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Chapter 5: Problem 30

A neutralization reaction between an acid and a base is a common method of preparing useful salts. Give net ionic equations showing how the following salts could be prepared in this way: (a) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4};\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{3} ;\) and \((\mathrm{c})\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\).

Short Answer

Expert verified
The equations are: 1. \(\mathrm{NH}_{3}\) + \(\mathrm{H}_{2}\mathrm{HPO}_{4}\) -> \((\mathrm{NH}_{4})_{2} \mathrm{HPO}_{4}\) 2. \(\mathrm{NH}_{3}\) + \(\mathrm{HNO}_{3}\) -> \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) 3. 2\(\mathrm{NH}_{3}\) + \(\mathrm{H}_{2}\mathrm{SO}_{4}\) -> \((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\)

Step by step solution

01

Equation for \((\mathrm{NH}_{4})_{2} \mathrm{HPO}_{4}\)

First, let's find the acids and bases for this salt. We see the salt is formed of \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{HPO}_{4}^{2-}\). Recognizing these ions, you might have used \(\mathrm{NH}_{3}\), ammonia, as the base and \(\mathrm{H}_{2}\mathrm{HPO}_{4}\), dihydrogen phosphate, as the acid. The equation would go as: \(\mathrm{NH}_{3}\) + \(\mathrm{H}_{2}\mathrm{HPO}_{4}\) -> \((\mathrm{NH}_{4})_{2} \mathrm{HPO}_{4}\)
02

Equation for \(\mathrm{NH}_{4} \mathrm{NO}_{3}\)

For this salt, we can recognize the ions as \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{NO}_{3}^{-}\). In this case, we could have chosen \(\mathrm{NH}_{3}\), as the base and \(\mathrm{HNO}_{3}\), nitric acid, as the acid. The equation would then go: \(\mathrm{NH}_{3}\) + \(\mathrm{HNO}_{3}\) -> \(\mathrm{NH}_{4} \mathrm{NO}_{3}\)
03

Equation for \((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\)

With this salt, the ions are \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{SO}_{4}^{2-}\). For this reaction, \(\mathrm{NH}_{3}\) would be the base and \(\mathrm{H}_{2}\mathrm{SO}_{4}\), sulfuric acid, would be the acide. Therefore, the equation is: 2\(\mathrm{NH}_{3}\) + \(\mathrm{H}_{2}\mathrm{SO}_{4}\) -> \((\mathrm{NH}_{4})_{2} \mathrm{SO}_{4}\)

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Most popular questions from this chapter

Determine which of the following react(s) with HCl(ag) to produce a gas, and write a net ionic equation(s) for the reaction(s). (a) \(\mathrm{Na}_{2} \mathrm{SO}_{4} ;\) (b) \(\mathrm{KHSO}_{3}\); (c) \(\mathrm{Zn}(\mathrm{OH})_{2};\) (d) \(\mathrm{CaCl}_{2}\).

In your own words, define or explain the terms or symbols \((\mathrm{a}) \rightleftharpoons(\mathrm{b})[] ;(\mathrm{c})\) spectator ion; (d) weak acid.

Thiosulfate ion, \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\), is a reducing agent that can be oxidized to different products, depending on the strength of the oxidizing agent and other conditions. By adding \(\mathrm{H}^{+}, \mathrm{H}_{2} \mathrm{O},\) and/or \(\mathrm{OH}^{-}\) as necessary, write redox equations to show the oxidation of \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to (a) \(\mathrm{S}_{4} \mathrm{O}_{6}^{2-}\) by \(\mathrm{I}_{2}\) (iodide ion is another product) (b) \(\mathrm{HSO}_{4}^{-}\) by \(\mathrm{Cl}_{2}\) (chloride ion is another product) (c) \(\mathrm{SO}_{4}^{2-}\) by \(\mathrm{OCl}^{-}\) in basic solution (chloride ion is another product)

Consider the following redox reaction: $$\begin{array}{r}4 \mathrm{NO}(\mathrm{g})+3 \mathrm{O}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \longrightarrow 4 \mathrm{NO}_{3}^{-}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq})\end{array} $$ (a) Which species is oxidized? (b) Which species is reduced? (c) Which species is the oxidizing agent? (d) Which species is the reducing agent? (e) Which species gains electrons? (f) Which species loses electrons?

Complete each of the following as a net ionic equation. If no reaction occurs, so state. (a) \(\mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}+\mathrm{CH}_{3} \mathrm{COOH} \longrightarrow\) (b) \(\mathrm{H}^{+}+\mathrm{Cl}^{-}+\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH} \longrightarrow\) (c) \(\operatorname{FeS}(\mathrm{s})+\mathrm{H}^{+}+\mathrm{I}^{-} \longrightarrow\) (d) \(\mathrm{K}^{+}+\mathrm{HCO}_{3}^{-}+\mathrm{H}^{+}+\mathrm{NO}_{3}^{-} \longrightarrow\) (e) \(\mathrm{Mg}(\mathrm{s})+\mathrm{H}^{+} \longrightarrow\)
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