Write the formula for the conjugate base of each acid. $$\begin{array}{l}{\text { a. HCl }} \\ {\text { b. } \mathrm{H}_{2} \mathrm{SO}_{3}} \\ {\text { c. } \mathrm{HCHO}_{2}} \\ {\text { d. } \mathrm{HF}}\end{array}$$

Short Answer

Expert verified
a. HCl -> Cl-; b. H2SO3 -> HSO3-; c. HCHO2 -> CHO2-; d. HF -> F-.

Step by step solution

01

Understanding Conjugate Base

The conjugate base of an acid is formed by removing a proton (H+) from the acid molecule. The formula for the conjugate base is thus the formula of the acid minus an H+ ion.
02

Identify the Conjugate Base of HCl

For hydrochloric acid (HCl), removing one H+ ion yields Cl-, which is the chloride ion. The conjugate base of HCl is Cl-.
03

Identify the Conjugate Base of H2SO3

For sulfurous acid (H2SO3), removing one H+ ion yields HSO3-, which is the hydrogen sulfite ion. The conjugate base of H2SO3 is HSO3-.
04

Identify the Conjugate Base of HCHO2

For formic acid (HCHO2), removing one H+ ion yields CHO2-, which is the formate ion. The conjugate base of HCHO2 is CHO2-.
05

Identify the Conjugate Base of HF

For hydrofluoric acid (HF), removing one H+ ion yields F-, which is the fluoride ion. The conjugate base of HF is F-.

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Key Concepts

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

Acid-Base Chemistry
Understanding the nature of acids and bases is a cornerstone of chemical science, particularly when venturing into the world of acid-base reactions. An acid is a substance that can donate a proton (or hydrogen ion, H+), while a base is a recipient of a proton. This proton exchange defines their role in chemical reactions.

The strength of an acid or base is determined by its ability to donate or accept protons. Strong acids, such as hydrochloric acid (HCl), dissociate completely in water to release a proton and form their conjugate base. On the other hand, weak acids do not dissociate completely and have a stronger bond with their proton, like formic acid (HCHO2).

The concept of a conjugate base is directly tied to this idea of proton donation. When an acid donates a proton, it transforms into its conjugate base, effectively lowering its potential to donate further protons.
Conjugate Acid-Base Pairs
Delving into the heart of acid-base reactions, we encounter conjugate acid-base pairs. When an acid loses a proton, it forms its conjugate base, and the proton that was lost can bond with another base, forming its conjugate acid. This relationship is essential for understanding the acid-base balance in chemical reactions.

For example, in the case of hydrochloric acid (HCl), after it donates a proton, it becomes the chloride ion (Cl-), while the proton can go on to form the hydronium ion (H3O+) with a water molecule. This transformation exhibits the dynamic nature of acids and bases within aqueous solutions.

Understanding conjugate acid-base pairs also helps with predicting the outcomes of reactions and the strengths of the acids and bases involved. In fact, the stronger an acid, the weaker its conjugate base, and vice versa; this concept is a reliable indicator of the acid-base properties of substances.
Chemical Formulas
Chemical formulas represent the composition of a substance, providing information about the types and numbers of atoms within a molecule. When it comes to acids, their formulas typically start with hydrogen (H), indicating the hydrogen ion that may be donated. Upon forming a conjugate base, the formula is altered by the removal of this hydrogen ion.

Identifying Changes in Chemical Formulas

Take sulfurous acid (H2SO3) as an example. When one hydrogen ion is removed, the formula changes to HSO3-, illustrating its new role as a hydrogen sulfite ion. This change symbolizes the acid's shift to its conjugate base form, a critical aspect of chemical reactions involving acids.

Chemical formulas therefore are not arbitrary; they convey important information about a substance's role and behavior in chemical reactions, such as its potential to function as an acid or base in the transfer of protons, highlighting the interconnectedness of chemical concepts.

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Most popular questions from this chapter

A solution of 0.23 mol of the chloride salt of protonated quinine \(\left(\mathrm{QH}^{+}\right)\) a weak organic base, in 1.0 \(\mathrm{L}\) of solution has \(\mathrm{pH}=4.58 .\) Find the \(K_{\mathrm{b}}\) of quinine \((\mathrm{Q}) .\)

Base Solutions For each strong base solution, determine \(\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}_{3} \mathrm{O}^{+}\right], \mathrm{pH},\) and \(\mathrm{pOH} .\) $$\begin{array}{l}{\text { a. } 0.15 \times 10^{-3} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}} \\ {\text { b. } 1.5 \times 10^{-3} \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}} \\ {\text { c. } 4.8 \times 10^{-4} \mathrm{M} \mathrm{Sr}(\mathrm{OH})_{2}} \\ {\text { d. } 8.7 \times 10^{-5} \mathrm{M} \mathrm{KOH}}\end{array}$$

Determine the \(\mathrm{pH}\) of each two-component solution. $$\begin{array}{l}{\text { a. } 0.050 \mathrm{M} \mathrm{KOH} \text { and } 0.015 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}} \\ {\text { b. } 0.265 \mathrm{M} \mathrm{NH}_{4} \mathrm{NO}_{3} \text { and } 0.102 \mathrm{M} \mathrm{HCN}} \\ {\text { c. } 0.075 \mathrm{M} \mathrm{RbOH} \text { and } 0.100 \mathrm{M} \mathrm{NaHCO}_{3}} \\ {\text { d. } 0.088 \mathrm{M} \mathrm{HClO}_{4} \text { and } 0.022 \mathrm{M} \mathrm{KOH}} \\ {\text { e. } 0.115 \mathrm{M} \mathrm{NaClO} \text { and } 0.0500 \mathrm{M} \mathrm{KI}}\end{array}$$

Autoionization of Water and pH Like all equilibrium constants, the value of \(K_{\mathrm{w}}\) depends on temperature. At body temperature \(\left(37^{\circ} \mathrm{C}\right), K_{\mathrm{w}}=2.4 \times 10^{-14} .\) What is the \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\) and \(\mathrm{pH}\) of pure water at body temperature?

Acid-Base Properties of Ions and Salts Determine whether each cation is acidic or pH-neutral. For the cat- ions that are acidic, write an equation that shows how the cation acts as an acid. \(\begin{array}{l}{\text { a. } \mathrm{NH}_{4}^{+}} \\ {\text { b. } \mathrm{Na}^{+}} \\ {\text { c. } \mathrm{Co}^{3+}} \\ {\text { d. } \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}}\end{array}\)

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