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(a) Give the conjugate base of the following Br酶nstedLowry acids: (i) \(\mathrm{H}_{2} \mathrm{SO}_{3},\) (ii) \(\mathrm{HSO}_{3}^{-}\) (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\), (ii) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\).

Short Answer

Expert verified
The conjugate bases of the given Br酶nsted-Lowry acids are: (a) (i) \( \mathrm{HSO}_{3}^{-} \), (ii) \( \mathrm{SO}_{3}^{2-} \) And the conjugate acids of the given Br酶nsted-Lowry bases are: (b) (i) \( \mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \), (ii) \( \mathrm{CH}_{3} \mathrm{COOH} \)

Step by step solution

01

Conjugate bases of Br酶nsted-Lowry acids

To find the conjugate base of a Br酶nsted-Lowry acid, we must remove one hydrogen ion (H鈦) from the acid species. (a) (i) The first acid is H鈧係O鈧, so we remove one H鈦 ion, which results in the conjugate base of the acid: \[ \mathrm{H}_{2} \mathrm{SO}_{3} \rightarrow \mathrm{HSO}_{3}^{-} \] (ii) The second acid is HSO鈧冣伝, so we remove one H鈦 ion, which results in the conjugate base of the acid: \[ \mathrm{HSO}_{3}^{-} \rightarrow \mathrm{SO}_{3}^{2-} \]
02

Conjugate acids of Br酶nsted-Lowry bases

To find the conjugate acid of a Br酶nsted-Lowry base, we must add one hydrogen ion (H鈦) to the base species. (b) (i) The first base is CH鈧僋H鈧, so we add one H鈦 ion, which results in the conjugate acid of the base: \[ \mathrm{CH}_{3}\mathrm{NH}_{2} + \mathrm{H}^{+} \rightarrow \mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \] (ii) The second base is CH鈧僀OO鈦, so we add one H鈦 ion, which results in the conjugate acid of the base: \[ \mathrm{CH}_{3} \mathrm{COO}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{CH}_{3} \mathrm{COOH} \] To recap, the conjugate bases of the given Br酶nsted-Lowry acids are: (a) (i) HSO鈧冣伝, (ii) SO鈧兟测伝 And the conjugate acids of the given Br酶nsted-Lowry bases are: (b) (i) CH鈧僋H鈧冣伜, (ii) CH鈧僀OOH

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Key Concepts

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

Conjugate Acid
In the Br酶nsted-Lowry acid-base theory, a conjugate acid is formed when a base gains a hydrogen ion (H鈦). When a compound acts as a base, it "accepts" the hydrogen ion to become the conjugate acid of that base. This concept illustrates the reversible nature of chemical reactions involving acids and bases.

Let's dive into the examples provided. Consider the base \(\mathrm{CH}_{3} \mathrm{NH}_{2} \) (methylamine). When it accepts a hydrogen ion (H鈦), it transforms into its conjugate acid:

\[ \\mathrm{CH}_{3}\mathrm{NH}_{2} + \mathrm{H}^{+} \rightarrow \mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \\]
The structure of methylamine changes with the addition of H鈦, and it becomes positively charged due to the extra proton. This is the essence of becoming a conjugate acid. The addition of an H鈦 ion signifies the switch from being a base to becoming its conjugate acid.

Another example from the exercise is the base \(\mathrm{CH}_{3} \mathrm{COO}^{-} \) (acetate ion). It becomes its conjugate acid, acetic acid (\mathrm{CH}_{3}\mathrm{COOH}), by accepting an H鈦 ion:
  • Gain of H鈦 by \mathrm{CH}_{3}\mathrm{COO}^{-}
  • Formation of \mathrm{CH}_{3}\mathrm{COOH}
These transformations are vital to understanding the dynamic response of substances in different environments, showcasing the versatility and reactivity of chemical species.
Conjugate Base
The concept of a conjugate base is crucial in understanding acid-base reactions in the Br酶nsted-Lowry theory. A conjugate base results when an acid donates a hydrogen ion (H鈦). This means that the acid has lost a proton and converted into its conjugate base counterpart.

For instance, take the acid \(\mathrm{H}_{2} \mathrm{SO}_{3} \) (sulfurous acid). Upon losing an H鈦, it forms the conjugate base \(\mathrm{HSO}_{3}^{-} \). The transformation can be represented by the equation:

\[ \\mathrm{H}_{2} \mathrm{SO}_{3} \rightarrow \mathrm{HSO}_{3}^{-} + \mathrm{H}^{+} \\]
By donating an H鈦, sulfurous acid decreases its positive charge, effectively turning into the conjugate base \(\mathrm{HSO}_{3}^{-} \). As a result, this conjugate base is ready to accept an H鈦 ion in another reaction, demonstrating its potential to revert back to its acid form.

Another example is with \(\mathrm{HSO}_{3}^{-} \). Upon losing another H鈦, it becomes \(\mathrm{SO}_{3}^{2-} \) (sulfite ion), following:
  • \mathrm{HSO}_{3}^{-} loses H鈦
  • Forms \mathrm{SO}_{3}^{2-}
The remarkable ability for acids and bases to switch between donating and accepting protons is a key highlight of the dynamic equilibrium between compounds, allowing for compensation and adjustment to pH changes.
Hydrogen Ion (H鈦) Exchange
The exchange of hydrogen ions (H鈦) is at the heart of the Br酶nsted-Lowry acid-base theory. The movement of these protons helps define the nature of acidic and basic reactions.

When a hydrogen ion is transferred from one compound to another, it determines whether a substance acts as an acid or a base. An acid donates an H鈦 ion, while a base accepts one. This exchange is what leads to the formation of conjugate pairs.

Consider the process of H鈦 exchange with \(\mathrm{H}_{2} \mathrm{SO}_{3} \):
  • As an acid, \(\mathrm{H}_{2} \mathrm{SO}_{3} \) donates an H鈦.
  • The remaining species, \(\mathrm{HSO}_{3}^{-} \), becomes the conjugate base.
This balance maintains the dynamic nature and equilibrium seen in many acid-base systems, allowing substances to adapt and perform various biochemical functions.

Similarly, when \(\mathrm{CH}_{3} \mathrm{NH}_{2} \) receives an H鈦, it transitions from a base to its conjugate acid, \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+} \):
  • H鈦 acceptance changes its nature from base to acid.
The constant exchange of H鈦 is fundamental, facilitating many of the vital chemical processes found in nature and industrial applications. By comprehending this exchange, we gain insight into how substances can switch roles, maintaining balance in chemical reactions.

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Most popular questions from this chapter

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is a diprotic acid. By using data in Appendix \(\mathrm{D}\) as needed, determine whether each of the following statements is true: (a) \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) can serve as both a Bronsted-Lowry acid and a Br酶nsted-Lowry base. (b) \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is the conjugate base of \(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\). (c) An aqueous solution of the strong electrolyte \(\mathrm{KHC}_{2} \mathrm{O}_{4}\) will have \(\mathrm{pH}<7\).

Write the chemical equation and the \(K_{a}\) expression for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with \(\mathrm{H}^{+}(a q)\) as a product and then with the hydronium ion: (a) \(\mathrm{HSO}_{4}^{-}\), (b) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\).

Calculate the pH of each of the following solutions \(\left(K_{a}\right.\) and \(K_{b}\) values are given in Appendix D): (a) \(0.150 \mathrm{M}\) propionic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{COOH}\right)\) (b) \(0.250 \mathrm{M}\) hydrogen chromate ion \(\left(\mathrm{HCrO}_{4}^{-}\right),(\mathbf{c}) 0.750 \mathrm{M}\) pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\right)\)

(a) Given that \(K_{b}\) for ammonia is \(1.8 \times 10^{-5}\) and that for hydroxylamine is \(1.1 \times 10^{-8}\), which is the stronger base? (b) Which is the stronger acid, the ammonium ion or the hydroxylammonium ion? (c) Calculate \(K_{a}\) values for \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{H}_{3} \mathrm{NOH}^{+}\).

(a) Write a chemical equation that illustrates the autoionization of water. (b) Write the expression for the ionproduct constant for water, \(K_{w} .(\mathbf{c})\) If a solution is described as basic, which of the following is true: (i) \(\left[\mathrm{H}^{+}\right]>\left[\mathrm{OH}^{-}\right]\), (ii) \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right],\) or (iii) \(\left[\mathrm{H}^{+}\right]<\left[\mathrm{OH}^{-}\right] ?\)

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