution. Writing a balanced reaction equation for this process: \[ \mathrm{H}_{2}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}^{-} + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5} + \mathrm{OH}^{-}(a q) \]
02

2. Analyze the given molecule as an acid

When the given molecule acts as an acid, it will donate a proton (\(\mathrm{H}^{+}\)) to the water molecules in the solution. Writing a balanced reaction equation for this process: \[ \mathrm{H}_{2}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}^{-} + \mathrm{H}_{2}\mathrm{O}(l) \rightleftharpoons \mathrm{HC}_{6}\mathrm{H}_{6}\mathrm{O}_{5}^{2-} + \mathrm{H}_{3}\mathrm{O}^{+}(a q) \]
03

3. Find the conjugate acid

A conjugate acid is formed when the given molecule acts as a base, i.e., it accepts a proton (\(\mathrm{H}^{+}\)). From the balanced reaction in step 1, we can see that the conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}\) is \(\mathrm{H}_{3}\mathrm{C}_{6}\mathrm{H}_{7}\mathrm{O}_{5}\).
04

4. Find the conjugate base

A conjugate base is formed when the given molecule acts as an acid, i.e., it donates a proton (\(\mathrm{H}^{+}\)). From the balanced reaction in step 2, we can see that the conjugate base of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}\) is \(\mathrm{HC}_{6}\mathrm{H}_{6}\mathrm{O}_{5}^{2-}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conjugate Acid and Base
Understanding the concept of conjugate acid-base pairs is fundamental in acid-base chemistry. In any acid-base reaction, an acid donates a proton (hydrogen ion, \(\mathrm{H}^+\)), becoming its conjugate base, while the base accepts the proton, becoming its conjugate acid. This can be visualized as a balanced equation with two conjugate pairs.

For instance, when \(\mathrm{H}_2\mathrm{C}_6\mathrm{H}_7\mathrm{O}_5^-\) acts as a base and accepts a proton from water, the resulting conjugate acid is \(\mathrm{H}_3\mathrm{C}_6\mathrm{H}_7\mathrm{O}_5\). Conversely, when \(\mathrm{H}_2\mathrm{C}_6\mathrm{H}_7\mathrm{O}_5^-\) acts as an acid and donates a proton, it forms the conjugate base \(\mathrm{HC}_6\mathrm{H}_6\mathrm{O}_5^{2-}\).
Chemical Equilibrium
Chemical equilibrium occurs when a chemical reaction and its reverse reaction proceed at the same rate. As a result, the concentrations of the reactants and products remain unchanged over time.

In the equation provided:
\[\mathrm{H}_2\mathrm{C}_6\mathrm{H}_7\mathrm{O}_5^{-} + \mathrm{H}_2\mathrm{O}(l) \rightleftharpoons \mathrm{H}_3\mathrm{C}_6\mathrm{H}_7\mathrm{O}_5 + \mathrm{OH}^{-}(aq)\]
The double arrows indicate that the reaction can go both ways, achieving equilibrium. The ratio of the product's concentration to the reactant's concentration at equilibrium is called the equilibrium constant \(K_{eq}\). It is crucial to recognize that changing conditions such as concentration, temperature, or pressure can shift the equilibrium, favoring either the forward or the reverse reaction.
Proton Transfer Reactions
Proton transfer is the hallmark of acid-base chemistry. Reactions involving the transfer of a proton from one species to another are called proton transfer reactions. They are also known as Br酶nsted-Lowry acid-base reactions.

An important aspect of these reactions is their reversible nature. In solution, molecules are constantly engaging in proton transfers, reaching a state of dynamic equilibrium. For example, when \(\mathrm{H}_2\mathrm{C}_6\mathrm{H}_7\mathrm{O}_5^-\) donates a proton, it does so through a reversible reaction. It can easily regain the proton under the right conditions, showcasing the dynamic interplay between acids and bases.

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Most popular questions from this chapter

(a) What is a strong base? (b) A solution is labeled \(0.035 \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}\). What is \(\left[\mathrm{OH}^{-}\right]\) for the solution? (c) Is the following statement true or false? Because \(\mathrm{Mg}(\mathrm{OH})_{2}\) is not very soluble, it cannot be a strong base. Explain.

(a) Which of the following is the stronger Br酶nstedLowry acid, \(\mathrm{HBrO}\) or \(\mathrm{HBr}\) ? (b) Which is the stronger Br酶nsted-Lowry base, \(\mathrm{F}^{-}\) or \(\mathrm{Cl}^{-} ?\) Briefly explain your choices.

Saccharin, a sugar substitute, is a weak acid with \(\mathrm{pK}_{a}=2.32\) at \(25^{\circ} \mathrm{C}\). It ionizes in aqueous solution as follows: $$ \mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q) $$ What is the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of this substance?

Although pure \(\mathrm{NaOH}\) and \(\mathrm{NH}_{3}\) have very different properties, their aqueous solutions possess many common properties. List some general properties of these solutions, and explain their common behavior in terms of the species present.

The amino acid glycine \(\left(\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}\right)\) can participate in the following equilibria in water: $$ \begin{aligned} \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons & \\ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-}+\mathrm{H}_{3} \mathrm{O}^{+} & K_{a}=4.3 \times 10^{-3} \\ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons & \\ { }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH}+\mathrm{OH}^{-} & K_{b}=6.0 \times 10^{-5} \end{aligned} $$ (a) Use the values of \(K_{a}\) and \(K_{b}\) to estimate the equilibrium constant for the intramolecular proton transfer to form a zwitterion: $$ \mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COOH} \rightleftharpoons{ }^{+} \mathrm{H}_{3} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{COO}^{-} $$ What assumptions did you need to make? (b) What is the \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) aqueous solution of glycine? (c) What would be the predominant form of glycine in a solution with pH 13? With pH 1?
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