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Chapter 1: Problem 22

Which of the following species is amphoteric? (A) \(\mathrm{H}^{+}\) (B) \(\mathrm{CO}_{3}^{2-}\) (C) \(\mathrm{HCO}_{3}^{-}\) (D) \(\mathrm{H}_{2} \mathrm{CO}_{3}\)

Short Answer

Expert verified
The amphoteric species among the given options is \(\mathrm{HCO}_{3}^{-}\).

Step by step solution

01

Understanding amphoteric species

An amphoteric species is able to behave as both an acid and a base. This means it has the ability to donate a proton (acid behavior), and accept a proton (basic behavior).
02

Analyzing \(\mathrm{H}^{+}\)

\(\mathrm{H}^{+}\) is able to accept an electron pair, however, it does not have a proton to donate, so it can't be an acid.
03

Analyzing \(\mathrm{CO}_{3}^{2-}\)

\(\mathrm{CO}_{3}^{2-}\) can accept two protons, acting as a base, but it can't donate a proton, so it can't act as an acid.
04

Analyzing \(\mathrm{HCO}_{3}^{-}\)

\(\mathrm{HCO}_{3}^{-}\) can both donate a proton, acting as an acid (forming \(\mathrm{CO}_{3}^{2-}\)), and accept a proton, acting as a base (forming \(\mathrm{H}_{2} \mathrm{CO}_{3}\)), hence it is amphoteric.
05

Analyzing \(\mathrm{H}_{2} \mathrm{CO}_{3}\)

\(\mathrm{H}_{2} \mathrm{CO}_{3}\) can donate a proton, acting as an acid (forming \(\mathrm{HCO}_{3}^{-}\)), but there is not available site for accepting a proton, hence it is not amphoteric.

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Key Concepts

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

Acid-Base Behavior
The concept of acid-base behavior is fundamental in understanding chemical reactions. Acids are substances that can donate protons, while bases are those that can accept protons.
Acid-base behavior gives us insights into how substances will interact in a solution:
  • Acids typically have a hydrogen ion (H+) to donate. When this ion is released, the substance becomes a conjugate base.
  • Bases can accept hydrogen ions. In doing so, they bond with the ion to form a conjugate acid.
These behaviors link directly to the role substances play in chemical reactions, influencing reaction rates, product formation, and acidity/alkalinity.
Various substances exhibit this behavior in different ways, and some can even show both acidic and basic properties, making them amphoteric.
Proton Donation
Proton donation is essential for understanding how acids behave in solutions. When a species donates a proton, it undergoes a chemical transformation:
A well-known example of proton donation can be seen when hydrochloric acid (HCl) dissolves in water. It releases a proton (H+), forming a chloride ion (Cl-):\[ \mathrm{HCl} \rightarrow \mathrm{H}^{+} + \mathrm{Cl}^{-} \]This ability to donate a proton is what classifies a substance as an acid.
In our exercise, some species like H2CO3 demonstrate proton donation by transforming into HCO3-, showcasing their role as acids.
Proton donation also affects the pH of solutions, influencing how acidic or basic the solution will be.
Proton Acceptance
Proton acceptance is the hallmark of basic substances. When a chemical species accepts a proton, it often forms a new compound or ion.
An easy example is ammonia (NH3) accepting a proton in water to form ammonium ion (NH4+):\[ \mathrm{NH}_{3} + \mathrm{H}^{+} \rightarrow \mathrm{NH}_{4}^{+} \]This behavior defines the substance as a base.
In the original exercise, carbonate ions (CO32-) showcase their ability to behave as a base by accepting protons and forming HCO3-.
Understanding proton acceptance helps in predicting how substances will react in acidic solutions and is critical for balancing chemical equations.
Chemical Species Analysis
Analyzing chemical species involves understanding their potential to participate in chemical reactions. This involves looking at their ability to gain or lose protons.
Let's analyze some of our specific species from the exercise:
  • HCO3-: This species can both donate a proton to form CO32- and accept a proton to become H2CO3. These dual properties make it amphoteric.
  • H+: Though it can accept electrons, it cannot donate a proton since it has none to begin with, limiting its role in reactions.
  • CO32-: This ion can accept protons but lacks the ability to donate them, acting primarily as a base.
  • H2CO3: It can donate a proton but cannot accept one, making it acidic but not amphoteric.
This analysis helps chemists determine how each species will behave in solution, guiding predictions about reaction outcomes.

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Most popular questions from this chapter

\(2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) \leftrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+\mathrm{F}_{2} \mathrm{O}(g) \Delta H=167 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}\) During the reaction above, the product yield can be increased by increasing the temperature of the reaction. Why is this effective? (A) The reaction is endothermic; therefore adding heat will shift it to the right. (B) Increasing the temperature increases the speed of the molecules, meaning there will be more collisions between them. (C) The reactants are less massive than the products, and an increase in temperature will cause their kinetic energy to increase more than that of the products. (D) The increase in temperature allows for a higher percentage of molecular collisions to occur with the proper orientation to create the product.

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which net ionic equation below represents a possible reaction that takes place when a strip of magnesium metal is oxidized by a solution of chromium (III) nitrate? (A) \(\operatorname{Mg}(s)+\operatorname{Cr}\left(\mathrm{NO}_{3}\right)_{3}(a q) \rightarrow \mathrm{Mg}^{2+}(a q)+\mathrm{Cr}^{3+}(a q)+3 \mathrm{NO}_{3}^{-}(a q)\) (B) \(3 \mathrm{Mg}(s)+2 \mathrm{Cr}^{3+} \rightarrow 3 \mathrm{Mg}^{2+}+2 \mathrm{Cr}(s)\) (C) \(\mathrm{Mg}(s)+\mathrm{Cr}^{3+} \rightarrow \mathrm{Mg}^{2+}+\mathrm{Cr}(s)\) (D) \(3 \mathrm{Mg}(s)+2 \mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3}(a q) \rightarrow 3 \mathrm{Mg}^{2+}(a q)+2 \mathrm{Cr}(s)+\mathrm{NO}_{3}^{-}(a q)\)

Which of the following expressions is equal to the \(K_{\mathrm{sp}}\) of \(\mathrm{Ag}_{2} \mathrm{CO}_{3} ?\) (A) \(K_{s p}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CO}_{3}^{2-}\right]\) (B) \(K_{s p}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{CO}_{3}^{2-}\right]^{2}\) (C) \(K_{s p}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CO}_{3}^{2-}\right]\) (D) \(K_{s p}=\left[\mathrm{Ag}^{+}\right]^{2}\left[\mathrm{CO}_{3}^{2-}\right]^{2}\)

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. Which of the following would create a good buffer when dissolved in formic acid? (A) \(\mathrm{NaCO}_{2} \mathrm{H}\) (B) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) (C) \(\mathrm{NH}_{3}\) (D) \(\mathrm{H}_{2} \mathrm{O}\)

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ If the \(\mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu}(s)\) half-reaction has a standard reduction potential of \(+0.52 \mathrm{V},\) what is the standard reduction potential for the \(\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}(s)\) half-reaction? (A) \(+0.76 \mathrm{V}\) (B) \(-0.76 \mathrm{V}\) (C) \(+0.24 \mathrm{V}\) (D) \(-0.24 \mathrm{V}\)
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