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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Most popular questions from this chapter

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) \(-0.24 \mathrm{V}\)

\(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu} \quad E^{\circ}=+0.3 \mathrm{V}\) \(\mathrm{Fe}^{2+}+2 e^{-} \rightarrow \mathrm{Fe} \quad E^{\circ}=-0.4 \mathrm{V}\) Based on the reduction potentials given above, what is the reaction potential for the following reaction? \(\mathrm{Fe}^{2+}+\mathrm{Cu} \rightarrow \mathrm{Fe}+\mathrm{Cu}^{2+}\) (A) \(-0.7 \mathrm{V}\) (B) \(-0.1 \mathrm{V}\) (C) \(+0.1 \mathrm{V}\) (D) \(+0.7 \mathrm{V}\)

$$\begin{array}{|c|c|}\hline \text { Time (Hours) } & {[\mathrm{A}] M} \\\ \hline 0 & {0.40} \\ \hline 1 & {0.20} \\ \hline 2 & {0.10} \\ \hline 3 & {0.05} \\ \hline\end{array}$$ Reactant A underwent a decomposition reaction. The concentration of A was measured periodically and recorded in the chart above. Based on the data in the chart, which of the following is the rate law for the reaction? (A) Rate \(=k[\mathrm{A}]\) (B) Rate \(=k[\mathrm{A}]^{2}\) (C) Rate \(=2 k[\mathrm{A}]\) (D) Rate \(=\frac{1}{2} k[\mathrm{A}]\)

$$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g) \mapsto 2 \mathrm{NOCl}(g) \Delta G^{\circ}=132.6 \mathrm{kJ} / \mathrm{mol}$$ For the equilibrium above, what would happen to the value of \(\Delta G^{\circ}\) if the concentration of \(\mathrm{N}_{2}\) were to increase and why? (A) It would increase because the reaction would become more themodynamically favored. (B) It would increase because the reaction would shift right and create more products. (C) It would decrease because there are more reactants present. (D) It would stay the same because the value of \(K_{e q}\) would not change.

Starting with a stock solution of 18.0 \(\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) , what is the proper procedure to create a 1.00 \(\mathrm{L}\) sample of a 3.0 \(\mathrm{M}\) solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) in a volumetric flask? (A) Add 167 mL of the stock solution to the flask, then fill the flask the rest of the way with distilled water while swirling the solution. (B) Add 600 mL of the stock solution to the flask, then fill the flask the rest of the way with distilled water while swirling the solution. (C) Fill the flask partway with water, then add 167 mL of the stock solution, swirling to mix it. Last, fill the flask the rest of the way with distilled water. (D) Fill the flask partway with water, then add 600 mL of the stock solution, swirling to mix it. Last, fill the flask the rest of the way with distilled water.

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