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In each reaction, identify the Bronsted-Lowry acid, the Bronsted-Lowry base, the conjugate acid, and the conjugate base. $$\begin{array}{l}{\text { a. } \operatorname{HI}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{I}(a q)} \\ {\text { b. } \mathrm{CH}_{3} \mathrm{NH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{CH}_{3} \mathrm{NH}_{3}^{+}(a q)+\mathrm{OH}^{-}(a q)} \\ {\text { c. } \mathrm{CO}_{3}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{OH}^{-}(a q)}\\\ {\text { d. } \mathrm{HBr}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{Br}^{-}(a q)}\end{array}$$

Short Answer

Expert verified
For each reaction: a. Acid: HI, Base: H2O, Conjugate acid: H3O+, Conjugate base: I-. b. Acid: CH3NH2, Base: H2O, Conjugate acid: CH3NH3+, Conjugate base: OH-. c. Acid: CO32-, Base: H2O, Conjugate acid: HCO3-, Conjugate base: OH-. d. Acid: HBr, Base: H2O, Conjugate acid: H3O+, Conjugate base: Br-.

Step by step solution

01

Identify the Bronsted-Lowry Acid

The Bronsted-Lowry acid is the species that donates a proton (H+) in the reaction. For each reaction, the acid is: a. HI, b. CH3NH2, c. CO32-, d. HBr.
02

Identify the Bronsted-Lowry Base

The Bronsted-Lowry base is the species that accepts a proton (H+) in the reaction. For each reaction, the base is: a. H2O, b. H2O, c. H2O, d. H2O.
03

Identify the Conjugate Acid

The conjugate acid is the species formed when a Bronsted-Lowry base gains a proton. For each reaction, the conjugate acid is: a. H3O+, b. CH3NH3+, c. HCO3-, d. H3O+.
04

Identify the Conjugate Base

The conjugate base is the species formed when a Bronsted-Lowry acid loses a proton. For each reaction, the conjugate base is: a. I-, b. OH-, c. OH-, d. Br-.

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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 acids and bases is fundamental in grasping Bronsted-Lowry acid-base theory. A conjugate acid is created when a base accepts a proton, whereas a conjugate base is formed when an acid donates a proton. In the given reactions, when the Bronsted-Lowry acid, such as HI, releases a proton, it forms its conjugate base, I-. On the flip side, when the Bronsted-Lowry base, like H2O, accepts a proton, it becomes its conjugate acid, which in this case is H3O+.

Remember that conjugate pairs are two species that differ by a single proton. In water's reaction with hydrogen iodide (HI), the water molecule (H2O) acts as a base and accepts a proton, which results in the formation of the hydronium ion (H3O+), now a conjugate acid. The remaining iodide ion (I-) is the conjugate base of HI. Every acid-base reaction includes these two pairs, and identifying them is crucial for perfectly understanding the acid-base processes.
Proton Donor and Acceptor
The Bronsted-Lowry theory categorizes substances based on their ability to donate or accept protons (H+ ions). A proton donor is an acid, while a proton acceptor is a base. For example, in the reaction of methylamine (CH3NH2) with water, CH3NH2 donates a proton and acts as the proton donor or Bronsted-Lowry acid. Water, receiving that proton, is the proton acceptor, making it the Bronsted-Lowry base.

This proton transfer is the heart of an acid-base reaction and can be reversible, leading to what we call equilibrium. The ability to recognize a molecule's potential as a proton donor or acceptor allows chemists to predict and explain the behavior of substances in various chemical reactions, which is a vital skill in chemistry.
Acid-Base Reaction
An acid-base reaction, according to the Bronsted-Lowry theory, involves the transfer of protons from an acid to a base. This reaction is visible in all the examples provided from the textbook exercise. In each one, a proton is moved from the acid to the base, resulting in new substances - the conjugate base and the conjugate acid. Take the reaction between hydrogen bromide (HBr) and water. HBr donates a proton to H2O, leading to the formation of hydronium (H3O+) and bromide ion (Br-).

The shift of a proton from an acid to a base is what characterizes these reactions and determines the pH level of the resulting solution. It's through this understanding that students can start to unravel the complexity of acid-base chemistry and build a foundation for exploring more advanced concepts in chemistry, like buffer solutions and titrations.

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Most popular questions from this chapter

Acid-Base Properties of Ions and Salts Pick the stronger base from each pair. \(\begin{array}{l}{\text { a. } \mathrm{ClO}_{4}^{-} \text { or } \mathrm{ClO}_{2}^{-}} \\ {\text { b. } \mathrm{Cl}^{-} \text { or } \mathrm{H}_{2} \mathrm{O}} \\ {\text { c. } \mathrm{CN}^{-} \text { or } \mathrm{ClO}^{-}}\end{array}\)

Without doing any calculations, determine which solution in each pair is most basic. $$\begin{array}{l}{\text { a. } 0.100 \mathrm{M} \text { in } \mathrm{NaClO} \text { and } 0.100 \mathrm{M} \text { in } \mathrm{NaF}} \\ {\text { b. } 0.0100 \mathrm{M} \text { in } \mathrm{KCl} \text { and } 0.0100 \mathrm{M} \text { in } \mathrm{KClO}_{2}} \\ {\text { c. } 0.0100 \mathrm{M} \text { in } \mathrm{HNO}_{3} \text { and } 0.0100 \mathrm{M} \text { in NaOH }} \\\ {\text { d. } 0.0100 \mathrm{M} \text { in } \mathrm{NH}_{4} \mathrm{Cl} \text { and } 0.0100 \mathrm{M} \text { in HCN }}\end{array}$$

Explain the difference between a strong acid and a weak acid and list one example of each.

Both \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{HS}^{-}\) are amphoteric. Write an equation to show how each substance can act as an acid, and another equation to show how each can act as a base.

Acid Strength and Molecular Structure Based on their molecular structure, pick the stronger acid from each pair of oxyacids. Explain your choice. $$\begin{array}{l}{\text { a. } \mathrm{H}_{2} \mathrm{SO}_{4} \text { or } \mathrm{H}_{2} \mathrm{SO}_{3}} \\ {\text { b. } \mathrm{HClO}_{2} \text { or } \mathrm{HClO}} \\ {\text { c. } \mathrm{HClO}_{2} \text { or } \mathrm{HBrO}} \\ {\text { d. } \mathrm{CCl}_{3} \mathrm{COOH} \text { or } \mathrm{CH}_{3} \mathrm{COOH}}\end{array}$$

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