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Chapter 14: Problem 19

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

Short Answer

Expert verified
The amphoterism of the two anions can be demonstrated through the following reactions: 1. Bicarbonate as an acid: \(\mathrm{HCO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{H}_{2}\mathrm{CO}_{3}\) 2. Bicarbonate as a base: \(\mathrm{HCO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+} + \mathrm{CO}_{3}^{2-}\) 3. Dihydrogen phosphate as an acid: \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{H}_{3}\mathrm{PO}_{4}\) 4. Dihydrogen phosphate as a base: \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+} + \mathrm{H}\mathrm{PO}_{4}^{2-}\)

Step by step solution

01

Reaction 1: Bicarbonate as an acid

In this reaction, \(\mathrm{HCO}_{3}^{-}\) (bicarbonate) will act as an acid, meaning it will donate a proton (H\(^{+}\)) to a base. We will use water as the base in this reaction. The balanced equation is: \[ \mathrm{HCO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{H}_{2}\mathrm{CO}_{3} \] Here, the bicarbonate ion (\(\mathrm{HCO}_{3}^{-}\)) donates a proton to water, forming hydroxide (\(\mathrm{OH}^{-}\)) and carbonic acid (\(\mathrm{H}_{2}\mathrm{CO}_{3}\)).
02

Reaction 2: Bicarbonate as a base

In this reaction, \(\mathrm{HCO}_{3}^{-}\) (bicarbonate) will act as a base, meaning it will accept a proton (H\(^{+}\)) from an acid. We will use water as the acid in this reaction. The balanced equation is: \[ \mathrm{HCO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+} + \mathrm{CO}_{3}^{2-} \] Here, the bicarbonate ion (\(\mathrm{HCO}_{3}^{-}\)) accepts a proton from water, forming hydronium (\(\mathrm{H}_{3}\mathrm{O}^{+}\)) and carbonate (\(\mathrm{CO}_{3}{^{2-}}\)).
03

Reaction 3: Dihydrogen phosphate as an acid

In this reaction, \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) (dihydrogen phosphate) will act as an acid, meaning it will donate a proton (H\(^{+}\)) to a base. We will use water as the base in this reaction. The balanced equation is: \[ \mathrm{H}_{2}\mathrm{PO}_{4}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{OH}^{-} + \mathrm{H}_{3}\mathrm{PO}_{4} \] Here, the dihydrogen phosphate ion (\(\mathrm{H}_{2}\mathrm{PO}_4^-\)) donates a proton to water, forming hydroxide (\(\mathrm{OH}^{-}\)) and phosphoric acid (\(\mathrm{H}_{3}\mathrm{PO}_{4}\)).
04

Reaction 4: Dihydrogen phosphate as a base

In this reaction, \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) (dihydrogen phosphate) will act as a base, meaning it will accept a proton (H\(^{+}\)) from an acid. We will use water as the acid in this reaction. The balanced equation is: \[ \mathrm{H}_{2}\mathrm{PO}_{4}^{-} + \mathrm{H}_{2}\mathrm{O} \rightleftharpoons \mathrm{H}_{3}\mathrm{O}^{+} + \mathrm{H}\mathrm{PO}_{4}^{2-} \] Here, the dihydrogen phosphate ion (\(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\)) accepts a proton from water, forming hydronium (\(\mathrm{H}_{3}\mathrm{O}^{+}\)) and hydrogen phosphate (\(\mathrm{H}\mathrm{PO}_{4}{^{2-}}\)).

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Most popular questions from this chapter

Arsenic acid \(\left(\mathrm{H}_{3} \mathrm{AsO}_{4}\right)\) is a triprotic acid with \(K_{\mathrm{a}_{1}}=5.5 \times\) \(10^{-3}, K_{\mathrm{a}_{2}}=1.7 \times 10^{-7}\), and \(K_{\mathrm{a}_{3}}=5.1 \times 10^{-12}\). Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right],\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right],\left[\mathrm{HAsO}_{4}^{2-}\right]\), and \(\left[\mathrm{AsO}_{4}{ }^{3-}\right]\) in a \(0.20-M\) arsenic acid solution.

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