Write the formula for the conjugate acid of each base. (a) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) (b) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\) (c) \(\mathrm{Cl}^{-}\) (d) \(\mathrm{F}^{-}\)

Short Answer

Expert verified
The conjugate acids are: (a) \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\), (b) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\), (c) \(\mathrm{HCl}\), and (d) \(\mathrm{HF}\).

Step by step solution

01

Identify the Base and Add a Proton

To find the conjugate acid of a base, add a hydrogen ion (proton, H+) to the base.
02

Write the Conjugate Acid for (a) \(\mathrm{CH}_{3} \mathrm{NH}_{2}\)

Add a hydrogen ion to the nitrogen in \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) to obtain its conjugate acid, \(\mathrm{CH}_{3} \mathrm{NH}_{3}^{+}\).
03

Write the Conjugate Acid for (b) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)

Add a hydrogen ion to the nitrogen in \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\) to obtain its conjugate acid, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\).
04

Write the Conjugate Acid for (c) \(\mathrm{Cl}^{-}\)

Add a hydrogen ion to the chloride ion \(\mathrm{Cl}^{-}\) to obtain its conjugate acid, which is hydrochloric acid, \(\mathrm{HCl}\).
05

Write the Conjugate Acid for (d) \(\mathrm{F}^{-}\)

Add a hydrogen ion to the fluoride ion \(\mathrm{F}^{-}\) to obtain its conjugate acid, which is hydrofluoric acid, \(\mathrm{HF}\).

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Key Concepts

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

Bronsted-Lowry Acids and Bases
The concept of Bronsted-Lowry acids and bases is fundamental to chemistry, defining an acid as any substance that can donate a proton (H+), and a base as any substance that can accept a proton. The proton, a hydrogen ion without its accompanying electron, is the currency of acid-base reactions. This definition expands the scope beyond just hydrogen ions and hydroxide ions, allowing for a broader array of chemical species to be classified as acids or bases. It's important to note that the process is dynamic; when an acid donates a proton, it becomes a base, often referred to as its conjugate base, while the base that accepts the proton becomes an conjugate acid.

For example, when ammonia (NH3) acts as a base and accepts a proton, it becomes the ammonium ion (NH4+), its conjugate acid. This concept is crucial as it led to a deeper understanding of chemical solutions and the reactions occurring within them.
Conjugate Acid-Base Pairs
Conjugate acid-base pairs are central to understanding the Bronsted-Lowry concept of acids and bases. In any acid-base reaction, the reactants (the original acid and base) and the products (the conjugate base and conjugate acid) are linked as pairs. The conjugate base is what remains of the acid after it has donated its proton, and vice versa for the conjugate acid after the base accepts a proton.

Acid-base pairs are like two sides of a chemical equation's coin. For instance, in the reaction between hydrochloric acid (HCl) and water (H2O), the HCl donates a proton to become Cl-, the conjugate base, while water accepts a proton to become H3O+, the conjugate acid. This pairing helps chemists predict the outcomes of chemical reactions and the strength of the acids and bases involved. Stronger acids have weaker conjugate bases, and conversely, stronger bases have weaker conjugate acids. Recognizing these pairs is essential for mastering acid-base chemistry.
Acid-Base Reactions
Acid-base reactions are characterized by the transfer of protons from one reactant to another. These reactions involve the formation of conjugate acid-base pairs, as seen in the reaction between an acid and a base. Acid-base reactions are also categorized as neutralization reactions, particularly when dealing with strong acids and bases, as they generally result in the formation of water and a salt.

The Bronsted-Lowry theory helps describe such reactions even when no obvious salt is formed, as in the case of ammonia reacting with water to form ammonium and hydroxide ions. Understanding these reactions goes beyond just recognizing the exchange of hydrogen ions; it's also about predicting the direction of reaction equilibrium, the strengths of acids and bases involved, and the pH changes in a solution.
Chemical Formulas
The language of chemical formulas is how we communicate the composition of chemical compounds. A formula uses letters and numbers as a shorthand to convey the types and quantities of atoms present. Hydrogen atoms, for example, are commonly represented by 'H', oxygen by 'O', chlorine by 'Cl', and so on. Subscripts are used to indicate the number of each type of atom within a molecule, and superscripts denote ion charges after an acid or base has reacted.

Understanding chemical formulas is crucial for studying acid-base reactions and creating conjugate pairs. For example, the formula for the conjugate acid of the base CH3NH2 is CH3NH3+, with the additional hydrogen marked as a subscript and the positive charge as a superscript. This ability to read and write chemical formulas is essential for communicating effectively in the language of chemistry and for succeeding in further chemical studies.

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Most popular questions from this chapter

Determine the \(\mathrm{pH}\) of each solution and classify it as acidic, basic, or neutral. (a) \(\mathrm{pOH}=12.5\) (b) \(\mathrm{pOH}=5.5\) (c) \(\mathrm{pOH}=0.55\) (d) \(\mathrm{pOH}=7.98\)

Four solutions of unknown \(\mathrm{HCl}\) concentration are titrated with solutions of \(\mathrm{NaOH}\). The following table lists the volume of each unknown \(\mathrm{HCl}\) solution, the volume of \(\mathrm{NaOH}\) solution required to reach the equivalence point, and the concentration of each \(\mathrm{NaOH}\) solution. Calculate the concentration (in M) of the unknown HCl solution in each case.

For each reaction, identify the Brønsted-Lowry acid, the Bronsted-Lowry base, the conjugate acid, and the conjugate base. (a) \(\mathrm{HI}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{I}^{-}(a q)\) (b) \(\mathrm{CH}_{3} \mathrm{NH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}{ }^{+}(a q)+\mathrm{OH}^{-}(a q)\) (c) \(\mathrm{CO}_{3}{ }^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_{3}{ }^{-}(a q)+\mathrm{OH}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)\)

Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\)for each solution. (a) \(\mathrm{pH}=1.76\) (b) \(\mathrm{pH}=3.88\) (c) \(\mathrm{pH}=8.43\) (d) \(\mathrm{pH}=12.32\)

Calculate \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\)for each solution. (a) \(\mathrm{pH}=8.55\) (b) \(\mathrm{pH}=11.23\) (c) \(\mathrm{pH}=2.87\) (d) \(\mathrm{pH}=1.22\)

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