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Chapter 14: Problem 21

Anions containing hydrogen (for example, \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} )\) usually show amphoteric behavior. Write equations illustrating the amphoterism of these two anions.

Short Answer

Expert verified
The amphoteric behavior of the given anions, \(\mathrm{HCO}_{3}^{-}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\), can be illustrated through the following equations: \( \mathrm{HCO}_{3}^{-} \): Acidic behavior: \[ \mathrm{HCO}_{3}^{-} \rightarrow \mathrm{CO}_{3}^{2-} + \mathrm{H}^{+} \] Basic behavior: \[ \mathrm{HCO}_{3}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{H}_{2}\mathrm{CO}_{3} \] \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\): Acidic behavior: \[ \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} + \mathrm{H}^{+} \] Basic behavior: \[ \mathrm{H}_{2} \mathrm{PO}_{4}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{H}_{3}\mathrm{PO}_{4} \]

Step by step solution

01

Identify acidic and basic behavior for \(\mathrm{HCO}_{3}^{-}\)

An acidic species can donate a proton (H\(^{+}\)), while a basic species can accept a proton (H\(^{+}\)). We will start with the \(\mathrm{HCO}_{3}^{-}\) anion, which is derived from carbonic acid (H\(_{2}\)CO\(_{3}\)). Acidic behavior: When \(\mathrm{HCO}_{3}^{-}\) donates a proton, it becomes CO\(_{3}^{2-}\). \[ \mathrm{HCO}_{3}^{-} \rightarrow \mathrm{CO}_{3}^{2-} + \mathrm{H}^{+} \] Basic behavior: When \(\mathrm{HCO}_{3}^{-}\) accepts a proton, it becomes H\(_{2}\)CO\(_{3}\). \[ \mathrm{HCO}_{3}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{H}_{2}\mathrm{CO}_{3} \]
02

Identify acidic and basic behavior for \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\)

Next, we will examine the \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) anion, which is derived from the dihydrogen phosphate ion (H\(_{2}\)PO\(_{4}^{-}\)). Acidic behavior: When \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) donates a proton, it becomes HPO\(_{4}^{2-}\). \[ \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \rightarrow \mathrm{HPO}_{4}^{2-} + \mathrm{H}^{+} \] Basic behavior: When \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) accepts a proton, it becomes H\(_{3}\)PO\(_{4}\). \[ \mathrm{H}_{2} \mathrm{PO}_{4}^{-} + \mathrm{H}^{+} \rightarrow \mathrm{H}_{3}\mathrm{PO}_{4} \] Now we have equations illustrating the amphoteric behavior of both anions.

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Most popular questions from this chapter

Place the species in each of the following groups in order of increasing acid strength. a. $\mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{S}, \mathrm{H}_{2} \mathrm{Se}\( (bond energies: \)\mathrm{H}-\mathrm{O}, 467 \mathrm{kJ} / \mathrm{mol}$ $\mathrm{H}-\mathrm{S}, 363 \mathrm{kJ} / \mathrm{mol} ; \mathrm{H}-\mathrm{Se}, 276 \mathrm{kJ} / \mathrm{mol} )$ b. $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}, \mathrm{FCH}_{2} \mathrm{CO}_{2} \mathrm{H}, \mathrm{F}_{2} \mathrm{CHCO}_{2} \mathrm{H}, \mathrm{F}_{3} \mathrm{CCO}_{2} \mathrm{H}$ c. \(\mathrm{NH}_{4}^{+}, \mathrm{HONH}_{3}^{+}\) d. \(\mathrm{NH}_{4}^{+}, \mathrm{PH}_{4}^{+}\) (bond energies: $\mathrm{N}-\mathrm{H}, 391 \mathrm{kJ} / \mathrm{mol} ; \mathrm{P}-\mathrm{H},$ 322 \(\mathrm{kJ} / \mathrm{mol} )\) Give reasons for the orders you chose.

Rank the following 0.10\(M\) solutions in order of increasing \(\mathrm{pH.}\) a. \(\mathrm{NH}_{3}\) b. KOH c. \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) d. \(\mathrm{KCl}\) e. HCl

An acid \(\mathrm{HX}\) is 25\(\%\) dissociated in water. If the equilibrium concentration of \(\mathrm{HX}\) is \(0.30 \mathrm{M},\) calculate the \(K_{\mathrm{a}}\) value for \(\mathrm{HX}\) .

Place the species in each of the following groups in order of increasing acid strength. Explain the order you chose for each group. a. \(\mathrm{HIO}_{3}, \mathrm{HBrO}_{3}\) b. \(\mathrm{HNO}_{2}, \mathrm{HNO}_{3}\) c. HOCl, HOI d. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{3}\)

Calculate the mass of sodium hydroxide that must be added to 1.00 \(\mathrm{L}\) of \(1.00-M \mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\) to double the pH of the solution (assume that the added NaOH does not change the volume of the solution).
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